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ESSENTIALS OF
PLANE TRIGONOMETRY
AND ANALYTIC GEOMETRY
ESSENTIALS OF
PLANE TRIGONOMET...
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ESSENTIALS OF
PLANE TRIGONOMETRY
AND ANALYTIC GEOMETRY
ESSENTIALS OF
PLANE TRIGONOMETRY
AND ANALYTIC GEOMETRY BY
ATHERTON
H.
SPRAGUE
PROFESSOR OP MATHEMATICS AMHEBST COLLEGE
NEW YORK PRENTICE-HALL, INC. 1946
COPYRIGHT, 1934, BY
PRENTICE-HALL, INC. NEW YORK
70 FIFTH AVENUE,
NO PART OF THIS BOOK MAT BB REPRODUCED IN ANY FORM, BY MIMEOGRAPH OR ANT OTHER MEANS, WITHOUT PERMISSION IN WRITING FROM THE PUBLISHERS. ALL RIGHTS RESERVED.
First Printing
Second Printing Third Printing Fourth Printing
June, 1934 April, 1936
September, 1938 October, 1939
Fifth Printing
April, 1944
Sixth Printing
August, 1946
PRINTED IN THE UNITED STATUS OF AMERICA
PREFACE purpose of this book
is to present, in a single the essentials of Trigonometry and Analytic that a student might need in preparing for a
THE volume, Geometry
study of Calculus, since such preparation is the main objective in many of our freshman mathematics courses. However, despite the connection between Trigonometry and Analytic Geometry, the author believes in maintaining a certain distinction between these subjects, and has brought out that distinction in the arrangement of his material. Hence, the second part of the book supplemented by the earlier sections on coordinate systems, found in the first part would be suitable for a separate course in Analytic Geometry, for which a previous knowledge of Trigonometry is assumed. The oblique triangle is handled by means of the law of sines, the law of cosines, and the tables of squares and square roots. However, the usual law of tangents and the r " formulas are included in an additional chapter, Supple-
mentary Topics." There is included in the text abundant problem material on trigonometric identities for the student to solve. The normal form of the equation of a straight line is derived in as simple a manner as possible, and the perpendicular distance formula is similarly derived from it. The conies are defined in terms of focus, directrix, and eccentricity; and their equations are derived accordingly. "
" are Transformation of Coordinates In the chapter discussed the general equation of the second degree and the types of conies arising therefrom. An attempt has been made to present rigorously, but without too many details,
the material necessary for distinguishing between the types
PREFACE
vi
by means of certain invariants, which enter into the discussion. Although this chapter may naturally be omitted from the course, it is well included if time
of conies
permits.
ATHERTON H. SPBAGUE Amherst College
CONTENTS PLANE TRIGONOMETRY CHAPTBB I.
PAGV
LOGARITHMS
3
1.
Exponents
3
2.
Definition of a logarithm
7
3.
Laws
4.
Common
of logarithms
8 10
5.
logarithms Use of the logarithmic tables
6.
Interpolation
13
7.
Applications of the laws of logarithms,
12
and a few 15
tricks II.
THE TRIGONOMETRIC FUNCTIONS Angles
21
9.
Trigonometric functions of an angle
21
23
11.
Functions of 30, 45, 60 Functions of (90 - 0)
12.
Tables of trigonometric functions
10.
III.
21
8.
25
...*....
26
SOLUTION OF THE RIGHT TRIANGLE
29
13.
Right triangle
29
14.
Angles of elevation and depression
30
IV. TRIGONOMETRIC FUNCTIONS OF 15. Positive
ALL ANGLES
....
and negative angles
35 35
16.
Directed distances
35
17.
Coordinates
36
18.
Quadrants
19.
Trigonometric functions of
37 all
37
angles
Functions of 0, 90, 180, 270, 360 as to 360. 21. Functions of varies from
40
20.
22. Functions of (180
23. Functions of
6)
and (360
0)
...
42 45 48
(-0) Vll
CONTENTS
viii
CHAPTBK
V.
PAQB
THE OBLIQUE TRIANGLE 24. Law of sines 25. Applications of the
26.
Ambiguous
27.
Law
51 51
law of sines
52
case
of cosines,
53
and applications
57
VI. TRIGONOMETRIC RELATIONS 28.
Fundamental
29.
Functions of (90
66 66
identities
30. Principal angle
+
71
0)
between two
73
lines
73
31. Projection 32. Sine
and cosine
+
of the
sum
of
two angles
33.
Tan
34.
Functions of the difference of two angles Functions of a double-angle
35.
(a
....
76
ft)
....
VII.
81
Product formulas
86
SUPPLEMENTARY TOPICS 38.
Law
39.
Tangent
78 79
36. Functions of a half-angle 37.
74
92 92
of tangents of a half-angle in terms of the sides of a
given triangle 40. Radius of the inscribed
,
.
.
.
94
circle
98
41. Circular measure of an angle
100
Summary of trigonometric formulas TABLE I: Logarithms to Four Places TABLE II: Trigonometric Functions to Four TABLE III: Squares and Square Roots
102
42.
107 Places 111
117
ANALYTIC GEOMETRY VIII.
COORDINATES
123
43. Position of a point in a plane
123
123 between two points 125 Mid-point of a line segment 46. Point that divides a line segment in a given ratio 126 129 47. Slope of aline 130 48. Parallel and perpendicular lines 44. Distance 45.
CONTENTS
ix
CHAPTER
pAOB
VIII. 49.
COORDINATES
(Con't).
Angle between two
lines
131
50. Application of coordinates to plane
geometry
.
IX. Locus
136
51. Definition and equation of locus
136
X. THE STRAIGHT LINE 52.
Equations of
53. Point-slope
140
lines parallel to the axes
140
form
140
56.
form Two-point form Intercept form
57.
General form of the equation of a straight
58.
Normal form
54. Slope-intercept
142
55.
143
59. Distance 60. Lines
from a
143 line
.
a point
152
through the point of intersection of two 157
THE CIRCLE
162
and equation of the circle General form of the equation of the circle.
61. Definition
62.
63. Circles
162 .
.
.
THE PARABOLA 65. Definition
and equation
168 172 172
64. Definition of a conic
of the parabola
172
Shape Equations of the parabola with vertex not at the
173
origin
176
THE ELLIPSE
181
68. Definition and equation of the ellipse
181
66. 67.
XIII.
164
through the points of intersection of two
given circles; radical axis
XII.
144
148 line to
given lines
XI.
132
of the parabola
70.
Shape of the ellipse Second focus and directrix
71.
Equations of the ellipse with center not at the
69.
origin
183 187
188
CONTENTS
x CHAPTER
PA<
XIV. THE HYPERBOLA 72. Definition 73. 74. 75. 76.
194
and equation
of the hyperbola
....
194 195
Shape of the hyperbola Second focus and directrix
199
199 Asymptotes Equations of the hyperbola with center not at the origin 202
XV. TRANSFORMATION OF COORDINATES
209
77. Translation of axes
78.
Rotation of axes
79.
Removal
of the xy
term
80. Invariants; classifications of types of conies
INDEX
.
.
209 213 215 219 223
PLANE TRIGONOMETRY
CHAPTER
I
LOGARITHMS Exponents. Since a knowledge of the theory of exponents is essential for a clear understanding of logarithms, we shall review briefly that theory. By 5 3 we mean 1.
5X5X5. By
a3
we mean a
By a w
,
m is
provided a
X
X
a
a.
we mean
a positive integer,
X
a
X
a
...
to
m factors.
call a the base and m the exponent. Consider the product of 5 3 X 5 4
We
.
By
this
we mean (5
X
5
X
5
X
5
...
5) (5
X
5
X
5
X
5)
or to seven factors
or 5 7.
m Similarly, a (a
X
a
an equals .
.
.
to
m factors) (a X
a ... to n factors).
Or: a
X
a
.
.
.
to
m+
n
factors
= a m+n
.
It is evident that this process gives the law:
of two or more quantities with a common base a quantity with the same base and an exponent equal equals
The product
to the
sum
of the exponents of the various quantities.
PLANE TRIGONOMETRY
4
Now
consider 5
I
'
58
By this we mean
is, the number 5 with an exponent equal to the difference of the exponent of the numerator and that of the denominator:
that
55
Or, in general, am
a
if
_
.
m is greater than n> <ji
n fi
=
"" 3
a
Xi X Xa
j*
then
.
.
.
to
m factors
.
.
.
to
n
factors
to
w
n
.
.
.
factors
Hence we have the law: The quotient of two quantities with a common base equals a quantity with the same base and an exponent equal to the difference of the exponent of the numerator and that of the denominator.
Now
consider 5! 56
'
Xft Xft
However,
if
the above law
is
,
to hold,
1 = 5- = 5It is quite apparent, therefore, that
5
X
5
...
to
5~ 2 does not imply
2 factors
LOGARITHMS (which
is
meaningless), but
a symbol for
is
5*'
Similarly,
is
if
ra is less
than
then
n,
a symbol for
Hence we
shall define a
m positive)
m when ,
m is greater than
(that
is,
to be equal to
J_ m a
'
In particular, consider
^
'
an
where
m equals n; that
is,
aT m a
By
'
the above law this equals
But we know that a*
am
=
1.
Therefore, for consistency, we define a to be equal to 1. Now let us see what we mean by a quantity with a First, what do we mean by the fractional exponent.
V?
symbol by itself gives have
We mean 3.
3*
or
that quantity which multiplied
Let \/3
=
3*
=
3*,
3,
and determine
x.
We
PLANE TEIGONOMETRY
6
Then 2x
=
1.
Therefore:
*-*. Hence we
define
M 3
to equal
vs. In general, we
define a m/n as follows
v^ =
=
a m/n
(V^)
:
m .
One fundamental law of exponents remains Take (5 3 ) 2 By our first law,
sidered.
3
(5
.
Similarly,
m (a
n )
=
=
2 )
(a
m
3
= =
a
=
3
(5 )(5
)
m )(a
)
... ^m+m-4-
__
From
to be con-
.
56
n
to
.
.
to
n term*
.
.
factors
nm
a mn
.
these computations
we have
the law:
// a quantity with a given base and exponent is raised to a power, the result is a quantity with the same base and an
exponent equal
The
to the
product of the two exponents.
application of
all
three laws
is
allowable for frac-
tional exponents, positive and negative, as well as for positive and negative integral exponents.
Problems 1.
(*
+
Find the value 8y
-
of:
(243)-*;
^8*; (- T -b)
3); 1~".
2.
Express with positive exponents only:
3.
Express without any denominator:
2x~H y*x*
3-VoT 2
'
H ;
(742);
LOGARITHMS 4. 5.
Solve: (a) Solve:
(a) 7* (6) 6.
(a)
21
= =
x* -
** -
4; (6)
(c)
27*
(d)
27 Z
49. 8.
= =
7
1.
27'
3.
(e)
9.
(/) 27*
=
4. J.
Solve: 10*
(6) 10* (c) 10*
= = =
1000.
(d)
10'
(e)
10*
(/)
10*
100. 10.
2" -
7.
Solve:
8.
Solve: 9
6 -2*
3 2x
-
= = =
+8 =
244
(g)
.1
(h) 10' (t) 10-
.01.
= = =
.001.
.0001.
.00001.
0.
X
3
10*
1.
+
27
=
0.
(Answer: x
= -2
or 3.) 9.
Simplify: '
10. Simplify:
(3
n+ 2
/
+
16-*
+
3
3
+
n )
5
^+ -
-s-
(9
3
(-2)-*. n+ 2 ).
Definition of a logarithm. Consider: 7 2 equals 49. Observe that 2 is the exponent of the power to which 7 2.
We
must be
raised to give 49. shall now, by this in write another form. expression definition,
means
the logarithm of 49 to the base 7 to be equal to expressed in symbols, Iog 7 49 2 Similarly, since 8 equals 64,
8,
if
2.
Or,
2.
=
2;
we have Iog 2 8
In general,
a
we have
logs 64
and since 2 3 equals
=
of
We define
6* equals
=
3.
N, we have
log*
N
-
x]
and the general definition:
The logarithm of a number N to the base
b is the
exponent
PLANE TRIGONOMETRY
8
power to which the base the number N.
of the
6
must be
raised to give
Example Find Iog 2
64.
= = = = =
Iog 2 64
Let
Then
2* 2*
x
.
Iog 2 64
Therefore:
x.
64,
26
,
6. 6.
Problems Find: 1.
logs 27.
5.
Iog 4 64.
2.
Iog 27 3.
6.
27.
3. logs
^.
Laws
11. logio 10. 12. logio 1.
7. Iog 27 9. 8. Iogi25 5.
4. logs 16.
3.
Iog 9
9. logio 1000. 10. logio 100.
13. log
.1.
14. logio .01.
16. logio .001.
16. logio .0001.
There are three important laws
of logarithms.
of logarithms which are immediate consequences of the three laws of exponents and the definition of a logarithm.
The
first
law
is,
given the numbers
(MN) =
M+
log&
M and N:
log*,
N.
Proof Let
log&
M
log b
N
Then
b* by
(Since
we
= x, = y. = M, = N.
are interested in the logarithm of the product
form that product.)
MN MN
Hence: or:
= b x by = b*^. -
;
In logarithmic form, logb
MN
=*=
=
x
+
Iog6
y
M + logb N.
MN, we
LOGARITHMS The second law
is
9
:
M=
log*,
logb
M - logb
JV.
Proof
As
before, let
logb
M
logb
N
Then
b
x
by
= x, = y. = M, = N.
M ~=
Hence:
6^ W'
N
or:
In logarithmic form,
= The
third law
is
log&
If
log &
N.
:
= p
logb
M
Proof Let
logb
Then
M b
Hence:
x
M
p
M
p
=
x.
= M. = b px
.
In logarithmic form, logb
px
= p
We
logb
state these laws as follows
Law
1.
numbers
M.
:
The logarithm of the product of two or more same base is the sum of the logarithms of the
to the
respective numbers.
PLANE TRIGONOMETRY
10
Law
The logarithm of
numbers to same base is the difference of the logarithm of the numerator and the logarithm of the denominator. Law 3. The logarithm of a number raised to a power is the product of the exponent of the power and the logarithm of the 2.
the quotient of two
the
number. It is quite apparent that law 3 handles the logarithm of the root of a number.
Problems 1.
Write in expanded form: 1
log&
2. 3.
Write in contracted form: 2 log& x + \ log y log xy*. Determine which of the following symbols equals 2 log& x: fe
2
(a) Iog6
z.
(6)
logbz
ft
2 .
(c)
Common logarithms. For numerical computation it desirable that a universal base be employed, and the
4. is
most convenient base to use
is the base 10. Logarithms with base 10 are called common, or Briggs, logarithms. (It may be stated, however, that in higher mathematics the base used is the irrational number e = 2.71828 .) It is understood that in this text the base is 10 unless other.
.
.
wise stated.
Let us construct a miniature table of logarithms by considering various powers of 10.
LOGARITHMS 1(T 2 10~
3
10~ 4
= = =
11
.01;
hence, log .01
.001;
hence, log .001
= -2 = -3 = -4
.0001; hence, log .0001
Now
suppose we wish the logarithm of a number not given here, say 386. Obviously since 386 is between 100 and 1000, the log of 386 is between 2 and 3 that is, 2 plus a fraction less than 1. This fraction, in decimal form, is given in the tables; it is found to be .5866. Hence: ;
log 386
=
2.5866.
Or: JQ2.5866
is
^
Now suppose we wish the log of 38.6. Since this number between 10 and 100, its log is 1 plus a fraction; from the we
tables
find the fraction to be the
above that ;
is,
.5866.
Hence log 38.6
From
=
these computations
clusions:
The logarithm
of a
same
as that noted
:
1.5866.
we draw the following connumber is composed of two
an integral part and a fractional or decimal part. the integral part, the characteristic, and the fracFor a given number, a shift in tional part, the mantissa.
parts,
We
call
the position of the decimal point changes the characteristic but does not change the mantissa.
The
characteristic of the log of 386 was found to be 2. apparent that the characteristic of the log of
It is quite
every number between 100 and characteristic of the log of every
1000
is
2; that
number with
is,
the
three digits
And it is not difficult to the left of the decimal point is 2. to see that the characteristic of the log of a number greater always numerically one less than the number of digits to the left of the decimal point. Now let us consider the logs of positive numbers less than 1. Consider log .00386. From our miniature table, since .00386 is between .001 and .01, log .00386 is between -3 than
1 is
PLANE TRIGONOMETRY
12
and
2; that
is,
3 plus .5866
or,
3 plus a positive fraction less than which we write as 3.5866. Hence:
=
log .00386
1,
3.5866.
The
characteristic is 3, a negative number and numerione more than the number of zeros immediately to cally the right of the decimal point. In general the characteristic of the log of a positive number less than 1 is negative
and numerically one more than the number
of zeros
immedi-
In the latter ately to the right of the decimal point. if we prefer, assume the decimal instance we may always, point to be in a convenient position, compute the characteristic of the log of that number, and then shift the decimal point to
its
proper position and revise the characteristic
by counting. 6. Use of the logarithmic tables. 726. The line from our tables is
Suppose we wish log
N|Q|l|2|3|4|5l6l7|8|9~ 72 |
8573
|
8579
I
8585
I
8591
I
8597
|
8603
|
8609
|
8615
I
8621
|
8627
The characteristic is 2. The mantissa is found as follows Look under N for the first two digits, 72. Then, since our third digit is 6, our mantissa is on the line of 72 and directly :
under
6.
It is .8609.
Hence: log 726
=
2.8609.
N
is Or, suppose we know that the log of a number 1,8591 and we wish to find N. We look for the number
corresponding to the mantissa 8591 and observe it is 723. must have Then, since the characteristic is 1, the number two digits to the left of the decimal point. Therefore:
N
N
=
72.3.
N
N
of finding a number when log is given the inverse of a we obviously process finding log. call the antilogarithm. In the process of finding an antilogarithm, the characteristic simply indicates the position
The process
is
N
LOGARITHMS of the decimal point.
13
And it is apparent that, for a positive
or zero characteristic, the resulting antilogarithm has one more digit to the left of the decimal point than the numerical
value of the characteristic; and that, for a negative characteristic, the antilogarithm has one less zero immediately to the right of the decimal point than the numerical value of the characteristic. Thus, if log equals 3.8591, then
N
N
=
.00723.
Problems 1.
Find the logarithms
of the following:
X
10~ 6
(a)
382.
(e)
382
(b)
3.82.
(/)
4.61.
(c)
38,200.
(g)
.000279.
(d)
382
(h)
.00963.
2.
X
10 6
.
Find the antilogarithms
.
(i)
100.
(j)
10 8
(fc)
(1)
of the following
.
.0243.
10~ 2
.
:
(a)
2.4829.
(d)
1.7050.
(jr)
(6)
1.4829.
(e)
2.8808.
(h)
(c)
0.4829.
(/) 1.9763.
(i)
3.9624. .8306.
2.0000.
The logarithms of all numbers of 6. Interpolation. three digits, preceded or followed by any number of zeros, can be found in our tables by the process described in shall now show how, by interpolation, we Section 5.
We
can find logarithms of numbers of four digits. Consider log 72.63. The tables give us log 72.60 and Hence we have: log 72.70. log 72.60 log 72.63 log 72.70
= = =
1.8609 (?)
1.8615
We
reason as follows: The number 72.63 is three-tenths of the way from 72.60 to 72.70. Hence, log 72.63 is threetenths of the way from log 72.60 to log 72.70; that is,
way from 1.8609 from 1.8615, we find
three-tenths of the tracting
1.8609
to
1.8615.
Sub-
that the distance
PLANE TRIGONOMETRY
14
between the two therefore
is
desired logarithm
is
:
1.8609
+ .3X6 units =
+ 1.8 units + 2 (approx.)
1.8609
= = Hence
The
6 units.
1.8609
units
1.8611.
:
=
1.8611.
=
2.8611,
log 72.63
Similarly, log 726.3
and
so on.
Find log 384.2. log 384.0
log 384.2 log 385.0
= = =
2.5843 (?)
2.5855
Log 384.2 is two-tenths of the way from 2.5843 The distance between them is 12 units. .2
Hence
X
12
= =
to 2.5855.
2.4
2 (approx.).
:
log 384.2
=
2.5845.
manner, interpolation is used in finding antilogarithms of numbers not given in the tables. Consider In
like
antilog 2.7463. nearest to 7463,
We look in the tables for the two mantissas
and 7466. Hence, disregarding temporarily the correct position of the decimal point,
we have
and
find 7459
:
antilog 7459 antilog 7463 antilog 7466
= = =
5570 (?)
5580
must be the same proportion of the way between 5570 and 5580 as 7463 is between 7459
As
before, antilog 7463
LOGARITHMS
15
and 7466. We first find what proportion this is. The distance from 7459 to 7466 is 7 units. The distance from 7459 to 7463
is
4 units.
Hence, 7463
four-sevenths of
is
way from 7459 to 7466. Hence, antilog 7463 is foursevenths of the way from 5570 to 5580. Therefore the
:
= = = -
antilog 7463
5570 5570 5570
+ f X 10 units + * units + 6 (approx.) units
5576.
Hence: antilog 2.7463
557.6.
Problems 1.
Find the logarithms
of:
(a) 3.286.
(d)
.0003428.
(g)
.01111.
(fe)
729.4.
(e)
82.37.
(h)
.3263.
(c)
68.43.
(/) 42.94.
(i)
.02438.
2.
Find the antilogarithms
of:
(a) 2.8531.
(d)
2.8906.
(g)
2.2375.
(6)
1.9276.
(e)
1.8660.
(h)
.3770.
(c)
4.6081.
(/)
1.0200.
(i)
.0964.
7.
Applications of the laws of logarithms, and a few
tricks.
Example
By
logarithms, find
1
:
(24.32)(6.431)
76.47
By
our
number
first
two laws
of logarithms, the log of the required
N may be expressed as follows: = log 24.32 + log log N log 24.32 log 6.431
=
1.3860
0.8083
6.431
-
log 76.47.
PLANE TRIGONOMETRY
16
Therefore: log numerator log 76.47
=
N N
= =
Subtracting,
log
Hence:
No
2.1943 1.8835 .3108 2.045.
tricks are necessary in solving the
above problem,
but suppose the problem reads: Find: (24.32) (6.431)
764.7
Then
= = =
numerator
log
log 764.7
log
The
N
2.1943
2.8835 (?)
solution will involve a negative number.
rectly,
Subtracting cor-
we have: log
However, there
is
N
=
T.3108.
an easier process for solving
this
problem. Write 2.1943
12.1943
as:
Then
log
Subtracting, Or, as above,
numerator = = log 764.7 = log N = log N
-
10.
12.1943
-
10
-
10
2.8835 9.3108 T.3108.
Anticipating negative characteristics and writing manner will be found a very useful practice.
in this
this point in the text
we
them
From
shall write negative characteristics
thus.
Suppose we wish
when
N\ equals
log
9.4762 Write log
10.
N
l
as:
9.3241
10,
and log
In this case, the following 19.3241
-
20.
is
JVi equals the solution:
LOGARITHMS Then
log log
Subtracting,
log
NI #j = -
9.4762
-
9.8479
-
19.3241
=
17
20 10 10.
Example 2 Find A^ 28.64 by logarithms.
By
our third law of logarithms, log
^28.64 =
log (28.64)
=
f log 28.64
K
log 28.64
Again no
tricks are necessary.
But consider the
fol-
lowing:
Find ^.0002864. log .0002864 _
Then
.
6.4570_
=
2.1523
-
10
- 3i
This answer is a bit confusing, however, since we are The trouble lies in left with a fractional characteristic. the fact that 10 is not exactly divisible by 3. Hence, the following is a better solution, since the answer can be handled more readily.* *
These two
results
may
2.1523
-
be shown to be the same, as follows: 34
-
2.1523
- 3.3333 - 10 - 3.3333 - 10.
12.1523
8.8190
PLANE TRIGONOMETRY
18
- - - --
_ .0002864 Wnte log _ as: ti
26.4570 _
-
30
Then 3
Similarly, consider the following: Solve: log JV
_ ~
"
4
-
Write 8.2869
and
10 as:
-
8.2869
10
4
38.2869
-
40,
so on.
Example 3 Find the amount to which $100
will
grow
in 10 years
if
the
compounded semi-annually at 6 per cent. the interest were compounded annually, we should have
interest is If
at
the end of one year: $100(1.06);
and at the end
of 10 years:
$100(1.06) If
10 .
is compounded semi-annually, we months
the interest
end
of six
:
$100(1.03);
and at the end
of
one year: $100(1.03)
that
2 ;
is,
/ $100^1
+
Hence, at the end of 10 years we
.06\ 2
) shall have: 110
or:
$100(1.03)
20 .
shall
have at the
LOGARITHMS Now,
letting x equal the
= = =
log x
19
amount, we have the following
log 100
2.0000
+ 20 log 1.03 + 20(.0128)
2.2560.
Therefore:
x
Our answer
=
180.3.
is:
$180.30.
Example 4 Solve: 28.62*
Taking
logs of
=
684.9.
each
side,
we have
x log 28.62
= =
x
=
log 28.62*
Therefore:
the following:
log 684.9
log 684.9. log 684.9
log 28.62
2 8356 '
=
~~
1.4567
=
1.947.
Problems 1.
Find: (2.382) (69.84)
(a)
(/)
(328.2) V.004691.
,
(1.286) (91.34)'
(4236) (.02438) ?i
2 N
8
(c)
(3.461)
.
V 98857
(7-241) (62.86)
3
296.3 2
(1.246)
98.77
'
(72.39)(1.006) -V/STJTS
2
:
PLANE TRIGONOMETRY
20 2.
Find the amount to which $6000
will
grow
compounded quarterly at 4 per cent. 3. Solve: 4.287* 52.39.
interest is
4. 5. 6.
Find: (12.48) Prove: Iog Prove: log a b
^=
N
log&
log& a
N =
-
Iog
1.
b.
in 5 years
if
the
CHAPTER
II
THE TRIGONOMETRIC FUNCTIONS 8. Angles. Suppose a line OX is revolved about the until it takes the position OP (Figure 1). An point angle XOP is then generated. We call OX the initial side of the angle, and OP the terminal side. The angle be 0. denoted the letter For the moment may by single we shall consider as acute.
initial side
Figure
Figure 2
1.
Trigonometric functions of an angle. Let us take any point on the terminal side and drop a perpendicular A right triangle to the initial side (extended if necessary). is then formed containing d (Figure 2), with legs x and y, 9.
and hypotenuse r. Obviously the lengths x, y, and r are determined by the position of the point chosen, but the ratios of any two of the quantities x, y, r are unique for a given 0. There are six such ratios, called the six trigonometric functions of an angle 6] and they are defined as follows
:
sine 6
cosine 6
tangent 8
= =
sin B
cos
= -y =
opposite side
r
hypotenuse
= -x =
adjacent side
r
hypotenuse
y x
opposite side
tan 21
~
7
:
adjacent side
PLANE TRIGONOMETRY
22
cosecant
=
CSC
secant 6
hypotenuse
~ V
opposite side
hypotenuse
= -
sec
adjacent side
cotangent
From
=
cot
x = - =
adjacent side
y
opposite side
these definitions and the law of Pythagoras, all an acute angle 6 can be found if any one
six functions of
function
is
known.
For example, suppose we are given .
sm
A
6
= -3 o
Since
construct a right triangle with y
Figure
Then
3.
X*
Therefore
_
= =
25-9
=
16.
r
2
y
:
Then we
x
=
,
= 3-
4.
have: .
sm
5
cos
fl
=
4 3
tan 4'
3 esc 3'
Figure
3.
sec
a
=
5 7-
4
cot
4 o
2
=
3 and r
=
5,
as in
TRIGONOMETRIC FUNCTIONS
23
Problems 1.
Given Figure 2
(a) (b) (c)
(d) 2.
(see
page 21)
Express x in terms of Express y in terms of Express r in terms of Express r in terms of
Given a right
cos 6
cdt 6 sin 6
esc 6
:
and y 01; r. and x or r. and x or y. and x or y.
triangle with hypotenuse
c,
opposite side
6,
adjacent side a; find all functions of 6 for the following:
(6)
a a
(c)
a
(d)
a
(a)
= = = =
3, 6 5, 6 1,
6
2, 6
= = = =
12, c 2,
c
5,
c
6.
Given cos Given esc Given tan
6.
Given cot
= = = =
7.
Given
=
3.
4.
sin
= 5. = 13. = \/5. = V29.
c
4,
if; find sin find tan
;
0.
0.
1; find cot 0.
A/3; find sin
0.
find sec
0.
-7=.;
V2
30, 46, 60. The functions of 30, from geometric considerations. can be found 60 and 45, We shall find first the functions of 45. If 6 = 45, then, from plane geometry, the right triangle in Figure 2 is = y. Hence we have immediately: isosceles, and x 10. Functions of
In the right triangle in Figure 4, x = 1 and y = 1; and we have r = \/2. Consequently we have: sin
45
=
1
\/2 1
oos45
X/2 tan 45
1
1
Figure
4.
PLANE TRIGONOMETRY
24
sec 45
= \/2 = \/2
cot 45
=
csc 45
We
30. If from plane geometry (Figure
shall next find the functions of
in Figure 5, then,
Hence
1
= 30, as = 2y. 2), r
:
and
In the right triangle in Figure 5, y = 1 and have x = \/3. Consequently we have: sin
=
30
r
=
2;
and we
-
V3
cos 30
2 1
tan 30 csc 30
=
sec 30
= -4^
VF Figure
6.
2
V3 V3
cot 30
It is quite evident from the above explanations that the functions of 60 can be computed from Figure 5; but since the 60 angle may come at the base, we have added Figure 6.
Consequently we have: sin 60
=
V3 1
cos 60
2
TRIGONOMETRIC FUNCTIONS
25
tan 60 csc 60
=
p.
Vs
sec 60
=
cot 60
=
2 -7=.
Va
The above eighteen functions are important because, since the functions are computed geometrically, no
/60 1
tables are
Figure
6.
necessary for computations with them and, hence, the functions furnish material for a host of
problems in many branches of science. We recommend that the student memorize the two basic, simple relations that follow
:
30
(1) sin
(2)
since
from these two
tions can be
tan 45
=
1
relations the remaining sixteen rela-
computed
readily.
Functions of (90 functions of 30 and 60, 11.
sin
csc
In a comparison of the observed that
8). it is
30
cos 30
and so on; that
= i
30
= = =
cos 60 sin 60
sec 60
the functions of 30 are the corresponding co-functions of 60. This conclusion suggests a relation is,
between the functions
of
sponding co-functions of
any acute angle its
We
6
and the
complement. have, in Figure
sin
= - =
7,
cos (90
-
0)
c
cos 6
=
c
6
Figure
7.
and so
on.
=
sin (90
-
6)
corre-
PLANE TRIGONOMETRY
26
Hence we have: Theorem.
Any
function of an acute angle 6 equals the
corresponding co-function of (90 0) equals the
(90 12.
0),
and any function
corresponding co-function of
Tables of trigonometric functions.
As
of
6.
illustrated in
Section 10, the trigonometric functions of 30, 45, and
can be computed geometrically. By more advanced methods the trigonometric functions of other acute angles have been computed and tables have been made, the use of which is similar to that of logarithmic tables.- Let us take two typical contiguous lines from our tables.
60
These two
us the sine, cosine, tangent, and 50', as well as the 10', 52, and 51 cotangent a characteristic 9 standing for logs of these functions, of
lines give
38, 38
and so on. The to 45, we from angles For down. top, working from and from the right 9
10,
tables are so arranged that, for read from the left and from the angles from 45 to 90, we read the bottom, working up. For
example, the sine of 38 is .6157, and the sine of 52 is Observe that the cosine of 52 is .6157, and the .7880. Sin 38 = cos 52 according to cosine of 38 is .7880. 38 and 52 are complementary angles. Section 11, since Similar results obtain for the tangent and the cotangent.
TRIGONOMETRIC FUNCTIONS
27
There are tables of secants and cosecants, but we shall not need them in our work. The function tables in this book are so arranged, with intervals of 10 minutes, that interpolation like interpolation in logarithms.
we wish
Sin 38
sin
38
4' is
Therefore
practically
Suppose, for example,
4'.
sin
38
sin
38
4'
sin
38
10'
= = =
.6157 (?)
.6180
way from
four-tenths of the
.6157 to .6180.
:
sin
38
4'
= =
+ +
.6157
.6157
.4(23) units
9 units
.6166.
The tangent is similarly handled. The cosine of an angle, however, and so
increases,
cos 38
is
is
handled a
decreases as the angle
bit differently.
Consider
4'.
cos 38
Cos 38
4' is
Therefore
cos 38
4'
cos 38
10'
= = =
four-tenths of the
.7880 (?)
,7862
way from
.7880 to .7862.
:
cos 38
4'
= = =
.7880
.7880
-
.4(18) units
7 units
.7873.
Observe that we have subtracted instead of added. The cotangent is handled similarly to the cosine. Problems 1.
Prove:
60 sin 30
(a) sin (6)
= =
2 sin 30 sin
cos
30.
60 cos 30
-
cos 60
sin
30.
PLANE TRIGONOMETRY
28
'-=? (d)
cos 30
= 2
(e)
2.
tan 30
cos
(c)
cos (90
-
(a)
sin 29
(6)
tan 53
5.
sin
S)
0.
=
^.
Find: 36'. 27'.
(c)
sin
62
13'.
(e)
cos 71
33'.
(d)
cos 40
42'.
(/)
cot 42
26'.
(c)
log tan 9
Find:
(a) log sin 19
is
6.
= \/3
cos
(6)
tan 30
Solve for 6 in the following equations:
(6)
4.
- ^ - ^^ 6QO
(a) sin 6
3.
-
tan 60
log cos 40
4'.
17'.
Find the angle
(a)
46'.
(e)
(d) log cot
48
4'.
(/)
whose
is
.5883;
and
sine
log sin 62 log cos 72 (6)
17'. 8'.
whose cosine
.4072.
6. Find the angle (a) whose log tan whose log cot is 1.6000.
is
9.7726
10;
and
(6)
CHAPTER
III
SOLUTION OF THE RIGHT TRIANGLE If we are given a right triangle and that one angle is 90 and we are given, in addition, either of the other angles and any side, or any two sides, the triangle is determined uniquely. It is the purpose of this chapter to show how the trigonometric
13.
Right triangle.
therefore
know
functions enable one to find the missing parts of a right We call triangle, when the above information is given. this process solving the right triangle.
Example
A
1
Given the right triangle ABC (Figure = 26 14'; solve the triangle.
B =
90
-
=
63
46'
=
sin
A
-
26
8),
with
c
=
14'
c .'.
a
-
= = = = =
A
c sin
100 sin 26
100
X
14'
.4420
44.20 sin
B
c .".
b
c sin
= = = Or,
B
100 sin 63 100
X
46'
.8970
89.70
we might have used
the following solution:
- = cos c
29
A
100 and
PLANE TRIGONOMETRY
80
.'.6
= = =
A
cos
100 cos 26
X
100
14'
.8970
89.70
Logarithms might have been used in this problem; however, having c = 100, as the multiplier, simplified the problem. Example 2 Given the right triangle ABC] find a when 70.
A = -
-
(90
=
20
=
tan
b
=
382.6 and
B =
B)
A
b .'.
With logarithms,
a
log a
log 382.6
log tan 20
Adding,
log a
/.
a
= = = = = = = =
b tan
A
log b
+
log tan
+
log 382.6
A
log tan 20
2.5828 9.5611
12.1439
-
10
10
2.1439 139.3
Example 3 Given the
right triangle
ABC]
find
A when
c
=
389 and a
=
202.
A = sin A =
sin
log
Subtracting,
= = log 202 = log 389 = log sin A A = .'.
c
log a
log c
- log log 202 12.3054 - 10
389
2.5899
9.7155
31
18
-
10
7
Angles of elevation and depression. The angle of elevation of an object above the eye of an observer is defined 14.
SOLUTION OF THE RIGHT TRIANGLE
31
which a
line from the observer's eye to the makes with a horizontal line. Roughly it is the angle through which the observer must elevate his eyes to
as that angle
object
see the object. Thus, in Figure 9, if the observer's eye is at and the object is at 0, the angle of elevation is 9.
E
Figure If,
Figure 10.
9.
on the other hand, the observer
is
above the object,
the corresponding angle that is, the angle through which the observer must depress his eyes is called the angle of Thus, in Figure 10, if the obserdepression of the object. ver's eye
is
at
E
and the object
0, the angle of depression
at
is
is 6.
Example
A
tower stands on the shore of a river 207.2 feet wide (Figure 11). The angle of elevation of the top of the tower
from the point on the other shore exactly Find 24'. opposite the tower is 44
207.2
Figure 11.
the height of the tower.
We are given (assuming the observer's eye E = 44 24', and EC = 207.2. We require h. h
=
on the ground)
tan 44 24'
207.2
Hence
log h log 207.2 log tan 44
Adding,
24'
log h
.'.
h
= = =
= = =
log 207.2
+
log tan 44
2.3164
9.9909 12.3073
2.3073 202.9
-
10
10
24'
PLANE TRIGONOMETRY
32
The height of the tower,
therefore, is 202.9 feet.
Problems Solve the following right triangles, where
1.
(a)
a
(b)
b
(d)
a a
(e)
c
2.
A
(c)
= = =
2234, 126.3,
A = A = = B = A =
1406, b 72.09,
2434,
36 58
19'.
(/)
c
44'.
(g)
c
2173.
(A) c
24 42
33'.
(i)
a
26'.
(j)
b
ladder 41.24 feet long
is
= = = = =
C = 90.
= = 60.23, B 3204, b = = 1.263, A = 2.304, A 149.3, a
so placed that
it
26.24.
68
43'.
2062.
80 22
will
14'.
46'.
reach a
window 34.62 feet high on one side of a street. If it is turned over without moving its foot, the ladder will reach a window 20.28 feet high on the other side of the street.
Find the width of the
street. 3.
A tower stands on the shore of a river 210.6 feet wide. The from a point on the other 40 52'. Find the height
angle of elevation of the top of the tower shore directly opposite to the tower of the tower. 4.
is
A rope is stretched from the top of a building to the ground.
The rope makes an
angle of 52 36' with the horizontal, and the feet Find the length of the rope. 70 high. building a of ladder 60.34 feet long rests against a wall at a 6. The top from feet the ground. Find the angle the ladder point 46.23 and the distance of its foot from the wall. with the makes ground, of hill the angles of depression of two a 6. From the top on milestones a straight, level road leading to the hill successive Find the height of the hill. (HINT: Let x = are 5 and 15. is
= the distance from the bottom to the height of the hill, and y nearer milestone; then eliminate y from the two equations representing the cotangent of 15 and 5, respectively.) From a
point on the ground the angles of elevation of the bottom and the top of a tower on a building are 64 17' and 68 43'. If the tower is 200 feet high, find the height of the 7.
building. 8.
tively.
Two flag poles are known to
be 60 and 40 feet high, respec-
A person moves about until he/find^ a position such that
the tip of the nearer pole just hides thkt of the farther. point the angle of elevation of the top of the nearer pole
At is
this
found
SOLUTION OF THE RIGHT TRIANGLE to be 35
10'.
Find the distance between the
poles,
33
and the
distance from the observer to the nearer pole. 9. A tin roof rises 4f inches to the horizontal foot.
Find the the roof makes with the horizontal. angle 10. From the top of a mountain, 2800 feet above a hut in the hut is found to be 38. Find the straight-line distance from the top of the mountain to
valley, the angle of depression of the
the hut. 11. Find the height of a tree which casts a shadow of 80 feet when the angle of elevation of the sun is 40. 12. From a ship's masthead 160 feet high, the angle of depression of a boat is 30. Find the distance from the boat to
the ship.
From
150 feet high, the angles of depresdue south of the observer, are 32 each sea, and 20, respectively. Find the distance between the boats. 13.
sion of
the top of a
cliff
two boats at
A
an equilateral triangle of perimeter Find the diameter of the circle. 16. Find the length of a chord which subtends a central angle of 64 in a circle whose radius is 10 feet. 14.
circle is inscribed in
90 inches.
From
the foot of a post 30 feet high, the angle of elevation of the top of a steeple is 64; and from the top of the post, the angle of depression of the base of the steeple is 50. Find the 16.
height of the steeple.
From one end
a bridge the angle of depression of an object 200 feet downstream from the bridge and at the water line 17.
of
From
the same point the angle of depression of an object at the water line exactly under the opposite end of the bridge is
is
23.
16.
Find the length
of the bridge,
and
its
height above the
river.
18. From a point directly north of an inaccessible peak, the angle of elevation of the top is found to be 28. From another point on the same level and directly west of the peak, the angle of elevation of the top is 40. Find the height of the peak if the
distance between the two points of observation is 6000 feet. The angle of eleva19. A tower stands on the bank of a river.
from a point directly opposite on the other bank is 55. From another point 100 feet beyond this Find point, the angle of elevation of the top of the tower is 28. the height of the tower, and the width of the river. tion of the top of the tower
PLANE TRIGONOMETRY
34 20.
which
A
tree stands
is
60 feet high.
upon the same horizontal plane as a house The angles of elevation and depression of
the top and the base of the tree from the top of the house are 40 and 35, respectively. Find the height of the tree. 21. The slope of a mountain 4000 feet high makes on one side an angle of 10 with the horizontal, and on the other side an angle of 12. A man can walk up the steeper slope at the rate of 2 miles per hour, and up the easier slope at 3 miles per hour. Find the
by which he will reach the summit sooner. minutes sooner will he arrive?
route
How many
CHAPTER IV
TRIGONOMETRIC FUNCTIONS OF ALL ANGLES and negative angles. In Sections 8 and 9, discussed acute angles and defined the six trigonometric functions of acute angles. In this chapter we shall be concerned with angles of any magnitude, positive or negaWe tive, and the trigonometric functions of these angles. 15. Positive
we
and negative angles. counter-clockwise a rotation, generated by as in Figure 12, we call angle 9 a positive angle.
first
If
distinguish between positive
an angle
6
is
initial side
initial side
Figure 12.
On the other hand, if an angle 6 is generated by a clockwise rotation, as in Figure 13, we call angle 6 a negative angle.
angle of 390, then, we mean an angle generated in a counter-clockwise direction, revolving about by making one complete revolution of
By an
OX
and moving 30 in addition, as indicated in Figure 14. There is a similar understanding
360
for negative angles greater numeri-
cally
than 360, the rotation being
clockwise.
Figure 14.
Directed distances. Just as in the previous section we made a distinction between positive and negative angles, so now we make a distinction between positive and negative 16.
35
PLANE TRIGONOMETRY
36
In general we call a line segment positive if it generated by a point moving from left to right, as AB Figure 15. We indicate such a line segment by AB.
distances. is
in
Figure 16.
the line segment is generated by a point moving from right to left, we call the line segment negative, and indicate If
it
by BA.
from
A
(Observe that AB means the segment generated and that BA means from B to A.) Thus,
to B,
in Figure 15,
BA = -AB. But
it is
desirable often to distinguish between positive and negative distances when the
direction
is
neither from left to right
nor from right to left. In that case, we choose a given direction as posi-
and consider the
tive, lgure
in Figure 16,
direction di-
'
if
Thus, rectly opposite as negative. the direction of the arrow indicates the
positive direction,
we have:
BA = -AB, or:
AB = -BA. It is
a
easy to prove that,
if
A, B, and
C
are three points on
line, then, regardless of the positions of
17. CoSrdinates.
BC =
AB
-4-
Let
X'X and
A, B, and C,
AC.
FT be two perpendicu-
P be any point in the plane For purposes of illustration, the point P is of these lines. taken as in Figure 17. Drop AP perpendicular to OX, and draw OP. The length OA is denoted by x, and is The length AP is denoted called the x, or abscissa, of P.
lar lines intersecting at 0.
Let
FUNCTIONS OF ALL ANGLES by
y,
and
is
37
The x and y
called the y, or ordinate, of P.
taken together, thus (x, y), are called the coordinates of P. F are called the Z'JT and y
F
azes of the coordinates. called the origin, and r the radius vector, of P. If
is
x
is
measured from
right,
we
left to
call it positive; if
from right to left, negative. If y is measured upwards, it is
positive
;
if
II
x
,
o IV
III
downwards,
For convennegative. ience we assume r always to be positive. IB. Quadrants.
Y
'
Figure 17.
The
axes divide the plane into four parts, called quadrants, numbered as in Figure 17. It is quite apparent then that the following arrangement holds for the signs of the x and y of a point in the quadrants indicated.
Thus the point (-2, 19. Trigonometric
3) lies in the
functions of
second quadrant. all
angles.
We
shall
define the trigonometric functions of angles greater numerically than 90 by a process exactly like the process used in
arriving at the definitions of the functions of acute angles. Take the vertex of a given angle at (Figure 18), and the Take any point P on the initial side along the re-axis.
PLANE TRIGONOMETRY
38
terminal side, and drop a perpendicular to the initial side, extended if necessary. This forms a right triangle, called the triangle of reference which, it should be noted, however, does not contain angle 6 unless it is acute.
We
define the trigonometric functions of
as follows
:
II
Observe that, when 6 is acute and therefore in the first quadrant, the above definitions reduce exactly to those given in Section
9.
X X IV
Y Figure 19a.
Figure
when
6
is
Y' Figure 196.
and Figure 196 illustrate the situation positive and in the third and fourth quadrants, 19a
respectively.
FUNCTIONS OF ALL ANGLES
39
It is quite apparent that, except in the first quadrant where x, y, and r are all positive quantities, some of the above functions may at times be negative. The following
table gives the arrangement of the signs of the functions of angle 6 in the quadrants indicated; the signs are derived from the table in Section 18 and the definitions in Section 19.
Example
The
sine of angle
is
I,
and the cosine
of
is
negative.
Figure 20, find the other five functions. Since sin is negative and is negative, cos third quadrant.
lies in
the
Q
Then, since
sin
==
5
we have:
20.
Therefore:
But, since Therefore:
lies in
the third quadrant, x must be negative.
x
= -4.
From
PLANE TRIGONOMETRY
40
Hence we have the
following: sin
=
-
=
4
5 cos
o 3
tan
6
= 7 4
esc e
=
sec
=
-f 5
7 4
cot 8
4 = o
Problems Find in what quadrants the following angles 214; -120; 750; -600; -423; 542; 6000. 1.
In the following problems, consider
positive
lie:
346;
and between
and 360. 2. 3.
4. 5. 6.
7. 8.
9.
10.
Given Given Given Given Given Given Given Given Given
cos sec
cot
tan 6 sin
tan 6 sin 6
esc
tan
f,
20. Functions of
angle
tan
negative; find
all
functions of
6.
= I sin negative; find all functions of 0. = T\, esc 6 negative; find cos 6. = ^, sin 6 positive; find cos 6. = ^, tan 6 positive; find cot 6. = |, 6 not in the first quadrant; find cos 8. = J, not in the fourth quadrant; find sec =4; find cos 0. = 1; find sin 0. ,
0, 90, 180, 270, 360.
very close to
Consider an
(Figure 21).
It is quite evident that,
with A, and hence x
0.
=
r,
= 0,
when
=
0.
= - =
-
=
r
r
and y
point P coincides Therefore:
8in0
cos
1
FUNCTIONS OF ALL ANGLES
41
=*--
tan
x
x
cseO y
(Infinity*)
sect)
x
= X~ =
Cot
X -
=
oo
v
consider an Similarly, ~iO angle 6 very close to 90
(Figure
When 0=
22).
Figure 21.
90, point P coincides with # = 0. Therefore:
and we have
J5,
=
1
cos 90
= - =
-
tan 90
= - = - =
sin
90
=
r
-
=
y,
and
r
= oo
x esc 90
- - =
1
y sec 90
x cot 90
Figure 22.
= - = - = y
Similarly, consider
cally,
and *
-,
y
but x
r is
By
that
0,
is
and x
point
=
esc
we mean
0,
very close to 180
P
that, as
(Figure
Hence and r are equal numeriand is therefore negative,
coincides with A.
r; for x
directed to the left
always positive.
infinity is,
an angle
= 180,
When 23). we have y =
y
Therefore: approaches zero, y approaches zero, and
increases without limit.
PLANE TRIGONOMETRY
42
sin 180
-
v -
-
r
=0
r
*-=!.-! r r X
X
r
r
=
r=
00
V
= -1 x COt 180
Figure 23.
r
= - =
=
oo
y Similar results obtain for the functions of 270, which are tabulated below; and it may readily be seen that the
functions of 360 tions of
are the
same
as the corresponding func-
0.
21. Functions of
as
varies from
to
360.
Let
us,
by means of the above table, study the development of the various functions of an angle as the angle varies from to
360.
First, consider sin 0.
Sin 6 starts at
when
is
0.
increases to +1, taking Then, as 6 increases to 90, sin increases from 90 to and 1. As all values between
FUNCTIONS OF ALL ANGLES 180, sin 270,
to
decreases from sin 6 decreases
increases from 270
This process repeats
to
to 0.
1
As
from
360,
sin
itself as d
43
increases
to
1.
increases
from 180
Finally,
from
-1
as 6
to 0.
continues to increase.
Figure 24.
The process may be illustrated geometrically. Let us consider a circle of radius 1, called a unit circle (Figure 24).
Then sin
= ^ =
= ^
Sin 6
may
be represented by the
y.
line
AP.
It is quite
apparent from Figure 24 that the development of sin 6 as 6 to 360 is as stated above. varies from Cos 6 behaves similarly. When 6 is 0, cos 6 is 1. As decreases from 1 to 0. 8 increases from to 90, cos As 6 increases from 90 to 180, cos 6 decreases from to increases As 6 increases from 180 to 270, cos 1. increases from 270 to 360, cos from -1 to 0. As to 1. increases from to 90, When is 0, tan is 0. As increases from Since the tangent function is tan increases from to <*> negative in the second quadrant and very large numerically for angles slightly larger than 90, and since it increases .
PLANE TRIGONOMETRY
44
without limit if we approach 90 through such angles, we oo say tan 90 equals Hence, as increases from 90 to oo to 0. 6 tan increases from As 6 increases from 180, oo 180 to 270, tan increases from to As 6 increases oo to 0. from 270 to 360, tan 6 increases from .
+
When esc
6 is
0,
esc 6 is
decreases from
oo
oo
As
.
to
increases
As 180, esc increases from 1 to oo oo to to 270, esc increases from from 270 to 360, esc decreases from .
When
from to 90, from 90 to increases from 180 1. As increases
increases
As
1.
.
As
1
to
oo
.
from to 90, 0, increases from 1 to oo As sec increases from 90 to oo to increases from 1. As increases 180, sec oo from 180 to 270, sec decreases from 1 to As increases from 270 to 360 sec decreases from oo to 1. When is 0, cot is oo As increases from to 90 cot decreases from oo to 0. As increases from 90 to oo As increases from 180, cot decreases from to 180 to 270, cot decreases from + to 0. As increases from 270 to 360, cot decreases from to - oo is
sec
is 1.
increases
.
.
.
,
.
<
.
The statement tan 90 = may be illustrated geometrimeans of circle a unit (Figure 25a). cally by <*>
tan
6
= AP
AP
1
Figure 250.
Figure
FUNCTIONS OF ALL ANGLES
P
is
45
determined as the intersection point of a tangent to
A and the terminal side of angle 6. It is quite apparent that, as approaches 90, the terminal side of with the tangent at A, and parallelism approaches recedes indefinitely (Figure 256).
the circle at
P
From the above discussion we see that the
sine and cosine 1 and +1, or equal to an angle are always between 1 or + 1 that the tangent and cotangent may take any values; 1 and +1, that the secant and cosecant are never between 1 and +1. but may take all other values, including 22. Functions of (180 In the 8) and (360 e). previous section, we observed that, as 6 varied from varied from to 1 to 0. It is quite to 90 to 180, sin if consider a number between and 1 we evident, then, in the first is an that there angle quadrant whose say an in there the second quadthat and sine is fangle is, also, is the relation between these What rant whose sine is f two angles? Our answer is: they must be suplementary; that is, the sum of the two angles must equal 180. We shall prove a general theorem regarding such angles. Theorem: of
;
.
sin (180
-
0)
=
sin
0.
Proof
Given the right (180
-
B).
triangle
Construct
ABO
Z.COX
(see Figure 26),
equal to
0.
M
Drop a perpendicular from P to OX at Letter triangles MOP and BOA are congruent. Hence we have:
y X
= -
y> -X'.
with
/.XOA =
Take OP = r = r'. Then the right .
as in Figure 26.
PLANE TRIGONOMETRY
46
The
sine of
/.XOA =
y
=
equals -> since y'
y and
r
sine
0.
-
(180
r'
is
0)
=
*y (see Figure 26).
y But -
r.
(see Figure 26) equals
r
Or, restated,
-
sin (180
6)
' .
.
sin (180
= ^ = r
r
-
sin
0)
=
-
=
sin
6.
0.
Figure 26.
Similarly,
-
cos (180
0)
=
=
-,
-cos
r
r
-
/.cos (180
0)
= -cos
0.
Similarly,
-
tan (180
0)
/.tan In
=
=
-,
(180
like fashion, esc (180 sec (180
cot (180
-j-
-
0)
0)
0)
0)
= --
-tan
0.
= -tan0. = esc 0, = -sec = -cot
0, 0.
Example Find: sin 150; cot 135. sin 150
cot 135
= =
sin (180
cot (180
- 30) = - 45) =
This
sin
30
=
-cot 45
1.
FUNCTIONS OF ALL ANGLES
47
Similar results obtain for the functions of (180 indicated in the following text:
Proof In Figure 27,
we have:
Figure 27.
Hence we have the
following:
sin (180
+
0)
=
~
=
cos (180
+
0)
=
~
=
tan (180
+
0)
csc (180
sec (180
cot (180 Similarly,
we have
=
+ + +
-_ x' 6) 0) 6)
=
=
= x
= -esc = -sec = cot 6
x 6
6
these results:
sin (360
cos (360
tan (360
-
0) 6} 0)
= -sin = cos 6 = -tan
6
=
sin
=
cos
tan $
+ 0),
as
PLANE TRIGONOMETRY
48
-
esc (360
sec (360
cot (360
From
0)
- -csc - sec
0)
-cot
6)
the above discussions
we may
state the following
theorem.
Theorem.
Any
in fact, of (n!80 to the
0) or (360 function of (180 n is a where 0), positive integer
same function of
0) is
equal with the sign depending upon the
0,
quadrant in which the angle
lies.
Thus: cos 240
(since
240
is
- cos (180 + 60) = -cos 60
in the third
quadrant and cosine
is
negative
there)
In the above proofs, was taken as acute. The theorem holds, however, regardless of the size of 0, as does also the information about to be obtained regarding a negative angle.
y
Figure 28.
23. Functions of
(
0) .
What relations hold between the
functions of a negative angle and the functions of the
corresponding positive angle?
Consider Figure 28,
FUNCTIONS OF ALL ANGLES r
x y
- r' = x = -y'
*---V
I/
sin
(-0)
0)
(
= -x =
x
r
r
cot
$
= +cos
= esc = sec ( 0) = -cot (-0) 0)
(
sec
-
_ sin
r
t/ v = - = -^v = -- = -tan
tan (~0) csc
H
r
r'
cos
49
Problems Example Find: sin
-
1
300. sin
-
= -sin 300 - -sin (360 - 60) = -(-sin 60)
300
sin
60
V3
as
"^'
Example 2 Find
all
-
2 sin 2
and 360
angles between 1 = 0. 3 sin
-
Hence:
sin
= =
or:
sin
0=1.
If
sin
2 sin 2
3 sin
+
1
then sin
If
then
Hence: 1.
In Problems (a) to Find:
the tables.
which
satisfy the equation:
+
= =
(2 sin
-
l)(sin
-
1)
=
0.
i,
i,
30 or 150.
0=1, = 90. = 30,
(I),
150, or 90.
the student
is
requested not to use
PLANE TRIGONOMETRY
50
Solve the following equations to find, for the and 360. letters, all values between 2.
C08
-- = -- =
(c)
tan 6
(d)
sin x H
(e)
tan x
(/)
3 sin 2 x
tan 6
2
(i)
(tan
sin 2 x
(1)
\/3 =
-
= x 6
C;)
2
=
-
(m) 6 cos x
0.
5 sin x
+
2
=
0.
0.
+
sin
-
x
3) (esc sin x.
= \/S 5 sin d
1
-
-
2 sin 2 x sin 2 ^
3.
sin x
+
(g) sin 2x 2 (h) 2 sin
(fc)
4.
-\
= 2)
0.
=
0.
sin x.
+
5 cos x
6
=
0.
+
1
=
0.
unknown
V THE OBLIQUE TRIANGLE CHAPTER
24. Law of sines. In Chapter III we considered the solution of the right triangle. In this chapter we shall be concerned with the solution of the oblique triangle, for
which we use two important laws giving relations between the sides and the angles of such a triangle. We proceed to the derivation of the first of these laws, called the
law of sines:
Law of Sines. The sides of a triangle are proportional to the sines of the angles opposite. Consider the triangle ABC in Figure 29. We wish to prove :
a sin
b
A
sin
c
B
sin
Figure 29.
C Proof
From
AB.
C, drop h perpendicular to
Then
sin
A =
and
sin
B =
sin
Dividing,
sin
A 5
->
a
h/a
a or: sin
Similarly,
it
may
a
h/b b
A
sin
B
be shown that a sin
A
sin 51
C
PLANE TRIGONOMETRY
52
or
B
sin
sin
C
Hence:
In Figure 29, all the angles were taken as acute. The law holds, however, if an angle is obtuse, as A in Figure 30. Proof
Drop h perpendicular
to
extended to
c,
M. sin
B =
a k
A =7
sin A
^"^
L.
M
by the
.
lgure
The
definition of the sine of
"
an angle
in the second quadrant.
rest of the proof is similar to the preceding proof.
25. Applications of the of sines,
any two
law of sines.
By
using the law
are enabled to solve a triangle if we are given angles and a side, as in the following
we
:
Example Given
A =
52
13',
B =
73
180'
-
24',
c
+
B)
=
triangle.
C = = .'.C a sin
179 60'
54 23' c
A a
sin
=
C
c sin
sin
A C
(A
-
125 37'
6293.
Solve
the
THE OBLIQUE TRIANGLE a = log c + log - 3.7989 log 6293 log
log sin 52
13'
log sin 54
23'
-
'.log a
.
.'.
a
13.6967
-
10
9.9101
-
10
9.8978
B
=
log 6293 log sin 73 24'
log sin 54
23'
=
'.log c
= =
log 6
C
5
log sin
C
10
c
C
log c
+
.'.c
C a = c =
log sin
3.7989
9.9815 13.7804
Hence:
log sin
3.7866
sin
= = =
.
A
6118
b sin
= =
sin
53
9.9101
-
10
10 10
3.8703
7418 54
23',
6118, 7418.
Problems Solve the following triangles:
A = 46 52', B = 64 43'. 406.2, 5 = 19 36', C = 80 52'. 6601, A = 50 32', C = 100. = 25 42', 5 = 40 19'. 32.04, A = A 46 10', B = 63 50'. 530,
5. c
= = = = =
26.
Ambiguous case.
2.
a a
3.
6
1.
4. c
26.32,
The other type of triangle handled
that for which there are given two by This sides and one angle opposite one of the given sides. case presents slightly more difficulty, however, because, the law of sines
is
with the above material, there may be no triangle possible, there may be one and one only, or there may be two triangles possible both of which contain the given material.
PLANE TRIGONOMETRY
54
This case is therefore known as the ambiguous consider a concrete illustration
Let us
case.
:
Example Given
A = 30,
we
In Figure 31, sides
AC.
a
=
75, b
=
100.
construct on
AX at A
an angle
of
30, with
On AM we lay off 100 units (6) from A, say center and (a = 75) as radius, we swing an arc.
AX and AM. With C as
The number
of solutions
depends on whether or not the arc cuts
AX and,
if
so,
In this particular at case, the arc cuts two points B f and B, both to the right of A. Since both triangles, ACB and where.
AX
-X
contain the given material, there are consequently two solutions.
Obviously there will always be two solutions when the length of a is numerically less than 6 and greater than the perpendicular h, dropped from C. Also, there will never be a solution if a is less than h. There will be but one solution, a right triangle, if a equals h\ and there will be but one solution, an isosceles triangle, if a is greater than h and equal to b. Finally, there will be but one solution if a is
greater than h and greater than b; for, although the arc at two points, one will be to the left of A and
will cut
AX
the triangle thus formed will not include A. The matter of finding h is very simple :
-
Since
=
sin
o
In Figure 31,
A,
h
b sin A.
h
100 50.
The above proof and explanation obtain when acute.
If
A
is
A
is
obtuse, there will be one and only one
solution provided a
is
greater than
6,
and then only.
THE OBLIQUE TRIANGLE
We may
55
summarize our findings as follows:
Case I. Given A, a, and 6, with A acute. There will be two solutions if 6 sin A is less than a, and a is less than 6. There will be one solution if 6 sin A equals a, or if a equals 6, or if a is greater than 6. There will be no solution if a is less than b sin A. Case II. Given A, a, and 6, with A obtuse. be one solution if a is greater than 6.
There
will
In solving a triangle, the student should realize that the only possible chance for there being two solutions is in the case when A is acute and a is less than b. If such is the case, the student should then find the relation between a and b sin A, and proceed accordingly. Example Solve the triangle; given a
This
Figure 32.
is
=
=
800, b
1200,
A =
34.
(See
obviously a chance for two solutions.)
C
B
B' Figure 32.
Using
log b
logs,
log sin
A =
=
log 1200
log sin
34
= =
A = = log 800 =
log 6 sin log a
(Since log a
We
first
is
3.0792 9.7476 12.8268
-
10 10
2.9031
greater than log 6 sin A, there will be two solutions.)
solve triangle
ABC in Figure 32. sin
B
sin
B
sin
A
PLANE TRIGONOMETRY
66 log sin
B
log 6
+
log sin .'.
C -
-
180
B =
57
+
C =
B) 88
C
sin
(A
:.
sin
=
log c
A -
log a
log
800
log sin 88
59'
= = = =
log sin 34
= =
log c c
.'.
9.9237
-
91
=
180
C'
A log sin
A
2.9031
9.9999
12.9030 9.7476
-
10 10 10
3.1554
1430
B = = 180 -
B'
= /ACS' =
57
1'
57
1'
-
180
Let
AB' =
c'
=
log 800 log sin 23
log sin
1'
= =
= =
34
log c
1
/. c'
&ABC:
sin
+ log sin
log a
122
c',
59'
+
59'
34) = 23
a_
sin C'
log
=
(122
c'
Hence, for
1'
C'.
.*.
/.
10
59'
We next solve triangle AJ5'C, in Figure 32. ZACB' =
-
1'
+ log sin C
log a
=
= =
A C'
log sin
2.9031
9.5922 12.4953
9.7476 2.7477 559.4
B =
57
1',
C c -
88
59',
1430;
-
10
10 10
A
1'
and
THE OBLIQUE TRIANGLE and, for
= C" = d =
AAS'C:
B'
122 23
57
59', 1',
559.4.
Problems 1.
Find the number of solutions; given:
A A A A A
(a) (6) (c)
(d) (e)
2.
30, 30, 30, 30, 30,
b
= =
6
==
b
200, a 400, 600,
= =
6 6
500, 500,
Find the number
A A A B B
(a) (b) (c)
(d) (e)
3.
= = = = = = = = = =
150, b 150, 6 150, b 150, b 150, c
= = = = =
a a a a
= = = = =
101. 100.
300. 500.
600.
of solutions; given:
200, a 200, a 200, a 200, a
400, 6
= = = = =
150.
300. 200.
300. 300.
Solve the following triangles:
(a) (6) (c)
(d) (e) (/)
= = = = C = A =
A A A B
59 140
32 47 62
Law
27.
65
= 7072, b = 7836. = 40.34, b = 30.29. 26', = 600.8. = a 464.7, 6 14', = = a 3015. b 3247, 46', = = a c 400. 375, 34', = 20.43, b = 30.32. 53', a 26',
a
a
of cosines,
and
applications.
There are two
other types of triangles to be considered; that is, triangles with three sides given, or triangles with two sides and the included angle given. These two types are handled by the
law of cosines.
Law
We
proceed to
its
derivation.
side of a triangle other two sides diminished equals the sum of the squares of the by twice the product of these sides and the cosine of the included
of Cosines.
The square of any
angle.
Given the triangle ABC, prove
:
in Figure 33,
We
wish to
PLANE TRIGONOMETRY
58
a2 52 C
2
Let us prove the a
2
a a2
=
2
6
26 C cos
2
2ac cos
A B
2
2a & cos
C
relation
first
2
2
+ C __ +c + b __
52
= =
:
+c 2
26c cos A.
C
Figure 33.
Proof to
A B at M.
-
AM MS
Drop a perpendicular h from C A2
Then and Equating,
h
2
= =
6
2
a
2
2 ,
2 .
we have:
-
- AM 2 2 a 2 = 6 + MS - AM MB = c - AM 2 M5 = c 2 + AM\ a2
or:
Then, since
and substituting,
Or:
MB
=
2
,
2
2
we have: a2
=
b
2
+
AM
But
-7
c
=
o
AM
or
Therefore:
62
a
2
=
&
=
-
2
cos A,
6(cos A).
+ A was
2
c
2c(AM).
2
2&c cos A.
taken as acute. We shall In Figure 33, angle that the law holds when A is obtuse, as in
now show Figure 34.
THE OBLIQUE TRIANGLE
59
C
Proof
Drop a perpendicular h As
h2
before,
and
A
Then a
2
-
MS
2
a2
Hence:
But
cos
from the rant.
2
= = = = = =
A =
6
2
a
2
6
2
6
2
6
2
6
CM)
(equal to
-
MZ M5 Ml
+M
,
2 .
2 .
2
-
+ (MA + + c + 2c(MA) + Ml - MZ 2
2
2
2
AM an angle
definition of the cosine of
However,
C to BA, extended.
from
2
in the second
quad-
since
AM then
= -MA,
MA
A =
cos
r~
MA
Therefore:
a2
Hence, substituting,
'
b cos
=
6
2
+
c
2
A.
-
26c cos
The other two relations may be derived in similar fashion
We may
apply the law of cosines as in the two examples
below.
Example Given a Since
=
4, 6
=
5, c
a2
=
= b
2
1
6; solve the triangle.
+
c
2
-
26c cos A,
PLANE TRIGONOMETRY *u then
+
.
A
cos
--
=
60
Similarly,
a8
2bc
+ 36-16
25
-
c'
45
=
/.
A =
cos
B =
3
=
-
n 7500
-41
a2
-
4
6-0
25'.
+ -C
2 __
52 >
2ac
+ 36 -
16 or
25
^
cos
C =
-
-1
40
.'.
As a check:
_9_
16*
log 9
log 16
10.9542
-
10
-
10
_
^2
1.2041
9.7501
55 ^2
46'. i
Jj2
2ao
+ 25-36 =
16 or
= ~
48
= log cos B log 9 = log 16 = = log cos B = /. B Similarly,
27
""
48
C =
A +5+C=
5
= -1 =
40
8
82
49'.
^
Prt
.1250.
180.
Of course, as soon as we had found A, we could have used B and subtracted (A + B) from 180 to find C. But with convenient numbers, as in this
the law of sines to find
example,
it is
fully as easy to proceed as above.
Example 2
=
Solve the triangle; given a Solving,
c
.'.
2
c
= = = -
20, 6
=
25,
+ 6 - 2ab cos C 400 + 625 - 2 500 2
a2
1025
C = 60.
-
525. 22.91.
500
$
THE OBLIQUE TRIANGLE
61
We
can now find A and B either by the law of cosines or by the law of sines. It is apparent from the above examples that the law of cosines is not particularly well adapted for the use of logarithms. There are formulas which are better fitted for logarithmic use, but it is the author's feeling that an intelligent use of the tables of squares and square roots combined with the law of cosines is fully as easy and does not involve
remembering a set of formulas and which are not particularly essential.
their
We
derivations,
shall illustrate
in the following:
Example 8 Solve the triangle; given a
We
have:
By
the table of squares,
cos
=
A =
20.63, 6 6
2
+
c
2
=
34.21, c
-
a2
=
40.17.
2bc
a2 2 fc
c
(The interpolation
is
2
= = =
425.6,
1171, 1614.
exactly the same as in logarithms.)
2359.4
2X40.17X34.21
Now we
can use logarithms:
log cos
A =
log 2359
-
log 2
log 2 log 40.17 log 34.21 log
denominator log 2359
log denominator log cos .'.
= = = = = =
A = A =
-
log 40.17
-
.3010
1.6039 1.5341
3.4390 13.3727
-
10
-
10
3.4390 9.9337
30
51'.
log 34.21
PLANE TRIGONOMETRY
62
We may proceed similarly to find B and C; or we may use the law of be easier in
sines.
The
latter procedure
would probably
this example.
Problems Solve the following triangles:
1.
a a a a a a a a
(a) (6) (c)
(d) (e) (/)
(0)
(h) (t)
b
0")
a
= = = = * = = = = =
= 7, c = 10. - 6, C = 60. = 8, C = 120. 5, 6 = 20, c = 25. b 12, = 28.43, c = 22.06. 19.62, 6 c 14.72, 25.39, B - 22 17'. = & 2032, 2491, c = 3824. = = 1.91. 6 2.63, c 1.32, =* = 135 27'. A c 3.96, 2.04, = = 47 43'. c 5 500.2, 423.1, 5,
b
4, b
2. Show that the area of a triangle may be written as one-half the product of any two sides and the sine of the included angle. 3. Show that the radius R of the circle circumscribed about a
triangle
ABC is given
by
2R sin
A
sin
B
sin
C
4. Show that the area of any quadrilateral equals one-half the product of the diagonals and the sine of one of the included
angles. 6. In a parallelogram, the sides are 6 and 15, and the smaller vertex angles are 50. Find the lengths of the diagonals. 6. A and B are points 300 feet apart on the edge of a river, If the angles CAB and is a point on the opposite side. are 70 and 63, respectively, find the width of the river.
and C
CBA
7. From a mountain top 3000 feet above sea level, two ships are observed, one north and the other northeast. The angles of depression are 1 1 and 15. Find the distance between the ships.
A tower stands on one bank of a river. From the opposite the angle of elevation of the tower is 61; and from a point bank, 45 feet farther inland, the angle of elevation is 51. Find the width of the river. 8.
THE OBLIQUE TRIANGLE
63
A cliff 400 feet high is seen
due south of a boat. The top observed to be at an elevation of 30. After the boat travels a certain distance southwest, the angle of elevation is found to be 34. Find how far the boat has gone from the first 9.
of the
cliff is
point of observation.
A vertical tower makes an angle of 120 with the inclined on which it stands. At a distance of 80 feet from the base plane of the tower measured down the plane the angle subtended by the tower is 22. Find the height of the tower. 11. Two persons stand facing each other on opposite sides of a pool. The eye of one is 4 feet 8 inches above the water, and 10.
that of the other, 5 feet 4 inches. Each observes that the angle of depression of the reflection in the pool of the eye of the other is
50. 12.
Find the width of the pool.
A flag pole stands on a hill which is inclined 17
to the hori-
From a
point 200 feet down the hill, the angle of elevation of the top of the pole is 25. Find the height of the pole. From 13. A tower 100 feet high stands on a cliff beside a river. a point on the other side of the river and directly across from the tower, the angle of elevation of the top of the tower is 35, and that of the base of the tower is 24. Find the width of the zontal.
river.
14.
A ladder leaning against a house makes an angle of 40
the ladder makes an angle of 75 length of the ladder. 15.
with
When its foot is moved
the horizontal.
Two forces
make an angle of 24 their resultant. 16.
Two men
18.
An
10 feet nearer the house, with the horizontal. Find the
one of 10 pounds and the other of 7 pounds Find the intensity and the direction of 42'.
a mile apart on a horizontal road observe a balloon directly over the road. The angles of elevation of the balloon are estimated by the men to be 62 and 76. Find the height of the balloon above the road. To find 17. Two points A and B are separated by a swamp. outside the is taken C the length of AB, a convenient point as follows: found are ACB swamp; and AC, BC, and angle AC = 932 feet, BC = 1400 feet, and ACB = 120. Find AB. sea.
A
on a cliff 200 feet above the surface of the hovering above him, and its reflection in the sea
observer
gull is
is
can be seen by the observer.
He
estimates the angle of eleva-
PLANE TRIGONOMETRY
64
tion of the gull to be 30, and the angle of depression of its reflection in the water to be 55. Find the height of the gull above the sea.
An electric sign
40 feet high is put on the top of a building. point on the ground, the angles of elevation of the top and the bottom of the sign are 40 and 32. Find the height of 19.
From a
the building. 20. A cliff with a lighthouse on its edge is observed from a boat the angle of elevation of the top of the lighthouse is 25. After the boat travels 900 feet directly toward the lighthouse, the ;
angles of elevation of the top and the base are found to be 50 Find the height of the lighthouse. respectively.
and 40, 21.
Two
trains start at the
same time from the same station
upon straight tracks making an angle of 60. If one train runs 45 miles an hour and the other 55 miles an hour, find how far apart they are at the end of 2 hours.
From the top of a lighthouse, the angle of depression of a at sea is 50; and the angle of depression of a second boat buoy feet farther out to sea but in a straight line with the 300 buoy from the top of the lighthouse is 28. Find the first buoy 22.
height of the lighthouse.
A
50 feet high stands on the top of a tower. of the tower, the angles of elevation of the top and the bottom of the pole are 36 and 20, 23.
flag pole
From an observer's position near the base
Find the distance respectively. the base of the tower.
from the observer's position to
A
lighthouse sighted from a ship bears 70 east of north. After the ship has sailed 6 miles due south, the lighthouse bears 40 east of north. Find the distance of the ship from the light24.
house at each time of observation. A 25. Two trees on a horizontal plane are 60 feet apart. person standing at the base of one tree observes the angle of elevation of the top of the second. Then, standing at the base of the second tree, he observes that the angle of elevation of one When the observer stands tree is double that of the other. half-way between the trees, the angles of elevation are complementary. Find the height of each tree. 26. Two points are in a line, horizontally, with the base of a tower. Let a be the angle of elevation of the top of the tower from the nearer point, and the angle of elevation from the far-
THE OBLIQUE TRIANGLE
65
ther point. Show that, if d represents the distance between the points, the height of the tower is
d
sin
a.
sin
sin (a
0)
A man on a
27. cliff, at a height of 1320 feet, looks out across the ocean. radius of the earth is assumed to be 4000 miles.) (The Find the distance from the man to the horizon seen by him. 28. Find how high an observer must be above the surface of the ocean to see an object 30 miles distant on the surface. 29. From the top of a building at a distance d from a tower, the angle of elevation of the top of the tower is a, and the angle Show that the height of the tower of depression of the base is /?. is
d sin (a cos
a.
+
g)
cos
30. If r is the radius of the earth, h the height of an observer level, and a the angle of depression of the observer's
above sea horizon,
show that tan a
=
An observer, due north, estimates 31. A balloon is overhead. the angle of elevation to be a. Another observer, at a distance d due west from the first observer, figures his angle of elevation Show that the height of the balloon above the observers to be p. is
d sin a sin
-
sin 2
/?
CHAPTER VI
TRIGONOMETRIC RELATIONS This chapter is concerned 28. Fundamental identities. with relations of the trigonometric functions of angles of various size and formation. In the present section we shall derive the so-called
fundaAlthough, in the Figure 35, angle under con-
mental
identities.
sideration
acute, any angle might have been used.
The
is
following relations
immediate consequences
are
of the
Figure 35.
the six trigonometric functions of an angle:
definitions
of
(1) esc e
= sin
(2) sec 6
=
(3) cot e
=
1
cos 1
tan
1
(4)
sin
CSC 1
(5)
cos sec B 1
(6)
The
first
relation
is
=
tan 6
cot e
proved as follows:
esc 6
= - = y
= y/r
66
-
sm
6
TRIGONOMETRIC RELATIONS The other
We
relations are
have
also
proved similarly.
^
:
(7)
,
The second
the
first,
cos
^
we
=
tan
=
cot e
cos e
(8)
To prove
67
sin 6
substitute
sin 6
y/r ~
~"
cos 6
x/r
x
and have
:
y
tall
(7.
proved in similar fashion. There remain to be discussed three other important
identities
is
:
(9) sin
(10) (11)
To prove
1 i
2
6
+
+ tan + cot
=
cos 2 e 2
2
6 e
= =
1
sec
2
esc
2
these three relations,
6 e
we apply the law
Pythagoras to the triangle in Figure 35
Dividing both sides of this equation by
of
:
r2
,
we have:
0y +eyor: sin 2 6
and dividing by y
,
cos 2
=
1.
2 by z we obtain the second we obtain the third.
Similarly, dividing 2
+ ,
relation;
Since these relations are of fundamental importance, the
student should memorize
all of
them.
object in proving an identity is to reduce both This may sides of the given relation to the same quantity.
NOTE The :
be done by working with the left-hand side alone, or with the right-hand side alone, or by working with both sides. In the last instance, we feel that the problem is aesthetically a bit more nicely done if the two sides are not combined;
PLANE TRIGONOMETRY
68
moreover, the practice of combining the sides frequently leads to errors in computation. Thus, suppose we wish to prove the following:
-2 = Squaring,
=
4
which
is
2.
we have true.
4,
Hence, we reason, the original relation
-2-2 is
true; but this conclusion
obviously absurd.
is
1
Example Prove the following identity:
-- --
cos 6
-
sin 6
=
1
sec 8 esc
0.
cos 6
sin
Since cos 2 6
+
=
sin 2
1,
the left-hand side becomes:
The
cos 8
sin 6
sin
cos 6
cos 2 6
+
sin 2 6
I
sin 6 cos 6
sin 6 cos
right-hand side becomes:
_ cos
sin
sin
cos
Example 2 Prove: (1
+
cot 2 0) cos 2
=
=
esc
cot 2
0.
Since 1
+
cot 2
2
0,
the left-hand side becomes:
esc
2
cos 2
=
-
sm 2 _ ~
COS 2 sin 2 6
cos 2
TRIGONOMETRIC RELATIONS which the right-hand side also equals.
Example S Prove: 1
+ sin
cos 6
cos 6
sin 6
1
We know 1
-
sin 2 B
=
cos 2
6.
That suggests multiplying both numerator and denominator sin 0). We have then: the left-hand side by (1 1
+
sin 6
1
_
-
sin
cos B (1
cos 8
2
of
6
sin 6)
cos 2 B cos 6 (1
sin 6)
cos 6
_ ~~
-
1
sin B
which the right-hand side also equals.
apparent from the above that there is no set rule to follow in proving identities; but, in general, a safe It is quite
rule
is
:
Reduce everything necessary,
make use
to
and
sines
cosines.
2 of the identity: sin 6
+
Then, wherever cos 2 6
=
1.
In later sections of the text where we are considering relations involving double-angles, half-angles, and so forth, it will generally be found desirable to reduce our quantities
to functions of a single angle.
Of course, any time the quantity
we may
substitute,
(1
+
tan 2
0)
appears,
first,
sec 2 6
and, then, 1
cos 2
forget that particular identity, the above method of reducing everything to sines and cosines will still hold. If
we
PLANE TRIGONOMETRY
70
one side of an identity to be proved is more complicated than the other, it is advisable to reduce the more complicated side first. Also, in general,
if
Problems Prove the following
+
1.
tan
2.
cos 6 tan 6
3.
(sin
identities:
=
cot 6
=
sec
sin
esc
6.
A + cos A) = 1 + A cos A) 2 = 1 (sin 2 2 = 1. (sin A + cos A)
6.
(1
4.
0.
6.
2
2 sin 2 sin
A A
cos A. cos A.
2
+
sec
-
7. sec 8
=
1
-
sec 8(1
+
tan 8 sin 9. sin X(l tan X) 10. cos tan esc 8. cos 6
2 A) = tan
cos
A) (I
=
cos sec
A
cos A.
6).
0.
+ cos X(l + cot X) = sec X + esc X. X = sin X sec X cot X. esc A cot A = esc A + cot A. = (1 + tan 0) + (1 + cot 0). tan (1 + 0)(1 + cot 0) + 1) + 1 = 0. (cos l)(cot +
X
11. 12.
13.
4
4
16. 17.
2
2
2
2
+ esc 0) = sin + cos + (1 - cos 0) = (sec (tan cot X cos X = cot X - cos X. (sin A + esc A) + (cos A + sec A) cos 0(sec
14. sin
15.
X
-
0.
sin 0)
2
-
18. sin 19. sin
22. rt
_
23.
24.
25
^ 26.
cos 3
-
sin
cos 2
2
+
(sin
tan 2
A -
cot 2
0.
cos 0)(1
sin
cos
0).
- sin 0)(1 + sin cos 0). 4 = 5 sec 5. 1 tan B 2 - cos 2 A) 2 = 1 - 4 cos 2 A + 4 cos A. (sin A tan A + tan 5 - = tan A tan B. cot A + cot #
20. cos 21.
cos
+
3
= = =
.
2
2
4
I)
2
2
2
4
2
2
2
3
-
sin 3
-
4
4
.
=
]
1
sin A + 5^4
1
+
tan 2
A
esc
sec
sec
(cos 2 sec 2
+
esc
(sec
1
=
A -
tan A) 2
+ cot A 2
01 --
tan tan
0+1
.
A =
7.
TRIGONOMETRIC RELATIONS 1
27.
+
2 cos
sm
=
tan
A
1
tan
A+
1
1
+
A 1 + cot A sec A tan A +
0.
2
cos 2
A
2
1
30.
2 cot
cot
1
1
29.
+
esc B
6
71
2
= sin A cos A. A + cot A sin A 1 + cos A = 2 esc A + 1 + cos A sm A sec A sin 3 A tan A sm A = 1 + cos A ,
-
tan
,
31.
32.
.
...
" 1
33.
,
1
+ .
.
,
sin 2
A +1+ ,
*
1
r-7 + 1 + cos 2 A
sec
2
A 1 1
A+
sin
tan
A
sec
sin ^ tan & 40. sin 3
v + X .
esc
2
A
cos A.
..
+ sec X. ,
-
+
t
1
sec 1
+
29. Functions of (90
cos
+
6).
For future work we wish
We shall take the functions of (90 6) in terms of 0. 6 as acute; the results obtained, however, hold when is of
+
any magnitude.
PLANE TRIGONOMETRY
72
In Figure 36, the angle XOB equals (90 0); the triwith and sides as in the figure. angle j/' appears r', x', From 0, take OP perpen-
+
OB, and of length
dicular to
=
f
Drop a perpendicular from P to OX at A; denote by r, x, and y the
r
r
.
sides of the right triangle
thus
formed.
Then,
the
two triangles are congruent and we have: r = r';# = y and j/ = x'. r
;
we
First
sin (90
+
(90
Figure 36.
+
=
0)
-,
= - =
r
shall find: sin
6).
cos
0.
r
Similarly,
= -- - -sin
cos (90
0.
And:
+ 0) + 0) + 0) + 0)
= -cot 0, = sec 6, = -esc B, = -tan 0.
Observe that, except for the
signs, the
tan (90 esc (90
sec (90
cot (90
exactly the same results obtain for the functions of (270
above
results are
as those obtained in Section 11.
Similar
Hence we
0).
have:
Theorem.
Any function
of (n90
6)
when n
is
an odd
positive integer is equal to the corresponding co-function of 0, with the sign depending on the quadrant in which the angle lies.
We
shall find particularly useful in Section lowing relation: cos (90
+
8)
-sin
0.
32 the
fol-
TRIGONOMETRIC RELATIONS
73
30. Principal angle between two lines. Consider the two directed lines AB and CD, intersecting at (Figure 37). There are various angles from the positive direction of
one
line to the positive
di-
rection of the other, such as those indicated by 1, 2, and 3
on the
figure.
A~*^
*D
Figure 37.
Of all such one angle which
is positive angles, there is call this angle the principal angle.
and less than 180.
We
is
""
y
\^
In Figure 37,
it
angle 2.
Consider the directed
31. Projection.
and the directed
line
respectively, drop
CD
line
segment
AM and BN
line
MN
segment
the projection of and is written proj<7z>AjB
M
Now,
N
Figure 38.
AB
if
meet
CD
angle
is
is
=*
called
is
AB on CD, MN.
extended to
and the principal denoted by 0, and if
AE is drawn perpendicular to BN, it is evident that BAE = 6 and that AE = MN. Hence, since cos
AB
From A and B, perpendicular to CD. The
in Figure 38.
~ = AE
AB
angle
>
then:
MN From
= AE = AB
the above explanation
on projection.
we
cos
0.
derive the
The theorem is true and the magnitude
direction of the lines
AB Theorem 1.
The projection of a
first
regardless of
theorem of
the
6.
cos B
line
segment on any line
equal to the product of the length of the line segment cosine of the principal angle between the lines.
and
is
the
PLANE TRIGONOMETRY
74
Consider the broken line OA,
OA, AB, and
OB
projc/>AJ5 projcz>OJ5
But
AB
(Figure 39).
Project
Then
on CD.
= MN, = NQ (NOT: QN), = MQ.
M Q = MN + tfQ.
Hence: proj C z>OA
From
this computation,
we have the second theorem on
This theorem may be extended for a broken projection. line of any finite number of parts. proj
OB =
proj
OA +
proj
AB
Theorem 2. The projection on any line OA, AB is equal to the projection of OB. 32. Sine
and cosine
present section,
two angles*
we shall derive formulas for a and /3 may be any given
and cos (a
+
Figure 40,
a and
]8)
of the stun of
of the broken line
;
j8
are taken as acute,
and
In the
sin (a
+
{$)
In angles. are of such
It may be less than 90. magnitude that their sum proved, however, that the formulas hold for angles of any magnitude. Consider axes of coordinates with angles a and j3 at the From any point P on the terminal origin 0, as in Figure 40. is
TRIGONOMETRIC RELATIONS
75
PA
to the terminal side of angle /3, drop a perpendicular to both axes to form angles as side of angle a. Extend
PA
in the figure.
The is
OAP
right triangle
the one upon which we focus our attention.
shall
The is
essence of our proof to project the sides of
this right triangle, first, on the z-axis and, then, on the
The
first
projection will give us cos (a /3) the second, sin (a /3).
?/-axis.
+
(90+*)
;
+
the
Projecting
directed
sides of the right triangle
OAP have,
on the
by
projoxOP
By
the
we
z-axis,
the second theorem on projection:
first
OP
= projoxOA
+ projoxAP.
projection theorem, this becomes:
cos (a
+ 0) = OA
cos
a
+ AP cos
(90
+ a).
Or, since
+ a) =
cos (90
-sin
a,
we have:
OP
cos (a
+
ft)
Dividing by OP, we have /
cos (a
+* = ,
ft)
= OA
cos
a
- AP sin
a.
:
(
cos a(
A\ J
Or, since
OP and
AP OP
-
/ AP\V
sin al
PLANE TRIGONOMETRY
76 therefore
:
cos (a
+ g) =
cos a cos
a
sin
ft
sin
g
In like fashion, we project the sides of the right triangle OAP on the y-axis and we have :
=
projorOP
projor&A
+
projorAP.
Substituting,
OP
-
cos [90
(or
+ g)]
OA
==
-
cos (90
+ AP cos a,
a)
or
OP
sin (a
+
= OA
0)
Dividing by OP, we have sin
(<*
+ g) =
+ AP cos a.
a
sin
:
sin
+
a ^
\
a
cos
^
J
Or, finally, sin (a
Tan ( + tan a and tan
33. of
tan (a
+
ft)
(5).
/3
is
=
A
sin
a
+
cos
+ +
sin (a
cos (a
tan (a
Therefore
+
cos
a
sin
sin
a
sin
of this last fraction
+
sin
a
cos
a
sin
by
(
cos
sin
a
sin
cos
a
cos
:
tan (a
+ g)
tan a 1
|8)
in terms
g)
a cos g cos a cos g
member
+
:
g)
sin
Dividing each we have:
sin
formula for tan (a
derived as follows
+
a
cos
+ tan
g
tan a tan g
cos
a
cos
/3,
TRIGONOMETRIC RELATIONS
77
Problems Example Find,
by using one sin 75
of the addition formulas, sin
75.
= =
sin
=
+
sin (45 sin
45
30)
+
cos 30
cos 45
30
J_ V2
= \/6
_
-T
-T
V6+V2 4
=
+
30), find cos 75.
1.
By
2.
Find tan 75.
3.
4.
Verify the relations for the functions of 90. Verify the relations for the functions of 180.
5.
Verify: cos (90
6.
Verify: sin (180
7.
Prove:
using (75
45
+ 0) = -sin + 0) = -sin
,^o
a
+
8.
If
9.
Prove:
+
= 45,
prove:
cot (a
11. 12.
cos
A + sin A~A sm A :
cos (1
+
tan
cot a cot
+ 0) =
cot a
+
)(!
+
ft
-
cot
tan 0)
=
2.
1 ft
tan
.
:
cos (a
13. If tan
find sin (a 14.
A)
0.
= \ and tan = J, prove: tan (a. + 0) = f If tana = and tan = A, prove: (a + j8) = 45 or 225. = n (assuming a and acute), If tan a = m and tan
10. If
prove
,
=
A\
tan (45
0.
a
+
=
+ 0) =
tan a
=
V(l+m )(l+n 2
=
f and tan
^ (assuming
2 )
a and
ft
0).
With the material
16. If
mn
1 ,
=
m m+
of
Problem
and tan
ft
=
1
tan (a
+
ft)
13, find cos (a
+ 0).
1
2m + = 1.
,
1
prove:
acute),
PLANE TRIGONOMETRY
78
two angles. We wish formulas for the sine, the cosine, and the tangent of (a j8). Using Roman instead of Greek letters, we 34. Functions of the difference of
rewrite
+ y) =
sin (x
Let x
=
and y =
a,
sin [a
+
(
sin
=
+
cos x sin y.
Then, substituting, we have:
/?.
ft)]
x cos y
a cos
sin
(
ft)
+
a
cos
sin
(
ft).
Or, since cos
(
cos
ft)
and sin
we have
(
=
ft)
sin
:
sin (a
j8)
= =
sin
a cos
sin
a cos
+
cos a
sin 0)
(
cos a sin
Similar results obtain for cos (a
and tan
0)
Hence we have: sin (a
0)
cos (a
)
tan (a
-
ft)
= =
=
sin
a cos
cos
a cos
tan a 1
+
cos a sin
+ tan
sin
a
ft
tan a tan
Problems 1.
Find:
(a) sin
15.
(6) cos 15. (c)
tan 15.
2. Verify:
(a) sin (180 (6)
cos (360
(c)
tan (360
(d) cos (270 3. If
tan a
=
- 6) = sin 0. - 0) = cos 0. - 0) = -tan 0. - 0) - -sin 0. |
and tan 3
=
J, find
.
tan (a
sin
(a
|8).
TRIGONOMETRIC RELATIONS tan
=
and tan # =
tan a
(z
+
1)
$ (assuming a and
(a-
cot
ft
acute), find
(z
1),
prove:
0)
Prove:
6.
(a) (6)
tan (A cot (A
cos (A (d) cos (A (c)
/
=
ft
0).
If
5.
= if and
tan a
4. If
cos (a
79
x
(a)
.
sin
/
,
(A
- 45) + cot - 45) + tan
(A (A sin (A sin (A
+ +
45) 45) 45) 45)
45) + - 45) + sin A cos A - 45) =
+
.
= 0. = 0. = 0. = 0.
x
a double-angle. We wish formulas for the sine, the cosine, and the tangent of a double-angle, say 2a. Hence, we proceed as follows: Replacing )8 by a in our addition formula, we have 36. Functions of
:
sin
2a
= =
sin (a sin
+
a)
a cos a
+
cos a sin a.
Or:
Similarly, cos 2a
= =
cos (a
+ a)
cos a cos
a
sin
a
sin a.
Or: cos 2 a
cos 2a
Then, since sin
we have
Or:
also
:
2
a
+
cos
2
sin
a =
2
1,
a
PLANE TRIGONOMETRY
80 Similarly,
2 tan a
tan 2a
tan 2 a
1
Example Find
3A.
sin
3A = = = = = =
sin
sin
(2A
+
sin
2A
cos
A)
A +
cos
2A
A cos A) cos A + 2 sin A cos A + sin A (2 sin
(1
2
sin 2
2 sin A(l 3 sin
A
4 sin
The above example may be made of the
3
A)
+
A
sin
sin
2 sin 2 A) sin
2 sin 3
A
A
A
2 sin 8
A
A
illustrates the extensive use that
addition formulas:
by
this device
the functions of any integral multiple of an angle
may
be
found.
Problems
By
1.
=
using (60
2
-
30), verify values for
sin
60, cos 60,
and tan 60. the values for the above functions of 90.
2. Similarly, verify
3A =
4 cos 3 A 3 cos A. Prove: cos By using the material in Problem 3, verify the value for
3. 4.
cos
90. Given
7.
A = f cos A positive; find tan 2A. Given cos A = -^-, tan A negative; find sin 2A. Given tan A = $, sin A negative; find cos 2A.
8.
Prove:
5. 6.
.
,
(a)
sin
,
2 tan
n = 2A .
.
sin
cos
1
2A
1
tan (A
to
1
-
1
+
+
cos 20
-
tan 2
tan 2
+ tan
B)
tan (A sin 26
A
A A 2 + tan A
+
1
+
B) tan cot 6 cot B
+
- B) = - B) (A
(A 1 1
tan 2A.
TRIGONOMETRIC RELATIONS
+
2A
tan
(e)
sec
A + cos A cos
2A
sec
sin
A
sin
A
81
esc
esc 26
(/)
2 cot
(g)
A cos
1
-
2A cos 2A
=
2 sin
sin
X
X+
2JL
sin
sec 2 6
w
sec 20
0')
1
(i)
-
2
1
(A)
sec
2
+ tan 2A = + tanX) + sin 2X = (11 tan X + +
sec
2A
tan
A
2
2
Prove:
9.
tan 2 A tan
+
(a)
1
(6)
sin
2A(tan
(c)
tan
+
(d)
cot
(e)
cos
(f)
sin
(g) cos
4
A +
cot
tan sin 4
sin
a
=
4X =
1
-
2a
A = cot
sec 2A.
A) =
2.
= 2 esc 20. = 2 cot 20. = cos 20. cos 2a) cos a.
(1
8 sin 2
X+
8 sin 4 X.
We
wish first a formula 36. Functions of a half -angle. take us Let for the sine of a half angle. sin
We
|
rewrite one of our formulas for the cosine of cos
Now
2X =
let v-
Then
1
a
-
2 sin 2
JC.
2X
thus
:
PLANE TRIGONOMETRY
82
Hence we have
:
cos
a =
1
2 sin 2 ->
=
1
cos a,
a =
1
cos
or:
2 sin 2 or: sin
2
a
2
Or, finally,
To
find
an expression
for the cosine of a half-angle, cos i'
we
rewrite one of the other formulas' for the cosine of
thus: cos
Again
2X =
2 cos 2
X-l.
let
X
=
OL
2
Then
2X =
a;
and we have: cos
a = 2
2 cos 2 -
a - =
9 **
cos 2
2 Finally:
=
1
1
cos 2 A
+ cos a, + cos
1,
2X
TRIGONOMETRIC RELATIONS
83
Next,
a
,
tan
COS
-
sin
z
=
a cos-
2
a
+
2
+
,1
\/
cosa 2
Or: COS
I
The formula
just
1
above
be simplified as follows:
may
cos
a
1
cos
a
1
+
1
a
1
cos
cos
+
2
cos a
cos a
a
cos a) 2
\ (1 + cos a) sin
+ +
cos a
1
(1
a
2
a
xs
1
Similarly,
+
cos
a
we multiply both numerator and denomi-
if
nator of 1
1
+
cos
a
cos
a
by 1
cos a,
1
cos
we obtain: sin
Hence we have
a
a
:
a
sin
a
1
cos
==
tan 2
1
+
cos
a
sin
a
a
PLANE TRIGONOMETRY
84
Example Prove:
sm A* = o2 sin .
.
We
-~
cos
2
2
might solve this problem as follows: .
2
sm
-
-=
22 cos
+ -
cos
/I
'
,
2
V
4
~
2
/I
V
+
cos
A
2
8in
o
= sn
.
this point the skillful student will recognize that the original relation is the formula (disguised a bit)
However, at
for the sine of a double-angle.
If
A =
2X,
we
let
then
A
Y* 2= X >
and the
relation sin
A, =
2
-
A
A cos
sin
2
2
becomes: sin
2X =
2 sin
X cos X.
Indeed, .
sm is
true for the
same
A = ~ 2
rt
2 sin
^.
4 ~
44 cos
reason.
In other words, if an equation assumes the form of a well-known formula and the angles have the proper relation,
TRIGONOMETRIC RELATIONS
85
one to another, the equation is true, regardless of the form Thus in which the angles are expressed. , a
=
cos 6
2 sin 2 -
1
be recognized as one of our formulas for cos 2a. The student should not be misled, however, into thinking that all the problems below are solved in a similar manner. They are not. The above fact was pointed out simply will
some instances
as being of use in
only.
Problems 1.
Given
3
A =
sin
tan
->
A
A
positive; find sin
o 2.
A
Given cos
2t
2
A
A
-> sin
negative; find tan
O 3.
Given tan
A =
A
>
12
2
not in the second quadrant; find
A
cos-4.
Given
5
A =
esc
A not in the fourth quadrant
->
4 6.
Given cot
4
A =
->
A
+ tan A
1
(6)
sin
+
tan
1
A - tan
1
+ tan
T
cot
esc csc
tan 2
A
A =
A.
sec
= \/l
1 (c) '
identities:
2
cos
A
=
+
A cot
2
(0/2) 2
A
A sin 2 sin 4 + sin 2 sin A + sin 2A
2
cos
-
C0t
sin
2
(8/2) 2 sin
tan
not in the second quadrant; find sin
Prove the following
(a)
find
2i
o 6.
A ;
:
A- cos 2A
A.
A 2
PLANE TRIGONOMETRY
86 1
(h)
+
tan (A/2)
(cot
tan
(A\ 1
+ 2
( j)
sin
(k)
tan
-
cot 2
2/
0/ -( cot
-6
2\
2
A ~
sin
)
A
tan
2 cot
A -
=
(m) cot ,
+
tan (45 \
x
,
(ft)
tan
(0)
1
37.
A ~
V 1
=
1
=
sin
1
/)
tan
0.
A
A
cot
1.
2
-
A =
2 cot A.
A + sin A 1 + cos A + sin A cot X tan - - tan 2 - = cos
1
We
Product formulas. sin
expressed as
P =a =
+ sin
-) 2/
2
A = 2
\
2/
2 2
-
-) = cot (45
tan
4 cot X.
2.
2i
(7)
2X j =
2
A +
tan
X
foot
J
a
/3,
and Q = a -
sin
P+
a cos
sin
+
Q =
sin
We
Q
proceed as follows: Let
Then:
/3.
+ + sin (a + sin a cos #
sin (a
cos a sin
=
wish a formula for
P+
product.
/3
0.
2 sin a cos
)
/3)
cos
a
sin
#
/3.
Here is our product; but it is in terms of functions of a and ]8, and we wish it to involve P and Q. Solving the original relations for a and /3 by addition and subtraction,
we have:
= .
P+Q __,
P-Q
TRIGONOMETRIC RELATIONS Hence we have sin
In
like
and
las
:
P+
sin
Q =
p
-|-
2 sin
Q
cos
p _Q 2
2
manner, we derive the other three product formubelow
collect the four
(1) sin
(2) sin
P+ P-
:
Q =
sin
Q =
sin
(3) cos
P+
cos
(4) cos
P
cos
The above formulas of
87
Q =
2 sin
P + - Q
2 cos
P+Q -
cos
P -- Q
2
2 sin
P - Q2
2
2 cos
Q =
P +- Q
P
2
cos
2
2 sin
P +- Q
P sin
Q -
Q
are particularly useful in the branch
mathematics called calculus. Example Prove
1
:
sin
3A
A = 2 cos 2A 3A =P, A = Q.
sin
Let
sin
A.
Then 2
2
3A - A
P- Q
and Then, substituting
in the second of the product formulas,
immediately: sin
3A
sin
A =
2 cos
2A
sin
Example 2 Prove: sin
cos
ZA 3A
+
sin
cos
= 5A
cot A.
A.
we have
PLANE TRIGONOMETRY
88 Let
Then and
- =
-
When we
substitute the
A.
product formula, the numerator
first
becomes: sin
= =
+ sin 5A
3A
When we substitute
2 sin 2 sin
4A 4A
cos
(A)
cos A.
the fourth product formula, the denominator
becomes: cos
3A
5A =
cos
2 sin
= = sin
cos
+
3A 3A
5A 5A
sin
cos
4A
2 sin 2 sin
__
2 sin cos
__
sin
=
4A 4A 4A
(A)
sin
2 sin 4A( sin
sin
A)
A.
cos -4 sin
A
A A
cot A.
Problems Prove the following 5
A +
3. sin
4. cos
A
cos
2A cos 2A cos 2A sin 2A sin
5.
sin
sin 3
cos
+ sin A = + cos A cos
+ sin B A + sin-
COB
A
cos
5
.
3A
tan
3A ~ 3A
,
,
2 sin 4 A cos A.
2A = 2 sin 3A sin A. + sin 4A + sin &A = 4 cos A cos 2A sin 3 A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A
4A 2A
4A.
A =
:
--
1. sin
2.
identities
cos
SB
sin
2
A 2
A + B
cos
sin
B A
cos
TRIGONOMETRIC RELATIONS cos sin
sin
cos sin
,_
10.
cos
6A 6A 7A 7A 5A 5A
-
cos 75
^ 12.
.
_
16.
A = + cos A - sin 15
tan A.
cot 2A.
sin
tan 2A.
- - +
= ~
cos 15
1
_
\/3
3A + sin A- = tan 3A. 2 cos 3A + cos A sin A + sin 3A + sin 5A + sin 7A = tan r~; ^T~~; rr~; ^rr cos A + cos 3A + cos 5A + cos 7A sin 6A + sin 4A sin 2A sin 8A = cot 5A. cos 2A cos 8A cos 6A + cos 4A sin 3 A + sin 2A + sin A = tan 2A. cos 3A + cos 2A + cos A -
2 sin
-
- --
-
+ 2fl) - 2 sin (A + B) + cos (A + 25) - 2 cos (A + 5) + sin (2A - 3jB) + sin 3 " an cos (2A - 3) + cos 3B sin 47 + sin 73 _ /cos 47 + cos 73 sin 4A - sin 2A _ 1-3 tan A tan A 3 sin 4A + sin 2A
sin "
cos
5A cos 5A
in sin
-
14.
+ sin + sin
4A = 4A 3A 3A
-
sin 75 JLl.
cos
sin
(A
cos
A A
an
*
2
2
A + sin A sin
,
tan
_
sin
sin
B B
+B -
A
2
A
.
Miscellaneous Problems
Prove the following 1.
sin
3. cos
4
(sec
identities:
A = sin 3A + sin (e - 120) +
5A
2. sin 9
4.
89
sin
A
2
sin
4
=
2 cos
tan A) (sec
2
A +
sin
2
2A.
sin (60
-
0)
1.
tan A)
*
1.
-
0.
V
-t-
;.
PLANE TRIGONOMETRY
90
+ cos = 1 3 sin 6 cos 0. + cos nQ sin (n + 1)0 = sin nO cos 6
6 6. sin 8
6. sin
2
2
4X = 8 cos 3 X sin X 4 cos X sin A + cos A T = tan 2A + sec 2A. sin A cos A - 1\ /sec /sin A92 9 2
7. sin
_ 8.
:
9. cot
(
3
tan
+
2
+
1
sec
14. cos 4
=
sin 4
__ ~~
(B
X
=
1
tan 0)(1
Q
1
A
=
-sin 20
1
__ """
sin
sin
.
2.
A
.
A
+ ~ cos X sin X
X 2 tan X + 2 sec
A
-
-
1
-
cot 3A cot A 3A sin 4A 2 77 = tan A. sm 4A ;
sin
2 cos
X+
sec 2
2
3A = 2
cot
_^
I
tan
sm 2A
23. cos
A) _
V
1
_-
:
2
22.
sin
cos
2 sin 2A
20.
21.
C
cos 30
tan
_
tan
^
-
2
==:
tan
1
A 2
2X cos
2X +
A
1
6
JST
X+2
cos --(2 cos
A
2
cot 8)
-sin
1
(C
sin
A
sin 30
18.
cos 2
cos
+ + sin 20 sin (A C) + 2 sin A + sin (A + C) sin (B C) + 2 sin 5 + sin (B + C) * + CQS A = (esc A + cot A) 1 cos A (1
17.
+
sin
C
-
A
A
0.
*
C)
cos 20(
2
15.
-
=
)
sin
C0t
sin J5 sin
sin 6
X+
A
2
cot
sin
+
1\
sec 07 sin 2
1-2
A
-
13. cos 6
cos
(
(9
\1
2X 2X
cos
1 -f sin
*
3
cot
.4
tan
,
)
sin 0/
A 2X + 1 + sin 2X sin (A ) sin A sin B 1
+
~
r
+
\1
10
6.
X.
sin
4 1).
+
5 sin
2X
A B
TRIGONOMETRIC RELATIONS e
cos -
4 sin
2
, 24. -
-
2sm sin
25.
~
8
+
A +
sm A
:
sin 26
cos cos
B B
=
sec
2
91
CHAPTER VII
SUPPLEMENTARY TOPICS* Law
In Section 27 when we were of tangents. of solution an oblique triangle by means the considering of the law of cosines, it was pointed out that there were 38.
additional formulas better adapted to logarithmic use but that we felt them to be unnecessary. However, since
some authorities prefer them, we shall derive such formulas and include them in this chapter. The first of these formulas is called the law of tangents.
We
shall proceed to its derivation.
Given an oblique triangle ABC, we have, from the law of sines,
By
a
From
the
becomes
*
sin
b
sin
A B
the theory of proportion, this becomes: a
Or,
a
first
+
b
sin
b
sin
A A +
sin
sin
B B
two product formulas, the right-hand
side
:
we have the law
of tangents:
This chapter may be omitted the material in his course.
if
the instructor does not wish to include
92
SUPPLEMENTARY TOPICS
93
use of the law of tangents, we can solve a triangle if two sides and the included angle are given, as in the example
By
below.
Example
ABC] given a = 2439, a - b = 1403 a + b = 3475 B = 180 - C = 180 - 38
Solve the triangle
A +
A+B
.
70-
(a
A -
'
.
=
log tan
-
=
log (a
a
-
+
log 1403
+
+
log tan 70
log tan
56'
3.1470 .4614
3.6084
3.5410 .0674
2
A -B =
49 26'
2
But, since
A+B =
70 56'
l
by
addition,
:.A =
52'
120
log (a
&
A - B log tan
7'.
:
b)
= log 1403 = log tan 70 56' = log numerator = log 3475
141
38
b
2i
=
=
C =
A+B
tan
tan
Using logarithms, we continue
8'
1036,
56'
6)
2
=
ft
22';
-
log 3475
+
6)
PLANE TRIGONOMETRY
94
and, by subtraction,
Side c
may now
.*.
B =
21
30'.
be found by the law of
sines.
Problems Using the law of tangents, solve the following triangles: 1.
a
2. b
3. c 4. 6.
a a
39.
= = = = =
= = = = =
= 16.92, C = 420.3, A 623.1, c = 33.93, B 53.28, a = 300.3, C 419.2, 6 = 70.34, C 60.66, b 28.43, b
Tangent
of
40
9'.
62
42'.
63
24'.
53
18'.
46
26'.
a half -angle in terms of the sides of a shall now derive formulas to be used
We
given triangle. in solving a triangle
Given the
triangle
when
three sides are given.
ABC.
Let
+b+
a
c
then: 6
+
c
a
a
+
b
c
a
_ s
By
the tan
-^
b
=
c
=
formula and the law of cosines:
A tan
cos
/T
2-Vf +
cos
A A c
1
+
26c
-
a
SUPPLEMENTARY TOPICS Or,
95
by substitution, tan ;
A =
121 2bc
-
\21 2bc
+
!
62
-
2
+
c
-
a2
b
c
2 2
+a -
2
a2
(6
4,(b (a
4
(b
+
2
c)
+ c)(a + b - c) + c + a)(6 + c - a) -
b
-
4
(2s) (2) (s
this expression
^
tan
A = 1(8 T; \^ 2
c).
a)
a)
s(s
To make
-
more symmetrical, we write
-
a)( /
s(s
6) (8 ^22
c) '
a)
or
A
1
2
-T^TV
tan
Now
j(s
-
a a)(s
6)(s
let
-
and we have:
Similarly,
a)(
-
&)(
-
c)
c)
it
:
PLANE TRIGONOMETRY
96
We shall next derive a formula for the area of the triangle ABC in terms of the sides. From Problem 2 in Section 27, we found: area
= -
be sin
= 7 4
(be)
A.
Or: 2
(area)
2
sin 2
A
l
-
cos 2
(6c) (l
+
cos 4)(1
2
A)
-
cos A).
Or: area
=
-
be
\(1
+
cos
cos A)(l
A)
2i
1
26c
1
bc
a)(b
(b
+c-
a) (a
+
6
-
c)(a
-
6
+
c)
j
\
*i
c).
Therefore
:
area
=
\s(s
a)(s
Solve the triangle AfiC; given a 2s s s - a s - & s - c
= = = = =
-<*)(
6)(s
=
100, 6
c)
=
360 180
80 60
40
-
&)(
-
c)
120, c
=
140.
SUPPLEMENTARY TOPICS log r
-
log (s
-
a)
+ log
-
(s
+ log
6)
(*
97
-
c)
-
log * j
log (s
log
(s
log (s
- a) = - 6) = - c) =
log s
log tan
^ =
= log 60 = log 40 = sum = = log 180 = 2 log r = .'. log r = log 80
log r
1.9031
1.7782 1.6021
5.2834 2.2553 3.0281 1.5141
log (s
a)
2i
^ .'.
= =
9.6110
=
22
A =
-
(11.5141
-
-
10)
1.9031
10
13'
44 26'
D
=
log tan
.'.
log tan
log r
log (s
= =
9.7359
=
28 34'
-
(11.5141
| B =
57
Q = -
log r
-
6)
10)
-
1.7782
10
8'
log (s
c)
2i
= =
.*.
= ^ C =
-
(11.5141
9.9120
-
39
10)
-
1.6021
10
14'
78 28'
Proof
A +B+C= =
44 26'
+ 57
179 62' 180
2'.
8'
+ 78
28'
PLANE TRIGONOMETRY
98
log area
=
-
log
(s
-
+
a)
-
log (s
6)
+ log
(s
-
c)
+
log s
= ^(7.5387) log area .".
area
= =
3.76985
5887 square units
Problems Solve the following triangles, and find the area of each:
2.
a a
1.
3.
a
4.
a
5.
a
40.
= = = = =
= 16.48, c = 30.24. 20.34, b = 266, c = 300. 144, 6 2743, 6 = 3201, c = 4002. = 400, c = 500. 200, & 42.81, &
=
=
22.03, c
30.22.
Radius of the inscribed
circle.
It is interesting to
note that the quantity (s
-
a a)(s
b)(s
c)
ABC
is actually the numerical associated with the triangle length of the radius of the circle inscribed in the given prove this result as follows: triangle.
We
Proof
From
plane
geometry, we know: area of
where
&ABC =
~r'-P,
f
r is the radius of the inscribed circle
and P
is
the perimeter
of the triangle.
P=
Since
we have:
area
=
2s, r'
s.
From the formula for area derived in Section 39, we have: area
= \0 r
(s
-
a)(s
a)(s
-
b)(s
b)(s
-
c)s 2
c)s
SUPPLEMENTARY TOPICS -
(s
/. r'
We
= =
s
a)(s
-
6)(*
-
99
c)
r.
r.
give below another proof that is independent of Consider the triangle with inscribed circle,
ABC
area.
and with
bisectors of the angles meeting at the center 41. Figure
C
Let
r'
be the radius of the ,
tan
A = /
AS
2
We
wish to prove:
AD =
Then
circle.
s
a.
Proof
From
plane geometry,
we have:
AD =
AF,
DB =
CE.
AD + DB + CF AD = s = s-
Then
s.
Hence:
or
by
=
substitution,
Therefore:
tan
2
But
since
we know
tan
r
2 /. r'
7
a
s
A =
r '
s
-
(BE a.
s
A -- =
(DJ5
r.
a
+ +
CF) CE);
BE, CF
in
PLANE TRIGONOMETRY
100
of an angle. Frequently it is in to calculus, express angles in desirable, particularly do so by means of circular units other than degrees. 41. Circular
measure
We
measure, as illustrated
by Figure
42.
Figure 42.
at
Consider angle ACB and circles (1) and C and with radii r\ and r 2 Since
(2),
with center
.
A'B'
AB we have: A'B'
AB 7*2
In other words, the ratio length of arc radius of circle
always the same for a given central angle. ratio the circular measure of an angle, and we Hence we have: radian.
We
is
number
of radians
in central angle
call
length of arc length of radius
call this
the unit a
SUPPLEMENTARY TOPICS
101
To find the size of one radian, we substitute in the above formula and then have :
1
length of arc
=
length of radius
Hence
:
=
length of arc
length of radius.
It is therefore evident that
a radian is a central angle subtended by an arc equal in length to the radius of a circle. There are as many radians in 360 as there are arcs of length r in a complete circumference. Since the circumference equals 2?rr, there are 2?r such arcs. Consequently there are 2ir radians in 360 or ;
:
Hence, to change degrees to radians, multiply the degrees
by 7T
180*
To change
radians to degrees, multiply the radians 180 7T
Thus: 60
=
60
=
radians
180
- radians. 3
Also: 27r
,.
radians
3
Since a radian
is
=
2ir
180
3
ir
,
degrees
equivalent to 180 3.14159'
a radian equals approximately 57
17' 45".
=
, rtrt0
120
.
by
PLANE TRIGONOMETRY
102
Similarly, one degree equals .01745
.
.
.
radian.
Problems
Change from degrees
1.
to radians:
(6)
60. 30.
(c)
120.
fa)
270. 240. 360.
(d)
45.
(h)
0.
(a)
(/)
\
(a)
(&)
(j)
(*) (/)
150. 300. 180. 90.
* -
t
to degrees:
\
(e)
M
v
WT
*
(.)
^
?
venience,
we have
of
4?r
T
V
(A) 47T.
(Z)
STT.
(a)
Summary
r
(*)
-
(d) 27T.
-
f\ %*
-
5?r
3?r
42.
(i)
334
Change from radians
2.
/
(e)
trigonometric
collected in
formulas.
summary
outline
For conform the
trigonometric formulas developed in the preceding text. 1.
Fundamental
Identities.
esc
A = sin
sec
A =
1
cos cot
A
A
A
1
A sin A tan A = cos A cos A cot A = sin A 2 2 sin A + cos A - 1 2 2 1 + tan A = sec A 2 2 1 + cot A = esc A tan
SUPPLEMENTARY TOPICS 2.
Addition and Subtraction Formulas.
+ (a +
sin (a
cos
ft)
ft)
sin (a
ft)
cos (a
ft)
= =
(-)
tan
3.
= =
a cos 8 + cos a sin a a cos ft tan a + tan ft
sin
sin
ft
cos
sin
ft
a cos cos a cos
cos
sin
+
j3
sin
a a
sin
sin
tan a
= __
tan # tan a tan
+
x
Double-Angle Formulas. sin
cos cos cos
= = 2a 2a = 2a =
2a
2 sin a cos a cos 2 a
sin 2
2
2 cos a
-
a
1 2
2 sin a
1
=
tan 2a
-
2 tan a 5
tan'
1
4.
103
a
Half-Angle Formulas. a sin- = cos -
11
.
a
cos
a
cos
a
+
=
=.+ 2 .
tan
cos
.
-
/^
"\1 +
a tan 2
a
tan -
sin 1
=
2'
+
1
.
cos
a
cos
a
cos
a
;
sin
a
*
5.
Product Formulas. sin
P+
sin
Q =
2 sin
P+ - Q 2
cos
P ~ Q &
PLANE TRIGONOMETRY
104
sin
P
sin
Q =
2 cos
P+ - Q
sin
-
cos
P+
cos
Q =
2 cos
P+ - Q
cos
P -- Q
cos
6.
Law
P
cos
Q =
2
P
i
u
P +Q sin r
P- Q sin
of Sines. a
A
sin 7.
62 c
9.
sin
C
Lau> of Cosines. a2
8.
B
sin
Law
2
= = =
62
a2 a
2
+ c - 26c cos A + c - 2ac cos J5 + 6 - 2ab cos C 2
2
2
of Tangents. tan
A -- B
a
-
6
2
a
+
6
A +
tan
Semi-Perimeter Formulas. an
A ^ "" 2
tan
B = 2
, tan
area
r 5
-
s
-
a
r
6
C 2
= ^s(s
Q
c
s
a
+6+
-
a)(
a) (s
-
c
&)(*
&)(
-
c)
c)
SUPPLEMENTARY TOPICS 10. Circular
105
Measure.
in this formula are interpreted: 6 = number of radians in a central angle, I = length of intercepted arc,
The terms r
=
length of radius. TT
11.
=
radians
180
Laws of Logarithms. log&
AB =
log&
A +
log&
B
A =
Iog 6
A -
log&
B
log*,
log&
12. Projection
D
An = n
logb
A
Theorems.
proj
AS +
proj
BC =
proj
= A5 cos projcz? AB The second theorem holds where between AB and CD.
is
AC
6
the principal angle
TABLE
I
LOGARITHMS TO FOUR PLACES
FOUR-PLACE LOGARITHMS
108
FOUR-PLACE LOGARITHMS
109
TABLE
II
TRIGONOMETRIC FUNCTIONS TO FOUR PLACES
FOUR-PLACE TRIGONOMETRIC FUNCTIONS
112
FOUR-PLACE TRIGONOMETRIC FUNCTIONS
113
FOUR-PLACE TRIGONOMETRIC FUNCTIONS
114
FOUR-PLACE TRIGONOMETRIC FUNCTIONS
115
FOUR-PLACE TRIGONOMETRIC FUNCTIONS
116
TABLE
III
SQUARES AND SQUARE ROOTS
SQUARES AND SQUARE ROOTS (Moving the decimal point one plaoe
in
N
requires a corresponding
two places in N*.)
118
move
of
SQUARES AND SQUARE ROOTS (Moving the decimal point one place
in
N
two places in
119
requires a corresponding
N
2
.)
move
of
ANALYTIC GEOMETRY
CHAPTER VIII
COORDINATES 43. Position of a point in a plane. directed distances, axes
discussed
The student
quadrants. sections
is
In Sections 16-18 of
advised to review these three
From them
immediately.
we
coordinates, and
it
is
quite
evident
that, for every point in a given plane, there is a unique set of two numbers called its coordinates; and that, conversely, for every set of two numbers, there is a unique point in the In this chapter we shall be concerned with the plane.
coordinates of various points and with the algebraic or analytic quantities which will express, in terms of the coordinates, certain geometproperties associated with
ric
the points. 44. Distance
The
points.
that
we
between two
first
shall derive
formula is
called
the distance formula; it expresses the length of the line
segment joining two points, in terms of their coordinates. Given the two points PI and P 2 with coordinates (#1, 1/1) and (x 2 i/ 2 ), respectively ,
Figure 43.
,
(Figure 43);
we
desire a formula that will express the
length PiP 2 After completing the right triangle we see that .
PiQ =
M iM
2
=
OM -
QP* =
X*
2
Similarly, 2/2
123
-
PiQP 2
2/1-
-
(Figure 43),
ANALYTIC GEOMETRY
124
Hence, by the law of Pythagoras,
= M(x* -
I
Therefore
we have
+
2
*i)
(2/2
the distance formula
-
:
-
(1/2
Example If Pi is (2,
-3),
andP 2
PaP 2 =
I
is (0, 4),
= V[0 -
then:
+
2
2]
= V(-2) 2 + = V53
[4
-
(-3)]
2
72
.
The points Pi and P 2 were taken in Figure 43 in most convenient positions. It is easy to show, however, that the formula is true regardless of the positions of PI and P 2 .
Problems 1.
2.
Plot the following points: (2, -3), (0, -4), (4, 0), (-6, 2). What can be said regarding the coordinates of all points
(6) on the t/-axis? (c) on the line through the the first and the third quadrants? (d) on the line origin bisecting x-axis and three units above it? and (e) on the line to the parallel
(a)
on the a>axis?
and four units to the left of it? Find the lengths of the sides of the triangle with the
parallel to the 2/-axis 3.
3). following points as vertices: (2, 1), (3, 4), (2, 4. Do the same for the triangle with these vertices:
(3,0), 5.
Show
vertices of 6.
(0, 4),
(-1, -6). that the points
an
(3, 4), (1,
-2), and (-3, 2) are the
isosceles triangle.
Show that
the points
(3, 2), (5,
-1), and (-3, -2) are the
vertices of a right triangle. 7. Find whether or not the following points are the vertices of
a right triangle: (-2, 8.
Show
0), (3, 5), (6,
-2).
that the points (-2, -3),
2) are the vertices of a parallelogram.
(5,
-4),
(4, 1),
and (-3,
COORDINATES
125
Show
that the following points are the vertices of a parallelogram, and find whether or not the figure is a rectangle: 9.
(-3, 10.
8)
(-7,
6),
(-3, -2),
Show that the points
(1, 0).
(0,
the vertices of a square. 11. Show that the points
-3),
(7, 2), (2, 9),
(-2,
(1, 4),
10),
and (-5,
and
4) are
(3, 0) lie in
a
straight line.
whether
or
not the points (0, -4), (3, (5, 2) lie in a straight line.
0),
12.
Determine
46. Mid-point of a line segment. Given the line segment PiP 2 with P the mid-point of this line seg,
ment (Figure
44); and the coordinates of PI(XI, j/i), of
We
P2(*2, 2/2), of P(x, y). wish to find the coordinates of
P
in terms of those of
As
in Figure 44, Then, since
Figure 44.
Pi and
P
2.
drop perpendiculars PiA,
P P = PP X
2,
we have, from plane geometry, AM = MB. But
AM
=
MB
=
x
xi,
and 2
~
x.
Substituting,
X or
2x
Hence: _+_ 2
PM, andP
2
5.
ANALYTIC GEOMETRY
126 Similarly,
I/I
Therefore
we have
+
2/2
the mid-point formula:
Example of the point
Find the coordinates and (6, 5). a;
=
x\
+ #2 =
1
~2
+
midway between (1,
+
6
=
2 2/2
3
46. Point that divides a line
3)
5
2
+5 segment
in
a given
ratio,
y
Figure 45.
Let us consider a line segment P\Pi (Figure 45), with a point on this line segment such that
P _ m n from P to P 2
distance from distance
PI
to
P
COORDINATES
127
where
m n is
any given
We
Section 45. of (xij
j/i),
Let us use the same coordinates as in wish to find the coordinates of P in terms
ratio.
of (#2,
2/2),
and
of
m and n.
In Figure 45, we know, by plane geometry,
PiP
AM
PP 2
MB
Hence, by substitution,
Solving for
z,
we have
:
Likewise:
+ m The above
constitute the ratio formula. In Figure 45, the segment PiPi is divided internally. If P lies on the segment extended, then we say the segment PiP 2 is divided externally. If the division is internal, the ratio
is
positive;
negative.
if
The work
the division will
external, the ratio is if the ratio is always
is
be simplified
considered as distance from
PI
distance from
P
to
to
P
P%
regardless of the position of P.
Example Given the segment joining of trisection nearer (2,
I).
(2,
1
-1) and
(8, 5); find
the point
ANALYTIC GEOMETRY
128 Call 2
=
-l)Pi, and
(2,
The
5.
(8,
5)P 2 hence, xi ;
m=
ratio is -; hence,
1,
-
n =
2, yi
- -1,
s2
8,
2.
Substituting in the ratio formula,
+ 2-2 +2
1-8 1
y
1
Hence, the required point
+ is
12
2 the point
3 (4, 1).
Example 2
The segment PiP 2 is extended half its Show that the ratio equals 3; that is,
length to
P
(Figure 46).
We have
Since
P2P
is
m
PiP
n
PP 2
"
Figure 46.
a unit,
PiP
3 units,
PP 2 = - 1
unit.
Hence:
m n
PP 2
= -3.
-
Problems 1.
A
triangle has the following points as vertices: A(0, 2),
5(6, 0), and C(4, (a) (6)
The The
from
A
from
C
6).
Find:
coordinates of the mid-point of BC. coordinates of the point two-thirds of the distance
to the mid-point of BC. of AB. (c) The coordinates of the mid-point of the distance two-thirds (d) The coordinates of the point
(e)
to the mid-point of AB. length of the median through B.
The
COORDINATES
129
Given the parallelogram with vertices at A (2, 3), JS(5, -4), C(4, 1), and D(-3, 2); show that the coordinates of the mid-points of AC and BD are the same, and hence that the 2.
diagonals bisect each other. 3. Prove by the mid-point formula that the points 10),
2), (5,
(6,
and
(3, 4)
(4, 12),
form a parallelogram,
Three consecutive vertices of a parallelogram are (3, 0), Find the fourth vertex. (5, 2), and (-2, 6). 6. The segment from ( Find the 1, 2) to (3, 4) is doubled. codrdinates of the new end point. 6. The center of a circle is (3, 4); one point of the circle is Find the coordinates of the other end of the diameter (6, 8). 4.
through this point. 7. Find the coordinates of the point that divides the segment from (0, - 1) to (6, 3) in the ratio 2:5. 8. Find the ratio in which the point (2, 1) divides the segment from (6, 1) to (0, 2). 9. Find the coordinates of the points that trisect the segment from (1, -4) to (3, 5). 10. Find the coordinates of the points that divide the segment (2, 3) and (4, 1) into four equal parts. Find the coordinates of the points that divide the segment from (1, 2) to (5, 8), internally and externally, in the numerical
joining 11.
ratio 3:2. 12. (3, 6).
A
is
If
(-1, 3), and B is AB is prolonged to
C, a distance equal to three times its length, find the coor-
dinates of C. 13.
Find
reached by
ment from
what
point is trebling the seg(2, 0)
to (3, 4).
The 47. Slope of a line. as is line defined of a slope the tangent of the angle between the line and the x-axis, the angle
from
AB]
/
gure
being measured in counter-clockwise sense Thus, in Figure 47, tan 0i is tan 0*, the slope of CD.
the z-axis to the line.
the slope of
Y
ANALYTIC GEOMETRY
130
now consider a line AB passing through two points
Let us
Pi and PI (Figure 48), the coordinates of which are (xi, y\) and (x 2 3/2), respectively. We wish to derive an expression ,
for the slope of the line in terms of the coordinates of the
two given
points.
Let us
call
the slope m.
In Figure 48,
Figure 48.
Hence
:
m=
tan
Thus we have the
=
tan
ZCPiP 2 =
~ = PiC
Vl
slope formula:
and perpendicular lines. If two lines are Let parallel, it is obvious that they have the same slope. us consider what relation, if any, exists between the slopes of two perpendicular lines. Given two perpendicular lines AB and CD, making angles 6 1 and 2 respectively, with the re-axis (Figure 49). Let mi = tan 0i, and ra 2 = tan 2 Then: 48. Parallel
,
-
0i
Therefore
-
90
+
2.
:
tan
0i
=
tan (90
+
2)
= -cot
2
-
= tan
2
COORDINATES
131
Hence, by substitution,
V
A
A
Figure 49.
That
two
lines are perpendicular, the slope of one is the negative reciprocal of the slope of the other. The converse
is
is, if
y
true also.
Angle between two Given two lines (1) lines. and (2) in Figure 50 we may define the angle which line (1) makes with line (2) as the angle through which (2) 49.
,
;
must revolve
in
counter-
clockwise sense to coincide with (1). In the figure, the
angle
is
a. 60.
Figure wish to express the tangent of a in terms of the slopes mi and m 2 of the given
We
lines.
Since
a
=
61
-
02)
=
02,
hence: tan a
-
tan
(0i
tan 1
0i
+ tan
tan 0i
tan
02 8
ANALYTIC GEOMETRY
132
Therefore,
by
substitution,
tan a
=
Problems Find the slopes of the
1.
lines
through:
-1) and (3, 5). (-3, -4) and (0, 6). (1, 1) and (4, 4).
(a)
(2,
(5) (c)
2.
Find,
points
lie
by
in the
slopes,
same
whether or not the following
sets of
straight line:
-6),and(l, -4). and (-3,4).
(a)
(4,2), (0,
(6)
(0,3), (9,0),
(c)
(1,1), (2,
-l),and(l, -3).
3. Find the tangent of the angle between the lines the slopes which are (a) 3 and ?, respectively; (6) 3 and f (c) 3 and 2. 4. Show, by slopes, that the points (7, 2), (0, 3), (2,9), and
of
;
5, 4)
(
are the vertices of a rectangle.
Three vertices of a parallelogram are (4, 1), (3, 2), and Find the fourth vertex. (NOTE: The student is ( 2, 3). to submit three possible solutions.) expected 6. Show, by slopes, that the following points form a parallelogram: (3, 0), (7, 3), (8, 5), and (4, 2). 7. A circle has its center at (3, 4). Find the slope of the 6.
2). tangent to the circle at (5, 8. The base of a triangle passes through Find the slope of the altitude.
9. (0,
10.
Show
(2,
and
1)
(3, 2).
that the diagonals of the square with vertices at
3), (7, 2), (2, 9),
and
(5,
4) are perpendicular.
Find whether or not the rectangle of Problem 4
is
a square.
An 50. Application of coordinates to plane geometry. in theorems is that of plane interesting problem proving geometry by means
of coordinates.
illustrate the procedure.
Two
examples
will
COORDINATES
133
Before we proceed, however, four considerations should be noted. First, to avoid a special case, we must use letters not numbers for coordinates. Second, we must use the most general type of figure for which the theorem is to
be proved. Third, since the geometric properties of the figure are independent of its position, we can employ the most advantageous position for our particular figure. Hence, if our problem concerns a rectangle, we shall take the rectangle with two sides along the coordinate axes and with a vertex, therefore, at Y the origin.
Finally, the coor-
and the relations between them add whatever dinates
further information
is
necesC(a,6)
sary to determine the type of figure.
Example
1
Prove that the diagonals of a In the are equal.
0(o,o)
rectangle
present example, we shall use the rectangle in Figure 51.
Observe that, having taken (o, 6), and then having given made the figure a rectangle.
We
Figure 61.
at
C
(o, o),
A
at (a, o),
the coordinates
We
and
B
at
we have OC = AB.
(a, 6),
wish to prove that
use the distance formula:
OC = V(a AB = V(o Hence
2
o) 2
a)
+ +
(6 (&
-
2
o) 2
OC = AB.
:
Example 2 Prove that the diagonals of a parallelogram bisect each other. shall prove this theorem by showing that the mid-points of the two diagonals have the same coordinates and, therefore,
We
coincide.
In Figure 52,
and
B at
(6, c).
we have taken one
vertex at
Then, since the figure
have the coordinates
(a
+
6, c).
is
(o, o),
A
at (a, o),
a parallelogram,
C
will
ANALYTIC GEOMETRY
134
We
The
use the mid-point formula.
point of
OC
coordinates of the mid-
are: (a
+ 6) +
o
+b
a
2
y
The
2
+C=
C
-
2
coordinates of the mid-point of
AB
are:
tf
jg(6,c)
Cat6,c)
o
+ c __*
c
2
5"
The two mid-points have the .same coordinates; hence, they coincide.
A(a,o)
__ =
Therefore,
OC and
X AB bisect each other. Problems
Figure 62.
Prove, by dinates, the following geometric theorems:
means
of coor-
The diagonals of a rectangle bisect each other. The medians of a triangle meet in a point that is two-thirds the way from a vertex to the mid-point of the opposite side. 3. The line joining the vertex of any right triangle with the 1.
2.
of
mid-point of the hypotenuse is equal to half the hypotenuse. 4. The line joining the middle points of two sides of a triangle equal to half the third side. 6. If the lines joining two vertices of a triangle to the middle points of the opposite sides are equal, the triangle is isosceles. 6. In any quadrilateral the lines joining the middle points of
is
the opposite sides and the line joining the middle points of the diagonals meet in a point and bisect each other. is the middle point of 7. In any parallelogram ABCD, if the side AB, the line and the diagonal AC trisect each other.
M
MD
8. The sum of the squares of the medians of any triangle equals three-fourths of the sum of the squares of the sides.
COORDINATES 9.
The
area of any triangle
is
four times the area of the
triangle formed by joining the mid-points of the 10.
The
135
sides.
lines joining the mid-points of adjacent sides of
any
rectangle form a rhombus.
The The
diagonals of a square are perpendicular. lines joining the middle points of the sides of any quadrilateral, taken in order, form a parallelogram. 13. The diagonals of a rhombus are perpendicular. 14. If the diagonals of a rectangle are perpendicular, the 11.
12.
rectangle 15.
angles 16.
A is
is
a square.
quadrilateral
whose diagonals
bisect each other at right
a rhombus.
The diagonals of an isosceles trapezoid are equal.
A
trapezoid whose diagonals are equal is isosceles. a quadrilateral bisect each other, the quadrilateral is a parallelogram. 19. The distance between the mid-points of the non-parallel sides of a trapezoid is half the sum of the parallel sides. 20. The sum of the squares of the sides of a parallelogram equals the sum of the squares of the diagonals. 17.
18. If the diagonals of
CHAPTER IX
LOCUS and equation of we considered geometric chapter 51. Definition
locus.
In the preceding
figures in which we were concerned with stationary points and their coordinates. we shall consider also the path described by a moving
Now
point.
Let us
(NOTE:
A
the path of a moving point a curve. one form of a curve.) Usually the coordinates of any general point on a curve will be represented by (x, y} for any particular point, subscripts will be used: (xi, yO, (x 2 2/2), and so on. If a point moves in such a way that it is constantly subject to a given condition that has been imposed, the path described by the point is defined as the locus of a point that moves subject to the given condition. The locus contains all those points, and only those points, that satisfy the given call
straight line is
;
,
condition.
The circle is a well-known locus; it moving so that its distance from a constant.
segment
is
is
the locus of a point
fixed point is always Likewise, the perpendicular bisector of a line the locus of a point moving so that it is always
equidistant from the extremities of the line segment. By the equation of a locus, we mean the equation which is satisfied by the coordinates of all those points, and only those points, that lie on the locus. The equation of a curve is
defined similarly. Thus it is evident that, if a point
on a
curve, its coordi-
nates must satisfy the equation of the curve;
and conversely,
lies
a point satisfy the equation of a curve, This idea is of fundathen the point must lie on the curve. mental importance and is applied constantly.
if the coordinates of
136
LOCUS
137
quite evident from the above that the intersection points of two or more curves are found by solving simultaneously the equations of those curves.
Furthermore,
it is
1
Example
Find the equation of the locus of a point moving so that distance from the point (2, 3) is constantly equal to 6. Call (x, y} the coordinates of the moving point. Then distance of
(x,
from
y)
=
3)
(2,
its
6.
Substituting in the distance formula,
-2) 2 + x2
or:
4x
+
x2
Therefore:
is
-
+ + 4
y*
-
y
4x
(t/
+
2
3)
=
+ Qy + 9 + 6y - 23
2
6,
= =
36. 0.
The last is the equation of the locus. It is evident that this the equation of a circle with center at (2, 3) and with radius 6. Example 2
Find the equation of the locus of a point moving so that always equidistant from (1, 4) and (3, 5).
As Then
before, call (x, y) the coordinates of the
distance of
=
(x, y)
from
distance of
(1,
(x, y)
moving
it is
point.
4)
from
(
3, 5).
Substituting,
V(x -
2
I)
Therefore:
This
is
+
- 4) 2 = V(x + 3) 2 + 8x - 2y + 17 = 0.
(y
(t/-5)
2 .
the equation of the perpendicular bisector of the line The student may, by (1, 4) and (3, 5).
segment joining
applying the mid-point formula, verify the fact that the midpoint of the given segment lies on the locus.
Example 8 Find the equation of the locus of a point moving so that its distance from the y-axis always equals its distance from the point
(
1, 4).
ANALYTIC GEOMETRY
138 Since
distance of
= x
substituting,
(#, y)
from
distance of
=
t/-axis
(x, y)
V(a?
+
+
(y
-
+ 2x +
17
2
I)
from
-
(
1, 4),
2
4)
.
Simplifying, therefore: 2 t/
8y
-
0.
Example 4
A
variable triangle has as vertices the moving point (x, y) t The area (0, 0), and the point (6, 0), as in Figure 53. always equals 10. Find the equation of the locus of (x, y).
the origin
A (6,0)
0(0,0)
Figure 63.
The area of
triangle
OPA
(Figure 53)
is
equal to
--OA-h - 6 y
=
10,
Find the equation of the locus of a point
P
moving
Since
OA =
6,
and h =
y,
or:
Therefore:
hence: area
3y
=
y
=
=
10.
10
Problems 1.
indicated in the following: (a) (6)
Distance from Distance from
(2, 4)
(3,
equals
5.
5) equals distance
from
(1,
-4).
as
LOCUS Distance from Distance from Distance from Distance from Distance from Distance from Distance from Distance from
(c)
(d) (e)
(/) (g)
(h) (i)
(j) (fc)
from
139
y-axis equals distance from (2, 1). a>axis is three times distance from 2/-axis.
three times distance from x-axis.
y-axis
is
origin
is 6.
(
equals twice distance from equals distance from x-axis.
2, 3)
(5, 3)
t/-axis
equals
(1, 1).
3.
o>axis equals 2. Square of distance from origin is equal to x-axis plus distance from ?/-axis.
sum
of distance
Sum of distances from
(3, 0) and (3, 0) equals 10. from (5, 0) and ( 5, 0) equals 8. of Difference distances (m) from Product of distances (2, 1) and ( 1, 3) equals 5. (ri) from Sum of distances of (o) (4, 2) and (1, 1) equals squares (Z)
10.
Ratio of distances from
(p)
(
3, 2)
and
(2, 7)
equals f
2. A variable triangle has as vertices the moving point the origin, and the point (4, 0) the area always equals 6. the equation of the locus of (x, T/). ;
3.
The ends
pendicular
of a line of variable length are on two fixed perFind the equation of the locus of the mid-point
if
The base
(2, 0).
5.
Find
lines.
the slope of
and
(x, y),
the area of the triangle thus formed is 20. of a triangle is AB, where A is ( 4, 0) and B Find the equation of the locus of the vertex P(x, y),
of this line 4.
.
1, 6).
of the hypotenuse of a right triangle are (3, 4) Find the equation of the locus of the vertex of
the right angle. 6. The base of a triangle is AB, where A is (4, 0) and B Find the equation of the locus of the third vertex (8, 0). (x, I/), if
if
AP is two units greater than the slope of BP.
The ends (
is
the median from
A
to
is
P
BP is always three units in length.
CHAPTER
X
THE STRAIGHT LINE 52. Equations of lines parallel to the axes. In the preceding chapter we considered equations of various curves
derived from various loci conditions. In this chapter we on the straight line, the simplest type of locus. We shall first consider lines parallel to the coordinate axes. Y In Figure 54, the line is parallel to the #-axis, and at a constant distance k from shall concentrate
AB
it.
The equation
of
AB will A
be:
since all points
nate
is
k will
AB, and points
lie
whose ordion the line
the ordinate of
on
AB
will
all
be
k.
Likewise, the equation of a line parallel to the ?/-axis and at a constant distance k from
Y' Figure 64. it
will
be
:
53. Point-slope form. Let us now consider a line passing through a fixed point and having a fixed slope that is,
a fixed direction.
We wish to find the equation of the
in terms of the fixed slope
and the coordinates
of illustration, consider the line parallel to either axis, as in Figure 55.
point.
By way
140
line
of the fixed
AB
not
THE STRAIGHT LINE
141
Call the fixed point (xi, j/0, and the slope m. Call (x, y) any other point on the line. Then, substituting in the slope formula (Section 47) ,
y
-
x
Therefore
we have
the point-
slope formula:
we wish
Conversely,
show that
to
points whose coordinates satisfy the above A" equation lie on the line AB ' through (xi, t/i) and with slope m. all
Figure 56.
Let (XD, 2/0) be the coordinates of any point satisfying the above equation (Figure 55). Then, substituting, 2/o
-
2/i
^
Or:
m But, by the slope formula, the expression 2/Q
XQ
is
the slope of the line
CD
~ Xi
through
(x Qj
t/
)
and
(xi, y\).
m equals this expression, m is the slope of the line CD. However, m is also the slope of the given line AB. Hence, and since both lines, AB and CD, pass through Since
(rti,
t/i)>
have the same slope w, they must coincide; y ) must lie on the given line AB. Therefore
since they each
and thus, (X Q) the above point-slope formula is the equation passing through (xi, y\) and with slope m.
of the line
ANALYTIC GEOMETRY
142
This formula is called the point-slope form of the equation a straight line, and is very useful in finding the equation of a line when its slope and a point on it are given.
of
Problems Write the equations of the following
lines,
and draw the
figures:
Parallel to the y-axis, and at a distance of 3 units from 2 units from to the o>axis, and at a distance of
1.
2. Parallel
3.
Through
(2,
Through
(4,
-3), and
(a) parallel to
OX;
(6) parallel
it.
it.
to
OF. 4. 2/
=
1),
and
(a) parallel to
x
=
3; (&) parallel to
6.
6.
Through (3, 4), and same line.
(a) parallel to
x
=
1; (6)
perpendic-
ular to the
The
x-axis; (6) the y-axis.
6.
(a)
7.
3), and with slope 2. Through (2, Through (1, 4), and with slope J. Through (1,5), and making an angle of 30 with the y-axis. Through (2, 0), and making an angle of 135 with the
8. 9.
10.
z-axis.
Through (3, 1), and parallel to the line passing and (-1,3). (2, 4) 12. Through (1, 6), and perpendicular to the line through (3, 2) and (4, -3). 13. Through (1, 1), and parallel to the line passing (4, 3) and (-1,3). 14. Through (3, 2), and parallel to the line passing (-2, 4) and (-2, 7). 16. Through (o, fc), and with slope m. 11.
through passing
through
through
The x-intercept of a line is 54. Slope-intercept form. defined as the distance from the origin to the point where the line cuts the x-axis.
The y-intercept is defined similarly. m and with t/-intercept 6.
Consider a line with slope The line then passes through point-slope formula, we have: y
6
}
(o,
= m(x
6).
o).
Substituting in the
THE STRAIGHT LINE Thus we have the
This formula
143
slope-intercept formula:
is
the slope-intercept form.
called
The
particular advantage of this formula is that it enables us to find the slope of a line immediately from its equation;
we do
solving for y and taking the resulting coefficient of x as the slope. 6 = 0, Thus, given: 3x 2y this
by
+
= ~fz
*/
+ 3.
Thus, comparing the coefficients of x in the example and in the formula,
we
see that
m=
f.
In this instance, the ^/-intercept is the resulting constant This fact is not of great importance, however, for we can always find the y-intercept by letting x equal zero and solving for y. Similarly, we might find the z-intercept
term.
letting y equal zero and solving for x. Given two points: (xi, yi) and 56. Two-point form. wish to find the equation of the line through (#2, 2/2).
by
We
these points. Call the slope m.
From
the slope formula,
m=
2/2
we know
that
~
Hence, substituting in the point-slope formula, we have: y
Thus we have the
_
yl
is
2/2
_
2/i
, {
X
_ XJ x
t
two-point formula:
56. Intercept form.
connection
-
Another formula interesting in
the one called the intercept form.
this
ANALYTIC GEOMETRY
144
We
are given
and the
two intercepts: the 6.
y-intercept,
(x\> yi) is (a, o),
and
(x a ,
Then,
^-intercept, called a;
above formula, Hence:
the
in
2/2) is (o, 6).
i
Thus we have the
y
o
b
o
x
a
o
a
intercept formula:
In 57. General form of the equation of a straight line. the preceding paragraphs of this chapter, we found various forms of the equation of a straight line; in every instance the equation found was an equation of the first degree in
x and
always be the case. such as two points, may be reduced to the case of a fixed point and slope, and then the point-slope formula may be applied. This procedure holds for all It is evident that
y.
such
will
Moreover, given conditions for a line two intercepts, or one point and slope
cases except
when
the line in question
is
parallel to the
form (x = k). equation consequently We now wish to show the converse that is, every equation of the first degree in x and y is a straight line. Consider the most general form of an equation of the
y-axis,
and
its
is
of the
;
first
degree:
where A, B, and divide
This
is
by
C are
constants.
If
B is not
B, and, after transposing terms,
of the
zero, we may we have:
form y
mx
+ b,
which we know represents a
line
with slope
m
and with
THE STRAIGHT LINE y-intercept
Hence, we have a
6.
line
145
^
with slope
and
JD
C
with ^-intercept
-5-
This solution, incidentally,
tells
us that the slope of the
line
C =
+ By +
Ax is:
coefficient of
x
coefficient of
y
j
and
gives us another method of finding the slope of a line equation is in the form
it
if its
Ax If
+
B equals zero and A
we have
By
+C =
0.
not equal to zero, dividing by A,
is
:
C
This equation
is
of the
form x
which represents a
fc,
line parallel to the t/-axis.
Hence we say: Every equation of the
=
first
straight line is represented by
degree,
and
an
every first degree equation
represents a straight line.
Example Find
1
the equation of the line passing
parallel to
3x
-
4y
+
2
=
through
(
1,
2),
0.
By solving for y, we first find the slope m of the given line. -4t/ = -3z - 2 3
3 -
1
and
ANALYTIC GEOMETRY
146
The
required line must have the same slope. slope formula, we have:
y
which reduces
-
2
-
+
|(*
Using the point-
1),
to:
3x
-
+
4
=
11
0.
Example 2 Find the equation of the line passing through 6 = 0. perpendicular to 2x + 3y
The
slope of the given line
(3,
4),
and
is
m---2 The
slope of the required line
is
--1 m
;
3 -
and therefore
j
Again using the point-slope formula, we have:
3x
or:
-
2y
-
=
17
0.
Example 8
The
or-intercept of a line is three times the ^/-intercept.
Find line passes through ( 6, 3). We use the intercept formula a Since the ^-intercept
is
+ ?b
1
equation.
-
three times the ^-intercept,
=
a
Hence:
its
+
36.
=
1.
The
THE STRAIGHT LINE
147
However, since the required line passes through (6, 3), (by the fundamental principle derived in Section 51) the coordinates ( 6, 3) must satisfy the equation of the line. Hence, substituting,
we have:
-63 '
b
b
= =
3.
=
1.
+ 3y =
3.
6
a =
Therefore: Substituting further,
- H
If
o
1
x
Therefore:
36
X
1.
Problems 1.
Find the equations
(a)
(2,0) and (-1,3).
(6) (c)
(d) (e)
(/)
2.
(a) (5) (c)
3.
and (4, -5). (-2, 6) and (2, -1). (0, 0) and (7, -3). (2, 4) and (2, 6). (3, -2) and (5, -2). (3, 3)
Find the slopes 3x 2x x
4.
of the following lines:
2y - 6 = 0. - 3 = 0. y 2 = 0. y
+ -
+
Find the equations
(a) ^-intercept (6)
of the lines passing through:
of the lines with:
3 and ^-intercept
re-intercept 2 and ^-intercept
Find the equation of the y = 3x + 6. Find the equation of the
7. 1.
line passing
through
(
line passing
through
(2,
2, 4),
and
1),
and
parallel to 6.
perpendicular to y
=
\x
+
2.
ANALYTIC GEOMETRY
148 6.
Find the equation of the line passing through 3x - 2y + 5 = 0. Find the equation of the line passing through (
(1, 1),
and
1,3),
and
parallel to 7.
perpendicular to x
= 0. = 0.
+y
6
Find: (a) the ^-intercept; (6) 3y - 6 the t/-intercept; (c) whether or not ( 1, 2) lies on the line. 9. A line passes through (3, 2), and its ^-intercept is 5. Find the equation of the line. 8.
Given 2x
-
10. A line passes through (2, 2. 4), and its ^-intercept is Find the equation of the line. 11. A line passes through (2, 4), and its x-intercept is twice its ^/-intercept. Find the equation of the line. 12. A line passes through ( 1, 3), and its ^-intercept is four times its re-intercept. Find the equation of the line. 13. Find the equation of the line with slope ^ and with ^-inter3.
cept 14.
Find the equation
of the line with slope
cept
2.
16.
A
and with
x-inter-
Y circle
is
tangent to
at (2, 0). 3x - 2y - 6 = Find the equation of the locus
p
of its center.
Normal form. There one other form of the
68. is
equation of a straight line that will prove valuable, particularly in finding the distance from a line to a point.
We
proceed to
its
Figure 66.
derivation as follows.
The position of a straight line is completely determined if we know its perpendicular distance from the origin, and the angle this perpendicular makes with the x-axis. Thus, Let in Figure 56, the line AB is determined by p and co. mx + & But since AB and OC the equation of AB be y :
are perpendicular,
m=
cos cot
tan
w sin
THE STRAIGHT LINE Since
tOEC =
149
Zco,
*
-
^ET sin
P Z.OEC
P w
sin
Hence, substituting, y
Therefore,
cos w
=
+
x
:
sm
co
p -r sin
co
we have the normal form: x cos
co
+
?/
sin
p =
co
We shall call p the normal, * and shall consider p as always positive, its direction being to the line
from the origin in
question.
If
the
line
passes through the origin so that p = 0, we shall
consider the arrow representing the direction of p as pointing in the first or
the second quadrant (Figure 57). Unless the line passes through the origin, to 360 co may vary from
Y
;
the line passes through the origin, co varies only from
Figure 67
if
line passes line as
co
Let us
= 20. now consider (1)
Ax
For example,
to 180.
through the origin,
co
'
= 200
if
the
yields the same
the general form
+
By
+C=
and the normal form (2) *
Historically,
(cos co)x
normal means
in this text to use the term in
its
+
(sin
common
w)y
-
p
=
or ordinary, but
geometric connection.
it is
convenient
ANALYTIC GEOMETRY
150 to see
what
if any, exists between the correspondx and y and the constant terms in the two (We assume that the two equations represent
relation,
ing coefficients of equations. the same line.)
It is evident that the equation of a line not affected by multiplying throughout by a constant. Thus,
is
+
Qx
represents the
same
4y
+
line as
+
3x
2y
Hence our problem may be if
=
14
+
=
7
0.
what constant,
stated: Find
any, which, multiplying equation
reduces
(1),
it
identi-
cally to equation (2).
Let us
call
the constant (kA)x
Comparing
this
+
Equation
(kB)y
+
(k)C
=
(1)
becomes:
0.
with
+
(cos co)x
we have
k.
p =
(sin o>)y
0,
the following three equations:
kA = kB =
cos w, sin
a?,
kC = -p. Squaring and adding the k 2 (A 2
Therefore
+B
2 )
first
=
two equations, we have:
cos 2
o>
+
sin
2
o>
=
1.
:
k
=
i -
We now
use the third equation to determine the sign of k. Since the quantity p must be positive, kC must be negative. Hence k and C must have opposite signs. If
C=
0,
we use
the relation
kB =
sin
o>;
for, if
C =
0,
the line passes through the origin, in which case o> varies to 180 and sin w remains always positive. only from
THE STRAIGHT LINE kB must then
Hence k and
be positive.
151
B
must have
like
signs.
We may sum up this discussion as follows To reduce the :
equation
Ax
+
By
+
C =
normal form, multiply throughout by
to the
1
and take the sign of the radical opposite to that C = 0, take the same sign as that of B.
of C.
If
Example Reduce to the normal form 3x
4y
:
5
=
0.
A = 3 B = -4 C = -5 Zx
-
4y
-
5
5
Problems Example Find the distance between the parallel lines (1) 5* - 39 - 0. (2) bx + I2y to the normal form, we have: lines the Reducing
+
12y
13
=
and
-89 The 1
distance from the origin to the first line that is, p\ equals and the distance from the origin to the second line that
unit;
p z equals 3 units. Hence the distance between the lines is p 2 minus pi, or 2 units. (NOTE: These lines are on the same side of the origin. If they had been on opposite sides, we should have added Pi and p 2 .)
is,
ANALYTIC GEOMETRY
152 1.
Reduce
to the
normal form, and give the distance from the
origin to the line in question: (a) (6) (c)
(d) (e)
(/) 2.
(a)
3x 3x 5z 3x 3x 2*
3x 2a;
(c)
3x
(a)
+ + -
+
-
10 * 0. 4y 10 0. 4y - 26 = 0. 12y 4y = 0. 0. by y - 0.
(0)
+
(h)
+ 3y +y -2
2x x
(f)
a?
0*)
*
-
(t)
y
+
y
=
(Z)
= =
y 5 6
Find the distance between the following
(6)
3.
-
-
4y 4y 4y
- 10 = -3 = 5
and and and
=
3z 4z 3x
-
Find the equations of the following
4y 8y 4y
4
=
=
0.
0.
0.
0. 0.
parallel lines:
-
25
+
10
1
= 0. = 0. == 0.
lines:
Distance from origin, 6; angle
made by normal and
o>axis,
Distance from origin, 5; angle
made by normal and
x-axis,
Distance from origin, 5; angle
made by normal and
x-axis,
Distance from origin, 4; angle
made by normal and
re-axis,
Distance from origin, 4; angle
made by normal and
x-axis,
30. (6)
60. (c)
90. (d)
150. (e)
225. Distance from origin, 5; angle made by line and -axis, 45. (g) Distance from origin, 5; angle made by line and x-axis, 60. f (h} Distance from origin, 6; slope, 2x 6 = 0. from Distance to 3y (i) origin, 5; parallel 4 = 0. from to # Distance + 2y (j) origin, 4; perpendicular
(/)
.
+
from a line to a point. We shall now employ the normal form to find an expression for the perpendicular distance from a line to a point, when we are given the equation of the line and the coordinates of the 69. Distance
point.
Let the equation of x cos
AB o>
(Figure 58) be:
+ y sin w
p
0,
THE STRAIGHT LINE
153
and the coordinates of P, (a?i, j/i). Let the perpendicular distance from AB to P be represented by d. Draw CD through (xi, t/i) and parallel to AB. Call the distance OE, Then, the equation of
p'.
CD
is:
x cos
co
+
y sin
= p
or, since p'
Z cos
+
co
2/
sin
-
P
=
p'
+
0, >
d,
co
(p
+
=
d)
0.
on CD, (xi, yi) must the equation of CD.
Then, since its
co
lies
coordinates
satisfy
Hence
Figure 68.
:
Xi cos
+
co
2/1
sin
co
+
(p
d)
=
0.
Therefore we have the formula for the distance from a line to a point :
d
=
Xi cos
co
+
sin
2/1
p
co
Thus, to find the perpendicular distance from a line to a point, reduce the equation Y line to the normal and substitute for x form, and y the coordinates of the
the
of
Taking the direction normal as a standard,
point. of the
we
shall consider a distance
from a tive
s
X
same
its
tive
if
its
direction
Thus, in Figure 59, di
is
is
normal
Directed distances/'
Section Figure 69.
a point as posiis the
direction
as that of the
" (see
line to
if
16);
and we
shall
consider the distance negaopposite to that of the normal. positive,
and d 2 and d 3 are negative.
ANALYTIC GEOMETRY
164
Example Find the distance from the -1). Reducing the equation
line
1
4x
Zy
+
=
4
to the point
(2,
of the line to the
4tx
-
3y
+4
=
-5
normal form, we have
:
0.
Hence, substituting, d
The negative from the
line
4x
=
>
sign indicates that the direction of the distance to the point (2, 3y -f 4 = 1) is opposite
to the direction of the normal.
Example 2
=
Find the equations of the lines parallel to the 0, and passing at a distance of 3 units from
line
3x
+ ty
5
(NOTE: This as treated the locus be following problem: "Find example may the equation of the locus of a point moving so that its distance 5 = from the line Sx + 4y always equals 3.") 5 = Since the distance of (x, y) from 3x ty 3, equals hence we have it.
+
:
(1)
(2)
^^
-
?iiLL_5
, _3
3,
.
5
Reducing the equations, we obtain the following two solutions: (1)
3x
(2)
3x
+ +
4y
-
20
y
+
10
= =
0, 0.
Example 8
A
point
moves
- 12 = ty tion of the locus.
3x
+
so that
Since the distance of
the distance of the two solutions:
it is
and 5x
(x, y)
(x,
-
always equidistant from the lines - 10 = 0. Find the equaI2y
y)
from 5x
from 3x 12y
+ 4y 10
=
12 0,
=
equals
hence we have
THE STRAIGHT LINE Zx
+ 4y -
(1)
12
_ ~
-
5x
12y
5
3a
+
10
13
-
4y
(2)
-
155
12
-
5x
_
12y
5
-
10
13
Or, simplifying,
-
+
(1)
7x
(2)
32x
-
CD
is
In Figure 60, 10 = 0.
53
y
-
3s
+
4y
5&y
= 0, 103 =
0.
=
12
0,
and
AB
is 5:c
I2y
(2)
v
/
I
-<-~-
Figure 60. -
4v
-
12
-
10
Let
and
=
<*,.
13
Then,
line
equation
(1)
is
the locus obtained by letting
Likewise, line (2)
56i/
-
53
=
.
Its
0.
by letting d\ =
-
4y
-
103
=
d%.
Its equation is:
0.
from the figure since, for points on and d 2 have the same sign; and for points on line (2), have opposite signs. The signs can be checked by com-
results are evident
line (1), di
and d%
+
is obtained
32*
di
d2
is:
7x
These
=
ANALYTIC GEOMETRY
156
paring the directions of d\ and dz with those of the normals
pi and
and
a
we wish
if
Hence,
2>2.
(2),
From plane geopaetry, bisectors of the angles 3x
+
ty
to distinguish between lines (1)
fairly accurate figure is necessary.
-
12
=
we know, also, that lines (1) and (2) are formed by the lines AB and CD, or
-
and 5x
12y
-
10
=
0.
Problems 10 = to each of the 4y and (-1, 3); (4, |). following points: (3, 2); (0, to each of the 2. Find the distance from 2x + y + 6 = following points: (1, 3); (-1, 2); (3, -1); (4, 0); and (0, -6). 6 = 3. Find the equations of the lines parallel to 2x 3y from it. at distance units a of 3 and 0, passing 4. Find the equations of the lines parallel to x y + 3 = 0, and passing at a distance of 2 units from it. 6. Find the equations of the bisectors of the angles formed by and x + 2y - 3 = 0. the lines 3x - 4?/ - 10 = 6. Find the equations of the bisectors of the angles formed 1 = 0. 6 = and 2x y by the lines x + y 2 7. A point moves so that its distance from the line x ty = always equals 5. Find the equation of the locus. 8. A point moves so that its distance from the line 3x + 4y 15 = always equals its distance from (1, 2). Find the 1.
Find the distance from 3x
-1);
(2, 0);
equation of the locus. 9. Find the equation of the locus of a point moving so that its distance from (2, 4) always equals three times its distance from 3 = 0. the line x - 2y 10. Find the equation of the locus of a point moving so that
+
its
distance from
the line 3x 11.
The
C(l, 4).
-
4t/
3)
(1,
+
5
=
always equals one-half
12.
Find the area
(c)
-2), B(4,
6),
and
Find:
(d)
(b)
distance from
0.
vertices of a triangle are: A(0,
The equation of the line joining B and The length of the side BC. The length of the altitude through A. The area of the triangle.
(a)
its
points: (2, -1),
C.
of a triangle with vertices at the following
(3, 0),
and (-2,
4).
THE STRAIGHT LINE 13.
+4
The
=
lines
form a
2x
~
4 = 0, x y Find the area.
+
y
triangle.
2
157
=
0,
and x
2y
A circle of radius 4 is tangent to the line 2x
0. Zy + 2 Find the equation of the locus of its center. 15. Find the equations of the lines parallel to 4x 3y + 5 = 0, and twice as far as that line is from the origin. 16. Find the equations of the lines parallel to 2x 3g/ + 1 = 0, and 2 units farther from the origin. 17. Find the equations of the lines parallel to x + Sy 6 = 0, and passing at a distance of 2 units from the point ( 1, 2). 18. Find the equations of the lines parallel to 2x + 5y 6 = 0, and passing at a distance of 3 units from (3, 4). 19. The base of a triangle is the line joining (2, 4) and ( 1,0). The area is 10 square units. Find the locus of the third vertex. 20. In Problem 19, if the triangle is isosceles, find the coordi-
14.
nates of the third vertex.
through the point of intersection of two given Suppose we are given the equations of two inter-
60. Lines lines.
secting lines
:
(1)
A& +
(2)
A 2x + B 2y
B.y
+d= +C = 2
0, 0.
Let us multiply the second equation by k, and add it What does the resulting equation to the first equation. represent?
We
wish to show that this equation
Aix
+ B iy +
Ci
+ k(A& + B
2
y
+C
2)
=
represents for each value of &, a straight line through the intersection of the two given lines.
equation is obviously that of a straight line, for the equation is of the first degree. Now, let us call the coordinates of the intersection point (a?i, T/I). Then, First, the
all
"
m 0" means equals zero identically, or vanishes identically the terms cancel out.
*
that
is,
ANALYTIC GEOMETRY
158
on
for the point (xi, yi) lies
line (1).
Also,
C2 = on
for the point (xi, y\) lies in the equation
Aix
+
Biy
0,
line (2).
Hence, substituting
+ Ci + k(A*x + B y + C 2
2)
0,
we have:
+
&
SE 0.
This solution proves that the point
Aix
+ B iy +
+
Ci
(x\ 9 y\) lies
+B
k(A 2 x
2
+C
y
2)
on the
=
line
0,
for its coordinates satisfy the equation of that line.
Example Find the equation of the of the lines 2x (1,
y
+
3
line
=
1
through the point of intersection + y 1 = 0, and the point
and x
-2).
Since the required line passes through the point of intersection 1 = 0, the line has an equation and x 3 = y y of the form of 2x
+
+
2x
-
y
+3 +
+
k(x
y
-
Since the required line passes through 2. must be satisfied by x = 1, andi/ = 2(1)
+2+3+ 7
or:
-
*(1
2k = k
Therefore:
-
2
=
1)
0.
2), this equation Hence, we have:
(1,
-
1)
=
0,
0.
= -
Thus, substituting the above value for
fe,
we have
equation:
or:
Therefore:
2x
-
y
4z
-
2y
+
3
+ ~(z + &
y
-
+ 6 + 7x + ly 11s + by - 1 = 0.
1)
=
0,
7
*
0.
the required
THE STRAIGHT LINE
159
Example 2 Find the equation of the and 2x 6 = y the line 3z 2y + 5 = 0.
line
+
of x
The
through the point of intersection 3
=
+
y
y
and perpendicular to
0,
required line has the form
x
-
y
-
6
+
k(2x
-
3)
=
0.
5 = 2y perpendicular to 3x must be the negative reciprocal of the slope of 3x
Since this line
Hence,
its
+
is
slope
may be expressed
0, its
2y
slope
m
=
0.
+5
:
We
next find the slope of the required line in terms of Rewriting, we have:
x(2k
+
+
1)
Then, since
y(k
=
slope
Therefore:
3fc
+
1
-
=
6
0.
2 k
=
2k
3 or:
2k
-
+1 *- 1 = 6k + 3.
2
-
1)
fc-1
hence:
2k
-
k.
5 7-
4
Hence, substituting for x
-
y
-
6
-
5
-(2x 4
2x
or:
we have
fc,
+ 3y +
the required equation:
+
y
3
=
-
3)
=
0,
0.
Problems Find the equation of the line passing through the inter1 = 0, and through and 2x + 3y section of x 2y + 3 = 1.
the point 2.
(2, -3). Find the equation
section of z
the point
(1,
+ y~2 = 2).
of the line passing
and 3z
-
y
+6
through the inter= 0, and through
ANALYTIC GEOMETRY
160
Find the equation of the line passing through the inter+ 2y + 1 = and 2x y 2 = 0, and parallel to
3.
section of x
3x
-
y
+
2
0.
Find the equation of the line passing through the interand x 6 = section of 2x y 2y + 1 = 0, and having a 4.
+
slope of f
.
Find the equation of the line passing through the inter= and 2x + y -~ 1 = 0, and perpendicofz t/ + 2 ular to a line of slope f 6. Find the equation of the line passing through the interand x 4 = 0, and perpendicusection ofx + t/ 3 = 2y lar to 2x + 3y - 5 = 0. 7. Find the equation of the line passing through the inter1 = and x section of 2x + y 3y + 5 = 0, and having 5.
section
.
x-intercept 2.
Find the equation of the line passing through the. inter2 = 0, and making an and x + y 6 = y the x-axis. with of 135 angle 9. Find the equation of the line passing through the inter4 = 0, and parallel to and Zx + y section of x 2y + 1 = 8.
section of 3x
the x-axis.
Find the equation of the line passing through the inter4 = 0, and parallel to ofx + 2/~2 = and 2x y
10.
section
the
7/-axis.
Given two intersecting lines: (1) A\x + B\y + Ci == 0, = 0, Find the value of k that will (2) A%x + B^y + C%
11.
and
make
C2) = represent: (a) line (1); 12.
=
0.
and
(6) line (2).
Given the lines A ix +
Biy
+
Ci
= OandA 2 z + B^y + C
2
What does Aix
represent
if
+
Biy
+
Ci
+
fc(A 2 x
+
B#j
the lines are: (a) parallel? and
+C = 2)
(b)
coincident?
Miscellaneous Problems
Find the distance between the parallel lines 2x and 4z - 8y + 1 = 0. f= 2. Find the equations of the lines parallel to 4x 3y and passing at a distance of 3 units from it. 1.
3
4i/
2
=
0,
THE STRAIGHT LINE 3.
The
vertices of a triangle are: (1, 3), (2,
161 1),
and
(0, 5).
Find: (a) (&) (c)
(d) (e) (/)
(0)
(h)
The equations of the sides. The lengths of the altitudes. The angles of the triangle. The area of the triangle. The intersection of the medians. The intersection of the perpendicular bisectors of the The intersection of the bisectors of the angles. The intersection of the altitudes.
sides.
4. Using the triangle in Problem 3, show that the points of intersection of the medians, of the perpendicular bisectors of the sides, and of the altitudes all lie in the same straight line.
CHAPTER XI
THE CIRCLE A circle is 61. Definition and equation of the circle. defined as the locus of a point P moving so that its distance from a fixed point is conThe
stant.
fixed
point is called the ewter} and the constant distance, the radius.
The coordinates of the moving point P we shall call (x, y)
;
the coordinates of the
center, (a, /3) and the radius, r, as indicated in Figure 61. ;
_x
We desire the equation of the circle.
Figure 61.
Since
hence
distance from
(a,
ft)
V(x -
+
(y
to (x, y)
=
r,
:
2
a)
-
or: (x
-
2
a)
+
(y
-
2
/3)
This equation is the equation of the circle with center at (a, j8) and with radius r. It is evident that, if the center is at the origin, the equation will be
Example Find the center and the radius of the equation
circle
with the following
:
x2
+
y
2
-
2x
+
162
4y
-
3
=
0.
THE CIRCLE We
163
wish to reduce this equation to the form
(z-a) 2 +(2/-0) 2 = Completing the square, we have
-
z2
2x
+
+
1
(x-
or:
+ +
1)
Hence we have:
+ 4 = 3 + 1 + 4, + 2) = 8. 2
(
a =
1,
/=
-2,
r or:
:
4y
y* 2
r 2.
=
\/8;
center:
(1,
radius:
Vs.
2),
Problems Find the center and the radius of each of the following circles, whose equations are: 1.
(a) (6)
+ y* - 4z + 2y - 3 = 0. + y + 6x - % - 5 = 0. + + x - 2y - 1 = 0. 3z + Sy + 60: - 4y + 2 = 0. 5z + 5y - 6x - 2y + * = 0.
x2 x2
2
2
2
(c)
(d) (e)
a:
z/
2
2
2
2
2. Find the equations of the following circles: (a) with center at (2, 3, 4), and 1), and radius equal to 4; (6) with center at ( radius equal to \763. Find the equation of the circle with center at (2, 1),
and tangent
to the y-axisi
Find the equation of the circle having its center at ( 1, 4), and passing through (3, 5). 5. Find the equation of the circle that has as a diameter the line joining (3, 1) and (2, 5). 6. Find the equation of the circle having its center at ( 1, 3), 4.
and passing through 7.
(2, 7).
Find the equation of the
circle
with center at
(5, 5),
tangent to the x-axis. 8. Find the equation of the circle with center at (1, 10 = 0. 4t/ tangent to the line 3x 9. Find the equation of the circle with center at (1, tangent to x
+
y
6
=
0.
and
3),
and
2),
and
ANALYTIC GEOMETRY
164 10.
Find the equation of the
to the line x 11. is
-
Prove by
+
circle
with radius
5,
=
6
and tangent
at (-4, 1). coordinates that an angle inscribed in a semi-circle
2y
a right angle.
62. General form of the equation of the circle. In Section 57, we proved that a straight line is always represented by an equation of the first degree in x and y, and
that the converse of this theorem is also true. We shall now discuss the corresponding theorem for the circle.
Expanding
(x-a)*+(y-0)* =
(1)
and
we have
collecting terms,
x2
+
This equation
y
2
-
may
(2)
z2
2ax
-
2$y
r
:
+
a2
+
2
-
r
2
=
0.
be written:
+
2 t/
+
Dx
+
Ey
+F -
0,
where
D = -2, E = -20, F =
a*
+
2 /3
-
r2
.
Moreover, D, E, and F are constants but not necessarily For example, the following is the equation of integers. the circle with center at (^, 1) and with radius 2:
or:
x2
+y 2
x
+
2y
-
V-
=
0.
Here
Since every circle has a center and a radius, and since a with a given center and a given radius has either the unique equation (1) or the same equation in a different
circle
THE CIRCLE form
(2), it follows
equation of the
that every circle
165 is
represented
by an
form
+
x2
2 t/
+ Dx +
Ey
+F =
That is, every circle is represented by an equation of the second degree in x and y, with the coefficients of z 2 and j/ 2 unity and with no xy term. Conversely, every equation of the form
+
x2
y
2
+ Dx +
Ey
+F =
For, completing the square,
represents a circle.
we have:
or:
This last equation
is
(x
Observe that,
if
quantity
of the
-
2
)
form
+
(y
-
$)*
the result
is
D +E -
4F
2
2
=
r2
.
to be a real circle, the
="',
4
must be
For,
positive.
imaginary
that
number; and
is,
r is
the quantity is negative, r is the square root of a negative
if
the quantity is zero, r is zero. Hence, for with that the equation imaginary r consistency, we say represents an imaginary circle and that the equation with r equal to zero represents a point circle. if
;
The
following example illustrates the principles given
above.
Example Find the equation of the (0, 2), (3, 3), and (-1,
points
circle 1).
passing through the three
ANALYTIC GEOMETRY
166
The
required equation will be of the form (2)
+
x2
+
2 i/
+
Dx
Ey
+F =
0.
Hence, since the three points lie on the circle, their coordinates must satisfy the equation of the circle. Thus we have: (3) (4)
9
(5)
1
+ 4 + D + 2E + F = + 9 + 3D + 3E + F = 0, + 1 - D + E + F = 0.
Solving these three equations for D, E, and F,
D= #=
4,
F =
-12.
+
y
2
-
6x
obtain:
-6,
Therefore the required equation
x2
we
is:
+
-
4y
=
12
0.
another method of solution, which does not depend on form (2). This method is illustrated in the
There
is
following:
Example
We shall first find the center. From plane geometry, we know the intersection of the perpendicular bisectors of the line segments joining any two pairs of points. By the method of the center
is
Section 51, the equation of the perpendicular bisector of the line
segment joining
(0, 2)
+
(y
and
-
(3, 3) is:
= V(x -
2
2)
2
3)
+
(y
-
2
3)
,
or: (6)
3x
+
y
-
=
7
0.
Similarly, the equation of the perpendicular bisector of the line
segment joining
(0, 2)
V* + 2
(y
and
-
2
2)
(1,
1) is:
- V(x
+
+y-
1
2
I)
or: (7)
x
-
0.
+
(y
-
2
I)
,
THE CIRCLE Solving (6) and
(7),
we
167
find:
x
=
3,
-2.
y
Hence the center
is at the point (3, the center to any one of the three points fore the required equation is: 2
(z-
3)
+
The
2).
distance from
found to be
is
+ 2) =
25,
-
=
5.
There-
2
(i/
or:
x2
+
y
2
-
6x
+
4y
12
0.
Problems 1.
points
Find the equations of the
through the following
circles
:
(a)
(2,1), (-1,3), (3, -2).
(6)
(1, 1), (0, 6), (2,
(c)
(-2,
1), (1,
-4),
-3). (3,
-1).
Find the equation of the circle having its center on the line 2y + 3 = 0, and passing through the points (1, 1) and
2.
x
-3). Find the equation of the circle with center at (3, 2), and 3 = 0. ty tangent to the line 3# 4. Find the equation of the circle with radius 5, and tangent at (1, 6). to the line 2x - y + 4 = 5. Find the equation of the circle tangent to the line x + y 2 = 0at(l,l), and passing through (2 4). 4 = 6. Find the equation of the circle tangent to 2x + y (0,
3.
?
and passing through (3, 4). Find the equation of the circle passing through (2, 4) and 3y + 6 = 0. ( 1, 3) and having its center on the line x 8. Find the equation of the circle tangent to 2x y + 6 = = and 2x y + 10 0, and having its center on the line x - 3y + 4 = 0. 3 = 9. Find the equation of the circle tangent toz + i/ = and having its center on the line 7 and x + y 0, 2x + y - 4 = 0. 10. Find the equation of the circle with center on the z-axis, and tangent to the lines y = 4 and x = 2. at
(2, 0),
7.
+
ANALYTIC GEOMETRY
168
y
Find the equation 2, and tangent to the
11.
of the circle
=
lines
=
2x
3y
with center on the 4 = and x y
line
6
0.
Find the equation of the circle that is tangent to the lines = 0, and x = 5. 0, y 13. Find the equation of the circle inscribed in the triangle 19 == 0, 4x + 3y whose sides are the lines 3x 17 = 0, &y 7 = 0. and x 14. Find the equation of the circle inscribed in the triangle whose vertices are (0, 6), (8, 6), and (0, 0). 16. Find the equation of the circle which passes through 6 = 0. (1, 7) and (8, 8), and is tangent to the line 3x + 4y 12.
x
=
63. Circles through the points of intersection of two given circles; radical axis. In Section 60, we considered
the equations of lines through the intersection point of two given lines. We shall apply the same treatment to
the
circle.
Given two
circles:
(1)
x^
x
(2)
We (3)
2
+ y^ + A x + B y + C = + y* + Aix + B*y + C = l
l
l
0,
2
0.
wish to show that the equation
+ yt + AiX + B^ + Ci + k(x + y* + Aix + BM +
xt
2
C2) =
represents, for all values of k (with one exception), a circle through the intersection points (real or imaginary) of the
two given
circles.
Collecting terms, 2
(4)
(x
+y
2
)(l
we have:
+ k) + x(A + kAJ + y(B + kB + d + fcC l
l
2)
2
=
0.
This equation is of the second degree and assumes, for all values of k except 1, the form of the circle equation Furdiscuss the exception later. shall We (Section 62). thermore, cles, it lies
an intersection point of the two on both circles, and equation (3) becomes:
if (#1,
yi) is
+
k
as o.
cir-
THE CIRCLE Hence
(xi, j/i) lies
on the
169
circle
represented by equation (3). the given circles are non-concentric but do not intersect in real points that is, there are no real values (x, y) that satisfy the equations simultaneously there will still If
be imaginary values that satisfy the equations simultaneously; hence, equation (3) will represent a circle through the imaginary points of intersection of the given circles.
the given circles are concentric, there are no values real or imaginary that satisfy the equations simultaneously; hence, equation (3) will represent a circle concentric with the given circles. If
(x y y)
An (4)
when k
interesting case arises
then becomes x(Ai
This is a a straight
- A2 + )
first
line,
1.
equals
Equation
:
- 5
y(B l
+
2)
d-C
2
=
0.
degree equation and, therefore, represents which is called the radical axis of. the two
circles.
It is quite
are in the
apparent that,
form
of (1)
and
the equations of the circles the equation of the radical
if
(2),
obtained by subtracting one equation from the other. Thus, if (1) is represented by
axis
is
51
and
(2),
=
0,
by 52 =
the equation of the radical axis
Si
0,
is
- S2 =
:
0,
or: *Si
=
$2.
It is further apparent that, if the circles intersect in two different real points, the radical axis is the common chord.
consider the three circles: S\ = 0, $2 = 0, and wish to show that the radical axes of the three 0. circles (taken in pairs) meet in a common point called the
Now,
$3 =
We
radical center.
ANALYTIC GEOMETRY
170
The equation 5 2 = Ois:
Si
(6)
The equation
53 =
and
of the radical axis of circles Si
- S2 =
0.
of the radical axis of circles
S2 =
and
=
and
Ois:
S2
(6)
The equation S 3 = Ois:
- Ss =
of the radical
Si
(7)
0.
axis of circles Si
- S3 =
we add equations
0.
(6), we have, by Section 60, a line through the intersection of (6) and (6). Hence, adding (6) and (6), we obtain:
However,
if
Si
But 0,
this result is the
- S3 =
same
radical axes
meet
in a
0.
as equation (7).
Si Sa = passes through the interin other words, the three Or,
Therefore, the line represented which is the third radical axis
section of the other two.
and
(5)
common
by
that
(7)
is,
point.
Problems 1.
(a)
x
2
-
+
Find the equations
of the following circles:
Through the intersections 2 y = 2y, and (3, -4).
(6)
Through the
2x
+%
-
6
=
of the circles x
intersections of x 2
0,
and
2
+y = 2
+y = 2
2x and
+y
25 and x 2
2
(1, 2).
(c) Through the intersections of the circle the line x 3). 2, and (4, y
a:
2
+y = 2
16 and
+
+ 6y
4# Find the radical axis of the circles x 2 + y 2 and x 2 + y 2 + 2x - y + 3 = 0. 12 2 x + y 3. Find the radical axis of the circles x* + y 2 2 = 0. and 3x + 3y + 2x 3y + 6 2 2 Gx - 4y 4. Find the radical axis of the circles x + y 2 = 2 1. and x + t/ 2 2 6. Find the radical center of the circles x y + 2x 2 2 2 2 = 4z + 6y - 3 1 0, and x + y y 0, x + y + x 2.
-
=
+
2
+9 y 0.
THE CIRCLE 6.
= =
0,
171
Find the radical center of the circles x* + y 2 x + y 2 + 4z - 2y - 3 = 0, and x 2 + y 2 - 2x
x2
+ 3y
-
4?/
-
2 6
0.
7.
8. 9.
In what case do two circles have no radical axis? In what case do three circles have no radical center? Give a geometric construction for the radical axis of two
non-intersecting circles.
Prove analytically that the radical axis of two circles is perpendicular to their line of centers. 11. Prove analytically that the radical axis of two circles of equal radii is the perpendicular bisector of their line of centers. 12. What is the radical axis of two circles of zero radius? 13. Using the material in Problem 12, prove that the perpendicular bisectors of the sides of a triangle meet in a point. 14. If P(x\j y\) is a point outside the circle 10.
(xshow that the length by the formula t
16.
of
two
2 )
+
(2/-0)
of the tangent
= V(*i-) 2
+
Using the result of Problem = and 2 = 0, circles, Si
a point circles,
P
moving and Si =
2
-r
from
2
P
(*/i-0) 14,
=0,
2
to the circle
-r
given
2 .
show that the
may
is
radical axis
be defined as the locus of
so that the tangents drawn = 0, are always equal. 2
from
P
to the
CHAPTER XII
THE PARABOLA 64. Definition of of a point
divided
The
P
by
a conic.
A conic is
defined as the locus
distance from a fixed point moving its distance from a fixed line is a constant ratio.
fixed point
so that
is
its
and the fixed line, the The constant ratio is positive and is called
called the focus;
directrix.
the eccentricity.
There are
three types of conies: (a)
The
parabola,
eccentricity equals (b)
The ellipse, when eccen-
tricity is less (c)
than
1.
The hyperbola, when
eccentricity
We
when
1.
is
greater than
1.
shall first consider the
parabola. 65. Definition
and equa-
A of the parabola. is defined as the parabola Figure 62. locus of a point P moving so that its distance from the focus equals its distance from the directrix. We now wish the equation of a parabola. Let us take the focus F on the x-axis, and the directrix r perpendicular to the #-axis at A (to the left of the tion
D'Y'
DD
half-way between A and F, and equidistant from DD' and F lies on the parabola. Let us call point V the vertex. Let us choose our origin of coordinates at the vertex, and let us call
focus), as in Figure 62.
The point
172
V
THE PARABOLA
AF
that
2a.
Then,
is,
the distance from the directrix to the focus
AV - VF the coordinates of the focus are directrix
173
a;
(a, o)
;
the equation of the
is
x
and the coordinates
of
P
=
are
a;
We may now proceed
(x, y).
to the derivation of the equation. Since distance from
hence
(a, o)
to
(x, y)
=
distance from
DD
r
to (x, y),
:
PF =
CP,
or:
\(z~a) 2
+^
=
a
+
.
After squaring the above equation and collecting terms,
we
have:
This
the required equation of the parabola. of the parabola. To obtain an idea of the of let us examine the resulting the parabola, general shape 66.
is
Shape
equation y
=
\4a#.
When x is zero, y is zero (counted twice). Also, since we have chosen a as positive, it is evident that x must be zero or positive; for, if x were negative, y would be imaginary. Hence the curve passes through the origin where the curve itself touches the y-axis, and all other points on the curve to the right of the j/-axis. Also, for every such value of x, of two values there are j/, equal numerically but with
lie
opposite sign. Thus we say that the curve is symmetrical with regard to the x-axis. We call the line through the Since x focus perpendicular to the directrix, the axis. and y may assume as large values as we wish, the curve
ANALYTIC GEOMETRY
174
may be said to extend indefinitely to
the right of the y-axis,
and to extend
indefinitely above and below the #-axis. The general shape of the curve is given in Figure 63 (a). In the above discussion, since the directrix was taken to the left of the focus, a was considered positive. If, however, the directrix is taken to the right of the focus, a will
-XX-
X'-
-XX
X
Y'
r' -4aj
4a
(c)
(d)
(W
(a)
-X
4ax,
Y'
a+
o
r'
Figure 63.
be negative, and the parabola will open to the left instead See Figure 63(6), on this page. The of to the right. of the parabola will still be equation
the coordinates of the focus,
(a, o)
;
and the equation
of the
directrix,
x If
we choose
=
the focus on the
a.
j/-axis,
the directrix below
the focus, and the origin of the coordinates at the vertex, the equation of the parabola will be :
and the parabola
open upward as in Figure 63 (c). If the directrix is above the focus, the parabola will open downward, as in Figure 63 (d). The line through the focus perpendicular to the axis of the parabola and terminating on the parabola is called the will
,
THE PARABOLA
175
length equals 4a. To illustrate, the coordinates of the ends of the latus rectum of the parabola lotus
rectum.
Its
2
y are
=
lax
Hence, substituting a for
(a, y).
y
=
2
x,
we have:
2
4a
,
or:
=
y
Thus the
2a.
Example Find the focus, the rabola y 2 Since a
rectum
total length of the latus
latus
=
8x.
is
positive, this equation is in the 2
y la
Then
a
or
Hence we have the
equal to 4a.
1
and the
directrix,
is
rectum of the pa-
form
lax.
= =
8, 2.
following:
focus: (2, 0) directrix : line (x
=
2)
latus rectum: 8
Example 2 Find the equation of the parabola with vertex at the origin
and with focus the point (0, 3). Since the focus is on the y-axis, the equation is in the form a
= =
2
=
x2
Here
Hence the equation
x
is:
lay. 3.
12y.
Problems 1. In each of the following parabolas, find the coordinates of the focus, the equation of the directrix, and the length of the
latus rectum: 2
(a) (b) (c)
y * y 2 y
=
y - -x. x2 -5y. lx 2 = 3$/. 2
8z.
(d)
-16*.
(e)
2z,
(/)
(g)
(h) (t)
x2 x2 x2
=
8y.
ANALYTIC GEOMETRY
176 2.
In the following, find the equation of the parabola,
vertex in each case
Focus, the point Focus, the point
(a) (6) (c)
(d) (e)
(/)
The
at the origin.
is
(2, 0). (0,
4).
Latus rectum 16; focus on the o>axis. Axis, the y-axis; and passing through (-4, = 5. Directrix, the line x Latus rectum 8.
2).
67. Equations of the parabola with vertex not at the
Y
We
origin.
have derived
the equation of a parabola with its axis the #-axis and
with
its
vertex at the origin,
as
We now
wish to find the
equation of a parabola with its axis parallel to the z-axis
and with point
its
vertex at the
(a, 0).
we must
first
To do
this,
derive a so-
called intrinsic property of
the parabola. Consider any point P on t fce para bola t/ 2 = 4az, as
Figure 64.
indicated in Figure 64. Drop PA perpendicular to the axis Call this distance di, and the distance VA of the parabola.
from the vertex
The
rabola y 2 (d 2 >
V
di)
=
to the foot of the perpendicular, d 2 are (d 2 , di). lies on the paSince .
P
coordinates of
P
the coordinates of the point must satisfy the equation
Therefore
4az,
we have
This equation
is
that
:
termed the property
of the parabola.
is,
THE PARABOLA
177
(NOTE This equation is true for any point on the parabola, and it is true regardless of the position of the parabola. The equation depends simply on the parabola itself not :
on
its position.)
Given the property equation
we
shall
now proceed
to find the equation of the parabola
represented in Figure 65.
Figure 65.
V
/3) and its on the parabaxis parallel to OX. Take any point P(x, y) Then, since ola, and drop a perpendicular to the axis.
Consider the parabola with
==
y
its
"~
vertex
at (a,
P
and
= therefore, substituting,
x
-
a,
we have:
(1)
This In
is
shall
have
the desired equation. like manner, if the axis
is
parallel to the y-axis,
:
(3)
(x-
-0).
we
ANALYTIC GEOMETRY
178
The reader
will
observe that, for a parabola of the type
-
(y
we have the
-
2
0)
following results
-
4a(z
a),
:
vertex:
(a,
focus:
(a
directrix:
x
axis:
y
(f)
+ a, 0) = a =
a
ft
latus rectum: 4a
Like results hold for the type (3)
(x
-
=
4a(i/
=
4a(*
-
4oz
2
a)
-
0).
Expanding the equation (1)
we have
(y
-
2
0)
This equation
may
this is the
The to
+ 4a =
0.
be written in the form: 2
(2)
may
a),
:
2
OX.
-
i/
+ Ay + Bx + C -
form always assumed
if
0;
the axis
is
parallel to
Similarly, the equation (3)
(x
(4)
x2
-
a)
-
2
4a(y
-
0)
be written
latter
is
+ Dx + Ey + F *
the form always assumed
if
0.
the axis
is
parallel
OK
It will be quite evident, if we reverse the above steps, that every second degree equation with only one squared term and with no xy term may be reduced to form (1) or (3),
and hence represents a parabola with
its axis parallel to,
or coincident with, one of the coordinate axes. in this category such an equation as 2 2/
=
4.
We include
THE PARABOLA
179
This equation represents the two parallel lines y 2 and 2, but we call the result a degenerate form of parabola y = as the equation is of the parabola type. 1
Example Given: y axis,
2
+ 2y
4x
=
+9
Find vertex, focus,
0.
and latus rectum.
Y D
\ *'
b'
Figure 66.
We wish
to reduce the equation to the form (y
(1)
-
0)
2
-
4a(x
-
a).
Completing the square, we have: y or:
2
- 4x - 9 + - 4(x - 2). a = 2,
+ 2y + 1 (y + I)
2
hence:
/?= -1. Hence:
vertex:
4a
a Therefore
we have
1)
(2,
= -
4, 1.
the following results: focus:
(3,
directrix:
x
1)
= +1
latus rectum: 4 axis:
y
1
1,
directrix,
ANALYTIC GEOMETRY
180
The following procedure is a good check: Draw the figure for the above results. First, plot the vertex; then, indicate, along the axis to the focus, the distance a from the vertex; then, the distance a from the vertex to the directrix. Example 2 Find the equation of the parabola with
and
its
focus at
(
its
vertex at
2, 3)
(
2, 6).
Since the x'a of the two points are the same, the parabola of the type (3)
(x
-
a)
= = =
2
a
Since
hence:
(x
+ 2)
2
4a(y
-
is
0).
3,
I2(y
-
3).
Problems In each of the following parabolas, find vertex, focus, direcand latus rectum:
1.
trix, axis, 2
(a) (6) (c)
(d)
y 2 y x2 x2
-
4t,
-
4z
=
0.
+ 2y + Sx - 15 = 0. 4z 3y + 6 = 0. - 2x + 2y - 5 = 0. - 24y + x + 40 = 0. 4*/ 2x - Sx - y + 12 = 0. 2
(e)
2
(/)
Find the equation of the parabola, given the following:
2.
(a) (&) (c)
(d) (e) (/)
Focus Focus
(g) (1,
1); directrix,
x
=
5.
4); vertex
(2, (2, -2). Vertex (3, 2) directrix, y = 7. Latus rectum 16; axis, y = 2; vertex (5, 2). Vertex ( 1, 3); axis, y = 3; passing through (3, 7). Axis parallel to the o>axis; and passing through (0, ;
-1), and
(2,
(3,
0),
(2, 2).
Axis parallel to the y-axis; and passing through
-l),and(-2, -4).
(2,
8),
CHAPTER XIII
THE ELLIPSE 68. Definition
and equation
of the ellipse.
defined as the locus of a point distance from a fixed point divided
is
fixed line is a constant less
than
1.
An
ellipse
P
moving so that its by its distance from a As in the case of the
parabola, the fixed point is called the focus, and the fixed
line,
the
directrix.
The constant less than call
1
we
the eccentricity, denoted
bye. In Figure 67, let us take the focus F on the z-axis, f and the directrix per-
DD
pendicular to the x-axis at and to the right of the
B
It is evident geometrically that the curve will cut the a>axis in two
focus.
points, focus,
A' and A, called the vertices one to the left of the and the other between the focus and the directrix.
Let us take 0, the origin of the coordinates, midway between A' and A. Let a represent the distance OA. We wish to obtain expressions for OF and OB in terms of a
and
e.
By
definition,
we have
the two relations
A'F (1)
A'B
FA 181
:
ANALYTIC GEOMETRY
182 or:
(3)
A'F =
6
A'B,
(4)
FA -
6
45.
Simplifying to obtain
Adding
(6)
and
a,
we have:
+ OJS),
(5)
a
+ OF
6(a
(6)
a
- OF =
e(05
(6),
we
-
a).
obtain:
2a
*
2e
OB,
or:
OB =e
Subtracting,
we have
:
OF =
06.
Then, the coordinates of the focus are tion of the directrix
(ae, o)
;
and the equa-
is
x
= a 6
We may now proceed
to find the equation of the ellipse.
D
F/A B
A'
(ae t o)
D' Figure 68.
Let P(x, y) represent any point on the distance from F to P distance from
DD'
to
P
ellipse.
SB 6.
Then
THE ELLIPSE
183
ur:
V(x~
2
06)
x
a/e
and: 2
V(:r
oe)
+y
2
ex
a.
Squaring the equation and collecting terms, we have: x*(l
For convenience, we
(This assumption
dividing
This
b 2,
by
-
e*)
+ y* -
a2
- aV.
let
is justifiable
we have
since e
is less
than
1.)
Then,
:
the required equation of the ellipse. Shape of the ellipse. Solving the above equation for y we obtain is
69.
:
,
y
=
-
Since for every value of x there are two values of y, equal numerically but with opposite sign, the ellipse is symmetrical with regard to the x-axis. If x is greater than a, or
if
x
is less
values (and If x equals
than
all
a,
the major axis;
a,
y
is
imaginary; hence x
may assume
a and +a. and including y equals 0. We call A' A (as in Figure 69) it is of length 2a and always contains the
values) between
focus.
Likewise, solving for
From
x,
we have
:
this solution it follows that the ellipse is also
metrical with regard to the y-axis; that y
sym-
may assume
all
ANALYTIC GEOMETRY
184
values between and including +b and 6, and that OE equals E'Q and equals 6, as in Figure 69. From these conclusions
it
follows that the curve
shape indicated. a
circle.)
We
(NOTE
call
E'E
:
If
a equals
and
is closed
6,
of the
the ellipse becomes
the minor axis;
it is
The
of length 26. intersection
of the
major and
the minor axes call
we
the center of
the ellipse. As in the parabola,
the latus rec-
tum
is
defined as
the line through the focus per-
pendicular to the
major axis and terminating on the
ellipse.
length
Its
is
26*
a
To
illustrate,
rectum
is ae.
as follows:
cated
the x coordinate of one end of the latus
Now,
to find the y coordinate,
we proceed
For convenience we have taken the form
indi-
by
In the above equation of the Hence: a
ellipse,
- aV
we
substitute ae for x.
a
Therefore the entire length of the latus rectum 2*-'.
a
UL is:
THE ELLIPSE
185
It is easily apparent that the line joining the focus to one end of the minor axis is of length a.
Y
A'
Y' Figure 70.
Similarly, 7/-axis
if,
and the
as in Figure 70, we choose the focus on the directrix perpendicular to the y-axis, and
then proceed as above, our equation will be
The coordinates
of the focus in this case are (o, ae),
equation of the directrix
is
The equation connecting
the quantities
the same as in the previous case, a2
Given an
ellipse of the
and the
-
aV
=
Example form
6
2 .
a, ae,
and
b
is
ANALYTIC GEOMETRY
186
find focus, directrix, vertices,
and
major and minor axes, latus rectum,
eccentricity.
This
ellipse is of
the form
= =
a
Hence:
6
From the we have:
= =
a 2e 2
relation
ae
a -
Hence:
=
Therefore
we have
4.
6 2,
a2 3. 2
=
=
25 >
3
ae
e
6
a
5,
06
=
7
3
5
the following results: focus:
(3, 0)
vertices:
(5,
major minor
axis:
10
axis:
8
j. directnx: -
x
eccentncity:
0)
25
=
3 5
32 latus rectum:
5
Problems In the following, find focus, directrix, eccentricity, major minor axis, latus rectum, and vertices. <
x2
" t/
2 a;
7+ a-2
2 i
2 3/
~
7. 1-
8.
16
W2
TO + To "
tt
9.
L
10-
x2 \
" y
2 -i
+ 3y - 4. 2x* + 5y - 10. 3x + y - 6. - 1. 2x + 2
z2
2
2
2
2
V
axis,
THE ELLIPSE 70.
Second focus and
directrix.
187
In the discussion in
Section 68, we chose the directrix to the right of the focus. It is evident that, if we choose the directrix to the left of the focus and choose the origin midway between A' and A, as before, the coordinates of the focus will be and the equation of the directrix,
Then, applying the definition
x
of the ellipse,
(
ae, o),
we have:
+ a/e
or, finally,
The above equation is when the ellipse had
From
these results
its
same one that we derived before, focus at (ae, o) and its directrix,
it
follows that every ellipse of the
the
form
has two foci with coordinates with equations x
ae, o),
(
and two
directrices
a -
=
e
Similar results follow for the ellipse of the form y* y __
x2
i
a
2>
I
7
__<>
i
*'
62
Problems Find the equation of each instance the center
is
of the following ellipses.
at the origin.
In each
ANALWIC GEOMETRY
188 1.
Foci at
2.
Foci at
(3, 0) (0, 4)
and (-3, and (0,
0); 4);
major axis minor axis
10. 6.
at (2, 0); one directrix, x = 4. 4. Latus rectum -^; major axis 10; foci on z-axis. 3.
One focus
5.
Vertices at
6.
One focus
7.
Distance between
8.
Foci on #-axis;
and
(0, 7)
at (0, 3)
;
7);
(0,
one
minor
directrix,
y
axis 4.
=
7.
foci, 6; minor axis 8. distance between foci,
6; latus
rectum
Answer:
9.
Foci on
t/-axis;
distance between foci, 4; eccentricity
f.
Answer:
10. Foci
on
z-axis; eccentricity
latus
rectum
Answer:
6.
12
16 f"
;
,_
-
1L
Foci on x-axis; passing through (2, 4) and 12. Foci on ^-axis; passing through (1, 4) and 13. Foci (4, 0) and (-4, 0); latus rectum 3.6. 14.
Foci on
t/-axis;
15. Directrices,
x
eccentricity f latus rectum 4; latus rectum 3. ;
4,
(
2v2).
(3, 2).
-J.
=
Equations of the ellipse with center not at the In Chapter XII we derived a so-called intrinsic origin. property of the parabola and applied that property to find the equation of a parabola with its vertex at (a, /8). We proceed similarly to find the equation of such an ellipse. Consider the ellipse (Figure 71) of the form 71.
a2
'
62
*'
point P, drop perpendiculars AP and major and the minor axes, respectively. Let
From any
BP to the AP = d\,
THE ELLIPSE
188
Figure 71.
and
BP =
d2
.
Then
the coordinates of
Substituting in the equation of the ellipse,
a2
+
b
2
P
are
(d*,
we have:
1
This equation is the required property of the ellipse. Let us now consider an ellipse with its center at (a, /8) and with its major axis parallel to the z-axis (Figure 72).
Take any point P and drop perpendiculars to the major and the minor axes. Now, we already have the above property equation:
But dt
a,
ANALYTIC GEOMETRY
190
Figure 72.
Therefore
:
(*
-
(1)
This
is
-
2 <*) O
(y
,
7
I
the required equation. this equation we observe
From
center:
(,
foci:
O
:
0)
(a
ae,
vertices:
(a
a, #)
directrices:
x
=
)
a -
a
e
Similar results obtain
if
axis parallel to the t/-axis.
we take an ellipse with its major The equation of the ellipse then
is:
(2)
(-
-
2 )
THE ELLIPSE If
we expand
of the
By
we
shall
have an equation
form
Ax 2
(3)
Here,
either (1) or (2),
191
A
+% + 2
+ Dy + E =
Cx
0.
B
and
have the same sign. reversing our steps and completing the square
of
(3), we may reduce this equation to either (2) or (1) except that the right-hand side may possibly be or 1. If we include point ellipses and imaginary ellipses in our
then we can say that every equation of the form (3), that is, every second degree equation, involving both x 2 and t/ 2 with coefficients of the same sign and no xy term represents an ellipse with its major axis parallel to, or coincident with, one of the coordinate axes. Of course, if A equals J5, the ellipse becomes a circle. classification,
,
Example Find center,
foci, directrices, vertices, axes,
and
latera recta of
the following ellipse:
4* 2
+ 9y 2
+
l&x
ISy
-
11
=
0.
Completing the square, we have: 4(z
2
-
4*
+ 4) + 9(2/ + 2y + 1) = 11 + - 2) + 9Q/ + I) = 36, 4(x - 2) (x (y + I) + __,!. 2
2
2
or:
__
2
2
This
last
equation
is
of the
(x
Hence we obtain the
-
a)
form
-
2
(y
ft*
_
following:
a b
ae
= -
3, 2,
!
e
_
* Va 2 ae
62
9
\/5'
= V5;
16
+ 9,
ANALYTIC GEOMETRY
192
a Therefore
we have
3
these results:
center:
(2,
1)
foci:
(2
-\/5,
(2
3,
vertices:
major minor
axis:
6
axis:
4
eccentricity:
-1)
-1)
V5 3 Q
latera recta:
3
directrices:
x
=
g 2
7=
V5
Problems Find center,
foci, directrices, vertices, latera recta, axes, of the eccentricity following: 1.
(*
-
2>
(g)
(0 (j) (fc)
-
2
2
(A)
64a;
-
50y
-
311
-
+ 5y + 6z 30y + 33 13a; + 4y - 26a; + 16y - 23 - 1% + 15 = 0. 3z + + 2j/ 6x + 4y + 9 = 0. + y - 2a; + 4y - 1 =0. 3s
2
2
a;
a;
2
2
2
V 2
2
-
0.
-
0.
0.
and
THE ELLIPSE 2.
Find the equation of the
193
ellipse with:
Foci at ( 2, 3) and (4, 3); major axis 10. Foci at (3, 5) and (3, 9) minor axis 8. 24. (c) Foci at (2, 4) and (8, 4) equation of one directrix, x (d) Center at (5, 1); eccentricity -J; major axis 14. (e) Latus rectum f; major axis 10; center at (3, 1). (/) Center at (2, 4); one focus at (8, 4); eccentricity (g) Axes parallel to coordinate axes; and passing through (0, 0), (1, 2), (1, -1), and (2,1). (a) (6)
;
;
.
.
3.
Given the
circle
z2
+y = z
64.
From
the circumference
Find the of this circle, perpendiculars are dropped to the re-axis. of these of the of the locus equation perpendiculars. mid-points 4.
Show
on the
that the distances to the foci from any point (x\ y\) 9
ellipse
*9
a2 are: a
+ ex\,
and a
ex\,
* + -=! o I
*> r2
and therefore that
their,
sum
is
the
constant 2a. 6. From the result of Problem 4, form a new definition of the ellipse. Then, with this definition, show how an ellipse may be constructed by continuous motion. (NOTE: Use two thumbtacks at the foci, and a piece of string of length 2a.)
CHAPTER XIV
THE HYPERBOLA equation of the hyperbola. An hyperbola is defined as the locus of a point moving so that its distance from a fixed point divided by its distance from a fixed line is a constant greater than 1. As in the ellipse, the fixed point is called the focus; the fixed line, the directrix; and the constant greater than 1, the eccentricity. 72. Definition
and
A F
O
Y'
D'
Figure 73.
We
proceed, as with the ellipse, to take the focus on the re-axis, and the directrix perpendicular to the #-axis but now to the left of the focus (Figure 73). Points A 1 and A (the vertices) will exist as before, and we take the origin
midway between them. 194
THE HYPERBOLA Now we
have, as in the
the following relations:
ellipse,
OF =
195
ae
coordinates of focus:
(ae, o)
equation of directrix: x
=e
Hence we have
also:
-
ae)*
x
and, finally,
x\l
But now, is
since e
is
-
e
2
a/e
2 )
+
y
2
=
greater than
Hence, we
negative.
+y
a2
1,
-
a 2e 2
.
the quantity (a 2
a 2e 2 )
let
and we have: This equation
is
the relation between ae and a and 6 for
The
the hyperbola.
final
reduction gives us the equation
of the hyperbola:
73.
for
i/,
Shape
of the hyperbola.
we obtain
Solving the above equation
:
As
in the case of the ellipse, the hyperbola is symmetrical with regard to the #-axis (Figure 74). If x is less than a
When x equals a, but greater than a, y is imaginary. values x assume (and all values) may y equals 0; hence, a. or equal to than or less to or than a, equal only greater This conclusion proves that there are no points of the curve
ANALYTIC GEOMETRY
196
between
(
a, o)
and
We call A'A,
(a, o).
as in Figure 74,
2a and, extended, conAs x increases positively, y increases both tains the focus. and Likewise, as x increases neganegatively. positively both increases positively and negatively. tively, y Hence, the transverse axis.
It is of length
the hyperbola extends indefinitely to the right of the j/-axis and above and below the o>axis; and it also extends indefinitely to the left of the t/-axis the x-axis.
and above and below
Y D
Similarly, solving for x,
x
=
we
obtain:
o
From
this equation it follows that the
symmetrical with regard to the
assume
all
values from
>
to
j/-axis,
+00.
shape indicated in Figure 74. In order to preserve symmetry, (Figure 74), joining
(o,
6)
and
is
also
and that y may
The curve
we
(o, 6),
hyperbola
call
is
of the
the line
E'E
the conjugate axis.
THE HYPERBOLA
197
The intersection of the transverse and the conjugate axes we call the center of the hyperbola. As in the ellipse, the latus rectum, perpendicular to the
Its length is 26.
transverse axis at the focus,
is
of length
26 2
It is evident that the line joining (a, o)
length
and
(o, 6) is
of
ae.
Figure 75.
In a similar manner, if we choose the focus on the y-axis and the directrix perpendicular to the t/-axis (as in Figure 75), and proceed as above, our equation will be:
The
coordinates of the focus are
of the directrix
is
-
!?
6
(o, ae),
and the equation
ANALYTIC GEOMETRY
198
Example Find focus,
vertices,
directrix,
eccentricity,
latus
rectum,
transverse axis, and conjugate axis of the hyperbola
~ This
is
of the
*'
form
~
^
= L b*
(NOTE: In the hyperbola, a greater than
= 16
9~
may
be
than b or equal to b or
less
6.)
a2
Hence:
6
2
ae
or:
= = =
a e
e
=
9,
16,
5,
a2
9
ae
5
O6
a
Hence we have the following
=
5
__ ~.
o
results:
focus:
(5, 0)
transverse axis: 6
conjugate axis: 8 vertices:
latus rectum:
(3,0) 32o
directrix:
x
=
Q ~
5 5
eccentricity:
o
Problems Find focus, transverse axis,
directrix,
eccentricity, axis of:
and conjugate
vertices,
latus
rectum,
THE HYPERBOLA
3.
x2 x2 x2
-
4 *'
!
~
1.
2.
3
74.
- 6. y* = -6. 2 2y - 8.
6.
y*
Vl
_ ~
7.
8.
- x - 1. 2x* - 3y = 9. z 2 - 3y - 9. y
2
2
2
2
*
f o *' 25
i
l'
6
199
~
**
_ ~
i
*'
24
Second focus and
ellipse, it is
directrix. As in the case of the evident that every hyperbola of the form
a
has two foci with coordinates with equations x
(ae,
o),
and two
directrices
a ~
-
e
Similar results follow for the hyperbola
_ a2
6
2
With every hyperbola there are associ76. Asymptotes. ated two lines of interest and importance called asymptotes. We shall derive the equations of these lines. Consider the hyperbola
a2 in the
-.
form
W Let
a:
i
&2
increase indefinitely.
1
As
'
2 a-
it
does, the quantity
ANALYTIC GEOMETRY
200
approaches
Thus y x
approaches
Hence, as x increases indefinitely, the hyperbola
_ a2
62
approaches the two straight lines
or, in
simpler form,
y
-x.
a
THE HYPERBOLA
We
call
201
these lines (Figure 76) the asymptotes of the
hyperbola a2
The equations
of the
6
2
asymptotes
may
be written:
Similarly, the equations of the asymptotes of the hyper-
bola 2 2/
X2
=
^""ft 2
1
will be:
a
V-
x
l
or
V
2
x2
a
2
62
Problems 1. In each of the following, find the equation of the hyperbola. In each case, the center is at the origin.
(a) (6) (c)
(d) (e) (f)
(g)
(h)
One One
one vertex at (4, 0). = x 2; one focus at ( directrix, 4, 0). Transverse axis 8; conjugate axis 10. One directrix, x = f transverse axis 20. One directrix, y = V; conjugate axis 6. Foci on o>axis; distance between foci 8; latus rectum *. Foci (0, 6) eccentricity 2. Directrices parallel to 2/-axis; distance between directrices, focus at
(5, 0);
;
;
between foci, 6. Latus rectum 4; slope of asymptotes, Foci on t/-axis; passing through ( 1, One directrix, x = i; latus rectum 6.
f; distance (i) (f) (jfc)
(I)
J.
1)
Eccentricity \/2; slope of asymptotes,
and 1.
(5,
3).
ANALYTIC GEOMETRY
202
(m) Foci on a>axis; passing through x2 y* (ri)
2.
Asymptotes,
4
0; foci
2
(3, 2)
on
and
(
\/6, 0).
y-axis; conjugate axis 6.
Prove analytically that an hyperbola cannot intersect
its
asymptotes. 76. Equations of the hyperbola with center not at the origin. Exactly as in the case of the ellipse, we shall
proceed to find the equation of the hyperbola with its center at (a, ]8) and with its transverse axis parallel to the #-axis.
Y
Consider the hyperbola
From any
point P, drop perpendiculars to the transverse
and the conjugate
Then the
axes.
coordinates of
Call
P
them
di
and d 2
are (d 2 , di).
derive the property of the hyperbola:
,
respectively.
Substituting,
we
THE HYPERBOLA
203
consider the hyperbola with its center at (a, $) its transverse axis parallel to the #-axis. From
Now,
and with
any point P, drop perpendiculars to the transverse and the conjugate axes.
Figure 78.
We
already have the property equation:
But x
-
y
Hence
:
(x
-
a)
2
(y
-
=
ft)
(1)
This
is
1
the required equation of the hyperbola. this equation we observe
From
:
center:
(<*,)
foci:
(a
ae
vertices:
(a
a,
y
ft)
ft)
ANALYTIC GEOMETRY
204
= a
x
directrices:
e
asymptotes: if we take the transverse axis paralThe equation of the hyperbola then is
Similar results obtain to the
lel
j/-axis.
:
-
(y (2)
a
If
we expand
2
(x
)
2
-
a)
2
2
6
either (1) or (2),
we obtain an equation
of
the form
Ax
(3)
Here,
A
and
B
+ By + 2
2
Cx
have opposite
+ Dy + E =
0.
signs.
reversing our steps, we may reduce equation (3) to either (2) or (1) except that the right-hand side may
By
possibly be
0,
the side equals 0, equation (3) represents lines, or a degenerate hyperbola.
If
two intersecting
Including this case, we that
may
tion of the form (3)
say, then, that every equa-
every second degree equation, both terms with opposite signs and no xy including squared term represents an hyperbola with its transverse axis parallel to, or coincident with, one of the coordinate axes, is,
Example Given an hyperbola 3z 2
-
Find center, focus,
form
of the 2
5*/
+
1
6z
+
-
2Qy
32
=
0.
directrix, eccentricity, vertices, transverse
conjugate axis, latus rectum, and asymptotes. Completing the square, we have:
axis,
3(z or:
2
+ 2x +
1)
-
2
5(2/
-
-
4y
+
3(s + I) 6(2, (x + I) _ (y 2
32
4) 2
2)
2
2
2)
-
15,
+3 -
20,
THE HYPERBOLA This equation
is
in the form
Hence we have the
"" 1 -
i
^2 following:
o2 6
oV
-
o2
2
+ 6*
= = =
5,
3,
5
+3
ft
e
ae
8;
\/g'
V^ rV5
=
(-1
V^,
2)
vertices:
(1
Vo,
2)
transverse axis:
conjugate axis:
2V5 2V3
directrices:
x
oe
=
a
we have
=
5
Q,
6
Therefore
205
these results:
center:
1,2)
(
foci:
S
=
1
/-
eccentricity:
\5 6
latus rectum: (x ^
asymptotes:
----
+-
1
I)
(y
- 2)
.
o
5
Example 2 Find the equation of the hyperbola with foci at (1, 2) and (7, 2), and with conjugate axis 4. Since the foci are on the line y 2, the equation will be of the form (x
a
a) 2
2
_ (y~fl* " 2 5
ANALYTIC GEOMETRY
206
=
Hence:
-
=
4,
AI
= 2. 2b = 4, 6 = 2; ae = 3, - 62 = 9 (3
Since
and
since
a2
therefore:
=
aV
4
,
=
5.
,.
Problems 1.
Find center,
foci, directrices, eccentricity, vertices, trans-
verse and conjugate axes, latera recta, and asymptotes of the following:
-1.
,
i.
- l) - (y - 2) 2 = 1. 9x" - 16y 2 - 54a; - 64y - 127 = 0. 6i 2 - 9y - 48x + 18y + 141 = 0. 2 20a; - 16y - 40X - 32y + 324 = 0. 2 2 4 16 + lOy - 9 = 0. 5y 2 - 16 = 0. x y + 2x + 6y z
(A) (t)
(x
2
(j)
2
(*)
(0
2
(TO)
(n)
i
2
-
y*
+ 2x =
0.
THE HYPERBOLA
207
.
Find the equation of the hyperbola with:
2.
(a)
Foci at
(3,
(b)
Foci at
(2, 2)
(c)
Center at
and (17, 1); transverse axis and (2, 8); conjugate axis 8.
1)
(2, 4)
(d) Vertices at
;
(8, 6)
one focus at
and
(8,
Eccentricity 3 center at
(/)
Transverse axis 16; foci on
-
__
(x
2
2)
one directrix, x = rectum *-. one focus at (8, 3).
(10, 4)
;
4.
4) ; latus
(e)
;
10.
3)
(2,
line
;
y
_
(y +
=
1;
asymptotes:
2
1)
.
o.
3. Find the equation of the hyperbola with axes parallel to the coordinate axes, and passing through (1, 0), (0, 2), ( 1, 2),
and
(3,
-1).
Show
4.
that the distances to the foci from any point
(x\, yi)
on the hyperbola
*!_, a2
are: ex\
+
and
a,
ex\
a,
6
2
and that therefore
their difference is
the constant 2a.
From
6.
the result of Problem 4, form a
new definition of the how an hyperbola (NOTE: Use two
hyperbola. Then, with this definition, show may be constructed by continuous motion. thumbtacks at the foci, and a piece of string.)
Two
6.
each
is
bolas.
7.
hyperbolas so related that the transverse axis of the conjugate axis of the other are called conjugate hyperFind the equation of the hyperbola conjugate to
Prove that two conjugate hyperbolas have the same
asymptotes. 8.
If e
and
e
1
are the eccentricities of
bolas, prove:
^-i. e
9.
Show
2
e'
2
that the foci of the hyperbola
two conjugate hyper-
ANALYTIC GEOMETRY
208
and those 0;2
+ y2
=
of its conjugate all
a2
lie
on the
circle
whose equation
is:
+ b*.
10. Show that the circle of Problem 9 meets either of the two hyperbolas on the directrix of the other. 11. Show that the line joining the focus of an hyperbola to the focus of its conjugate, passes through the point of intersection of the directrices of the two hyperbolas. Find 12. An hyperbola with a equal to b is called equilateral. the eccentricity of such an hyperbola. 13. A latus rectum of the hyperbola
!_!_i
a2
2
fe
extended by the amount k so that it just reaches an asymptote. that k is equal to the radius of a circle inscribed in the triangle formed by the asymptotes and the line x = a.
is
Show
CHAPTER
XV
TRANSFORMATION OF COORDINATES In Chapter
77. Translation of axes.
XI while discussing
the equations of circles, we saw that the equation of a circle with its center at the origin was x2
+y =
whereas, for the circle with
2
its
r
2 ;
center at (a,
/3),
the equation
was
(x-ar +
(y~W
=
r\
It is quite evident that the equation of the circle with its center at the origin assumes a simpler form than that
not at the origin. Similar conclusions obtain for the ellipse and the hyperbola, and also for the parabola if we consider the vertex in place of a
assumed when the center
is
center.
Therefore, given any curve with the center or some other element not at the origin, it would be quite convenient if we could move the coordinate axes in such a way that the origin would coincide with this point, and the coordinate axes would coincide with the axes of the curve, if such axes existed. This procedure is possible. There are
two motions involved
:
translation, or
axes parallel to themselves;
and
moving the coordinate
rotation, or turning the
axes about a point. We shall first consider translation, which involves changthe directions of the axes. ing the origin without changing We are given two sets of axes (Figure 79) the original and OF, with origin at 0(0, o) and a new set, CX set, We are given, also, a at C(h, k). with and :
f
OX
OF',
;
origin
referred to the original axes point P, whose coordinates 209
ANALYTIC GEOMETRY
210 are x
and y = AP, and whose coordinates referred = CB and y = BP. We wish to the relations between the primed and unprimed
to the find
OA
new
axes are x
f
f
quantities*
A
D Figure 79.
We have
:
OA = OD
+ DA,
or:
=
x
h
+ x'.
Also:
AP = AB +
BP,
or:
=
y
k
+ y'.
Hence we have the equations
of translation:
It is quite evident, then, if we are given the equation of a curve referred to any set of axes, that the equation of the same curve referred to new axes parallel to the given axes and with origin at (h, k) is obtained by replacing f k. x with x h and y with y'
+
+
Example Find the equation of the x
2
+ y*
1
circle
-
6x
+ 2y -
6
-
TRANSFORMATION OF COORDINATES
211
referred to parallel axes through (3, -1).
Since
x
and
y
we x'
= =
+3
x'
-
y'
substitute in the circle equation
+ 6x' + 9 + y' -
2
+
2
20'
-
1
and obtain:
+y = >*
2
-
ftr'
we have
After simplifying this result, z'
1,
+ 2y' -
18
-
6
-
0.
+ 4k -
4
=
0.
2
:
16.
Example 2
Remove
the
first
x2
degree terms from the following:
+y 2
2x
x = y =
Since
and
+%x' + h y' + k,
then, substituting in the above equation, x'
+ 2hx' + h + y'* + 2ky + k 2
2
(NOTE: In order for the must equal zero.)
first
0.
we
obtain:
2
f
-
=
4
-
2x'
2h
+
f
y
degree terms to vanish, their
coeffi-
cients
Collecting the
first
x'(2h f
y (2k
Hence:
2k
+4
k
We
= =
2
0,
0;
1,
-2.
we have:
+ y' + 1 + 4 x' + y' 2
or, finally,
= =
2),
+ 4).
-
h
flj'2
-
2ft
or:
Substituting again,
we have:
degree terms,
2
2
-
=
9.
8
-
4
0,
might have solved Example 2 by another procedure,
indicated in the succeeding text
:
ANALYTIC GEOMETRY
212
Remove
the
first
Example S degree terms from the following:
x2
+ y* -
+ 4j/ -
2x
4
=
=
9.
0.
Completing the square, we have:
-
(*
+
2
I)
Since
x
and or
y x
and
y
= =
1
+2 = = h
hence, as before,
+ 2)
(y
fc
x'
f
y
+
y'
= -
x
2
r
,
1
2,
1,
-2.
Problems Find the coordinates
(2,
of the points (1, 3), to referred axes through (1, 2) parallel 3). (3, 2. Find the coordinates of the points ( 1, 2), (3, 2). (x, y) referred to parallel axes through (3, 1.
5),
and
2),
and
6 == Find the equation of the line x referred to 2y 2). parallel axes through (2, 2 4. Find the equation of the curve x* + y 4z + &y 12 = referred to parallel axes through (2, 3). 2 2 5. Find the equation of the curve x + 2y + 2x - 12y + 17 3.
=
referred to parallel axes through ( Find the equation of the curve 3z 2
6.
=
1, 3).
%
2
6x
referred to parallel axes through (1, 2). 7. By translation of axes, remove the terms of
from: z 2
+
2 t/
-
2z
+ 4y -
3
=
l&y first
25
degree
0.
8. By translation of axes, remove the constant term and the 4z term in x from: x 2 Sy + 12 = 0. 9. Find values of h and k that will remove the constant term 4 = 0. Are the values obtained unique? from: 2x 3y
+
10.
By
translation of axes, derive the following equations:
+ ,
(6)
(x
-
a)
2
=
4a(y
-
|8).
TRANSFORMATION OF COORDINATES 78. Rotation of axes.
213
We now wish to change the direcWe do this
tions of the axes without changing the origin. by a rotation of the axes through an angle .
OX
and OF be an and OX' and OF', a new Let
set,
with
the
original set of coordinate axes;
angle
through which the original axes must be rotated, about the common origin 0, to coincide with the new axes (Figure 80) Let .
P
be a point whose coorreferred
dinates
to
original axes are x
and y
the
= OA
A
= AP, and whose
E
Figure 80.
coordinates referred to the new axes are x' = OB and y from B to AP and OX.
r
= BP.
Drop perpendiculars
Then:
= OE - AE = OE - DB = OB cos = x cos
f
BP
sin
r
y sin
.
Similarly:
= AD + DP = EB + DP = OB sin + BP cos = x sin + y' cos <. <
<
f
Hence we have the equations
of rotation:
ANALYTIC GEOMETRY
214
1
Example
By a rotation of axes, remove the term in y from 3x '+ 4y
10
:
=
0.
we have:
Substituting equations of rotation, 3(x' cos
y' sin <)
$
+ 4(s' sin + y' cos
-
10
0.
Or: a;'
(3 cos
+ 4 sin
<)
+ #'(4 cos
3 sin
<
f
(NOTE: For the term in y to vanish, Hence: zero.) 4 cos 3 sin
cos
tan
=
10
=
0.
must equal
0,
4 *"
or:
Then:
-
its coefficient
sin
)
3
= -4
<
o Therefore:
sin
=
<
4 ->
5 cos
=
3
o
Hence, substituting further, we obtain:
5x'
or:
-
10 x'
or:
= =
0, 2.
Problems
By a rotation of axes, remove the term in x from: 3x = -6 0. 2. By a rotation of axes, remove the term in y from: 5x + -7 = 0. 3. By a rotation of axes, remove the term in y from: x* + 1.
-
2z 4.
-
2y
=
3x
2
0.
After the axes are rotated through
of the line
y
-
2y
+6
=
30,
find the equation
0.
After the axes are rotated through 45, find the equation - 10 = 0, of the line 3x 3y 6.
+
TRANSFORMATION OF COORDINATES 6.
215
After the axes are rotated through 90, find the equation
y = 25. 7. After the axes are rotated through 45, find the equation 2 of the circle x 2 y = 25. 8. After the axes are rotated through 45, find the equation of the curve xy = 6. 9. After the axes are rotated through 45, find the equation of the curve xy = 4. 10. Remove the xy term from xy = 10, of the circle x
+
2
2
+
:
Removal of the xy term. In the four preceding chapters, we discussed the various types of second degree 79.
equations containing no xy terms, and the curves arising We propose to show here that there is a transformation which will always remove the xy term from the most general equation of the second degree; for
from these types.
example
:
Ax 2
+
Bxy
+
Cy
+
2
Dx
+
Ey
+F
=
0.
Thus, having reduced the equation to one with which we are familiar, the procedure enables us to handle completely the general second degree equation. We now proceed to the finding of this transformation.
We
are given
Ax 2
+
Bxy
+
Cy
+ Dx + Ey + F
2
=
0.
Substituting in the equation the following values:
= =
x y
we have, x'\A r
-\"X
cos
+B
y'(B cos
+y' (A sin +x'(D cos
+y'(E
cos
+F =
0.
x' sin
y' sin <,
$
+ y' cos
<,
after collecting terms,
2
2
x' cos
2 <
4)
<
cos
B sin B sin
2
sin
+ E sin D
2
+
C
sin
2
<)
+ 2C sin cos cos + C cos
<j>
2
<)
sin <)
4>)
2A
sin
cos <)
ANALYTIC GEOMETRY
216
f
r
(NOTE In order for the x y term must equal zero.) Hence:
to vanish, its coefficient
:
B(cos
2
-
sin 2
-
4>)
(A
-
C)2
sin
cos
0.
But, from Section 35, cos 2
sin
4>
2
=
cos 20,
=
sin 2<.
and 2 sin ^ cos
Therefore
<
:
B
=
cos 2<
-
(A
C) sin 20,
or: sin
cos
B
20 ~~ _ 20
A
C
Hence we have the following transformation
In problems generally, we need the values for sin and cos <, in order to substitute them in the rotation formula. Such values are easily found from a variation of the halfangle formulas (Section 36)
=
sin
;
that /I
1
cos
is
:
-
cos
+
cos
2
Before proceeding to an example,
let
us
make a few obser-
vations concerning the above material. In the first place, the translation and the rotation formulas of Section 78 are of the first degree; hence, when they are substituted in an equation, the degree of the equation is certainly not raised.
Moreover, the degree
is
not lowered;
for, if it
substituting further to restore the original axes
were, then
would yield
TRANSFORMATION OF COORDINATES
217
of lower degree than the original degree. Thus the degree of an equation remains unchanged by a transformation of coordinates.
an equation
Therefore
it
follows that,
tan 2 ,
by the transformation
.
B __,
the general equation of the second degree
Ax 2
may
+ Bxy +
Cy
2
+ Dx + Ey+F =
Q
be reduced to the form A'x'*
+
C'y'*
+ D'x' +
where A' and C" cannot both be
E'y'
+ F' =
0,
zero.
Y
X'
this equation However, we have previously shown that of degenerate, including some conic, type always represents we Hence say: Every second imaginary, or point conies. and every conic is repreequation represents a conic,
degree sented by
A 81.
of the second degree. the ellipse in Figure typical situation is illustrated by
an equation
ANALYTIC GEOMETRY
218
In Section 80, we shall show how the type of conic may be determined by certain relations among the coefficients A, B, and C. 1
Example
Remove Here,
Hence
the xy term from: 5z
A =
B =
5,
4xz/
-4, and C =
tan 2
:
2
+ 2y
2
=
6.
2.
-4
4
5-2
3
4 sin 2
Therefore:
;
5 cos
(Since
we assume
second quadrant.)
3
20 =
5
to be acute in this example,
Then:
sin
=
we have:
Substituting in the original equation,
or:
x'*
This equation
is in
+
2
6*/'
=
the form of an
6,
ellipse.
Example 2
Remove Here,
Hence: Therefore: or:
the xy term from: xy
A *
0,
B =
1,
=
and C
=
0.
tan 20
=
-
fc.
=
20 = 90, = 45.
o.
20
is
thus in the
TRANSFORMATION OF COORDINATES Substituting in the original equation,
or:
(*'
~
r
y')(x
or:
x'
This equation
in the
is
form
of
2
219
we have:
+ y') =
*,
-
2k.
y
12
=
an equilateral hyperbola with
the coordinate axes as asymptotes.
Problems
Remove
the xy terms, and determine the type of curve in each
Given:
of the following. 1.
2.
3. 4. 6. 6.
7. 8. 9.
10. 11. 12. 13. 14. 15.
xy = 4. 2xy = -7.
- 3y 2 - 4 = 0. x2 3xy 2 2 5x 8 = 0. 5y Gxy 2 = 2 2. x + 4xy + y 2 = 10. 3x 2 - 3xy - 5x - 3y 6 = 0. x2 xy - 4z - 2y + 8 x 2 - xy + 2 2 2x = 0. x 2xy + 2y 2 2 4 = 0. 2x 7/ zi/ 2 = 0. t/ 2xy + 2x 2 20 = 0. 8z 2 + 12xy 17t/ 2 2 x 2xy + y + 2x + 6y == 0. 2 3x 4^ + 6z + 4y - 1 = 0. 2 - 6y 2 ~ 30 = 0. x 24xy
+
+
-y
+
+
V
=
0.
+
+
+ + +
In this 80. Invariants; classifications of types of conies. section we shall find how to determine at a glance, by certain relations
the three coefficients A, B, and C, by the equation
among
the type of conic represented
Ax 2
+
Bxy
+
Cy
2
+ Dx + Ey + F
=
0.
Consider the above equation in connection with the rotation formulas: x
y
x' cos
=
f
x sin
y sin <,
1
+
y' cos
.
ANALYTIC GEOMETRY
220
Substituting in the original equation,
AV2 +
we
obtain:
-
+ CV + D'x' + E'y' + F' 2
B'x'y'
0,
where, as determined in Section 79,
A = A cos 2 B' = 5 cos C' = A sin r
(1) (2)
2<^>
2
(3)
+ B sin
- (A - B sin
+ C sin
cos
C) sin cos
2
20
+ C cos
2
and D' E" F'
= D = E = F
+ E sin
cos
D
cos
sin
(These last three relations are not necessary in our particular problem.)
We
shall first find
some
interesting relations that exist
between A, B, C and A', ', C'. Adding equations (1) and (3), we obtain: A'
+
= A (cos 2
C'
+ sin
2
0)
+
C(cos
+ sin
2
2
0),
or: (4)
A'
+
C = A 7
+
C.
Observe that the relation between the primed quantities, A' + C', is equal to the same relation between the unprimed We call A + C an invariant. quantities, A + C.
We (3)
now derive two from (1), we have: A' - C' = (A - C)(cos shall
other invariants.
-
2
sin
2
Subtracting
0)
+2
20
+ 5 sin 20.
sin
cos 0,
or: (6)
A'
-
C"
=
(A
-
C) cos
Observe that (A C) is noi an invariant. But if we 2 B' to the left-hand side of the equation, square (5), add and then add the value of B"2 from (2) to the right side of the equation,
we
obtain:
- C') 2 + /2 = - C) 2 cos 2 20 + 2B(A - C) sin 20 cos 20 + J5 2 sin 20 (A + (A - C) sin 2 20 - 2B(A - C) sin 20 cos 20 + 5 2 cos 2 20,
(A'
2
2
TRANSFORMATION OF COORDINATES
221
or: (6)
(A
1
-
+B /2
2
C')
(A
-
C)
2
+ B\
B2 invariant: (A - C) 2 Finally, square (4), and then subtract from (6).
+
Thus we have the
- 2A'C + r
The
is
resulting equation
A' 2
.
- 2A'C - C' 2 = A - 2AC + C + B - A 2 - 2AC - C 2
+
C' 2
B' 2
-
A' 2
f
2
2
2
,
or: (7)
Hence
B
2
Moreover,
B' 2
-
4^4 C is it is
=
4A'C'
2
-
4AC.
an invariant.
the invariant
mine the various types
of
B
2
conies.
4AC
that will deter-
Consider the trans-
formation
B applied to the general second degree equation. We know, = 0. Hence, the resulting equa79, that B'
from Section
tion of the second degree (8)
and
(7)
A'x'
is
:
+ CV + 2
2
D'x'
+ E'y + F' = f
0;
becomes: (9)
B 2 - 4AC =
-44'C'.
either A' or C' is zero, (8) represents a parabola. 2 4AC = 0. Again, from (9), we know that B However, if A' and C' have the same sign, (8) represents an ellipse. 2 4AC is negative. However, from (9), we know that JS have C" A' and if (8) represents a signs, opposite Finally, 4AC that B 2 know we from (9), hyperbola. However,
Now,
is
if
positive.
The
process
is
also reversible.
Hence we have: Parabola: Ellipse:
Hyperbola:
B 2 - 4AC = B 2 - 4AC < B 2 - 4AC >
ANALYTIC GEOMETRY
222
In each of the above formulas, it should be understood that degenerate and imaginary cases are included.
Example
- 4xy 2y + x y - 3 = 0. Classify: 3z = -4, and C = -2. = Here, A 3, B B 2 - 4AC - 16 - 4(3) (-2) Hence: = 16+24 2
2
= Therefore the form
is
40.
an hyperbola.
Problems Classify: 1. 2.
3. 4. 5. 6.
7. 8.
9.
10.
- 7x 3y - 10 4y~ 2xy 2 6 = 0. x + xy y + y 2 = 9. + 4xy 2y + 3xy - y 2 2x - y - 4 = 0. 2 5 - 0. y 4xy 6 = 0. 2x y 3xy 3x 2 + 3i/ 2 - a: - 2y - 4 = 0.
3z 2 x2 2x 2 x2 2x 2
-
+
+
+
+
+ +
+
O + 2^/) - 4z. = 4. (x + 21/) 2x + y xy + 3y 2
2
2
3
=
0.
0.
INDEX
INDEX [REFERENCES ARE TO PAGE NUMBERS] Abscissa, 36
Conies, 172
Addition theorems, 76 Ambiguous case, 54 Angle:
definition of, 172 determination of types of, 221
negative, 35 of depression, 31 of elevation, 30
types of, 172 Conjugate axis of hyperbola, 196 Conjugate hyperbola, 207 Coordinate axes, 37 rotation of, 209, 213 translation of, 209, 210 Coordinates, 37 transformation of, 209
positive, 35
Cosecant:
between two
131
lines,
bisectors, 156 definition of, 21
functions
of, 21, 22,
38
Anti-logarithm, 12
definition of, 22, 38
Asymptote, 199
variation Cosine:
Axis:
of,
conjugate, 196 major, 183 minor, 184
variation Cosines:
of coordinates, 37 of parabola, 173
law of, 57 Cotangent:
radical,
definition of, 21,
169
Curve, 136 condition that a point
on, 136
Directrix, 172 of ellipse, 181 of hyperbola, 194 of parabola, 172
Center: of circle, 162 of ellipse, 184 of hyperbola, 197
Distance
Characteristic, 11
two
circles,
162 general equation of, 164 standard equation of, 162 through intersection of two 168 through three points, 165 unit, 43, 44 Circular measure, 100 Common logarithms, 10 Compound interest, 18
:
between two parallel lines, 151 between two points, 123 from a line to a point, 153 of a line from the origin, 148
169 of
lie
Depression, angle of, 31 Directed distance, 35
Briggs logarithms, 10
common
38
43
variation of, 44
Base: of a logarithm, 7 of a number, 3
Chord
of,
definition of, 22, 38
transverse, 196
radical,
44
169
Circle,
Division of line segment, 126 Double angle formulas, 79, 80 circles,
Eccentricity, 172 of ellipse, 181 of hyperbola, 194 of parabola, 172 Elevation, angle of, 30
225
INDEX
226 Ellipse, 181
Half-angle formulas, 82, 83
184 criterion for, 221 center
Hyperbola, 194 asymptotes of, 199 center of, 197 conjugate, 207 conjugate axis of, 196 criterion for, 221
of,
definitions of, 172, 181, 193
general equation of, 191 imaginary, 191 major axis of, 183 minor axis of, 184 point, 191 property of, 190 standard equation of, 183, 190 vertices of, 181
Equation
:
degree, 144 normal, 149 second degree, 215, 219 trigonometric, 28, 50
definitions of, 172, 194, equilateral,
185,
208
general equation property of, 202
of,
standard equation 203, 204 transverse axis
first
of,
197,
196
:
ellipse,
191
number, 165, 169 point, 169
of ellipse, 181 of hyperbola, 194
of parabola, 172 Fractional exponents, 5
Infinity, 41 Initial side, 21
Functions
Inscribed circle, 168 radius of, 98
:
trigonometric, 21, 22, 38 of:
30, 45, 60, 23 90, 180, 270, 360, 40-42
90 - 0, 25 90 -f 0, 72 n90 0, 72 180 - $, 46 n!80 0, 48 -0, 49 + ft 76 a - ft 78 2, 79
Intercept form of straight line, 144 Intercepts on the axes, 142 Interpolation, 13 Intersections of two curves, 137 Intrinsic property: of ellipse, 189 of hyperbola, 202 of parabola, 176 Invariant, 220
Latus rectum: of ellipse, 184 of hyperbola, 197
83
Fundamental
195,
66 fundamental, 66, 67 Imaginary circle, 165
Focus, 172
82,
of,
Identities,
Exponents, 3
f
204
vertices of, 194
Equilateral hyperbola, 208
Functions
207
degenerate, 204
identities, 66,
General equation: of circle, 164 of ellipse, 191 of first degree, 144 of hyperbola, 204 of parabola, 178 of second degree, 215, 219 of straight line, 144
67
of parabola, 175 Law of cosines, 57 Law of sines, 51 Law of tangents, 93
Laws Laws
of exponents, 3, 4,
6
of logarithms, 8, 9, 10
Locus, 136 equation of, 136 Logarithms, 7 base of, 8 Briggs, 10 common, 10
INDEX Logarithms laws
227
Rotation of axes, 209, 213 formulas for, 213
(cent.):
10 of functions, 112-116 of numbers, 108, 109 to base e 10 use of, 12, 15 of, 8, 9,
Secant:
38
definition of, 22,
1
variation
of,
44
Second
directrix: of ellipse, 187 of hyperbola, 199
Mantissa, 11
Second focus:
Negative angles, 35 Negative direction, 36 Normal, 149 Normal form, 149
of ellipse, 187 of hyperbola, 199 Signs of the functions,
Oblique triangle, 51 Ordinate, 37
Sine: definition of, 21, 38 variation of, 43
Origin of coordinates, 37
Sines,
Slope
Parabola, 172 axis of, 173 degenerate, 179 general equation
of
of of,
property of, 176 standard equation 177 vertex of, 172
178 of,
173,
Parallel lines, 130, 145, 151 Perpendicular lines, 130, 146 Principal angle, 73 Projection, 73
theorems on,
51
of,
:
definition of, 129 of a line through
criterion for, 221
Property
law
39
73, 74
:
of ellipse, 189 of hyperbola,
202 of parabola, 176 Pythagorean law, 22 Quadrants, 37
174,
two points, 130 two parallel lines, 130 two perpendicular lines, 131
point-slope form, 141 slope-intercept form, 143 Solution of triangles, 29, 51
Straight line, 140 general equation, 144 intercept form, 144 normal form, 149 parallel to an axis, 140 point-slope form, 141 slope-intercept form, 143 through the intersection of two lines,
157
two point form, 143 Tables: of logarithms of
numbers, 108, 109
of
of
logarithms
trigonometric
functions, 112-116
formulas, 95 Radian, 100 Radical axis, 169 Radical center, 169 Radius: of circle, 162 of circumscribed circle, 62 of inscribed circle, 98 vector, 37 Ratio formula, 127 Removal of xy term, 215, 216 Right triangle, solution of, 29 Roots, square, tables, 118, 119 r
of squares
and square
roots, 118,
119 of trigonometric functions,
112-
116
Tangent: definition of, 21, 38 of a half-angle in terms of the sides of a triangle, 94
of the angle between two lines, 132 variation of, 43
Tangents, law of, 93 Terminal side, 21 Transformation of coordinates, 209
INDEX
228
Translation of axes, 209, 210 Transverse axis of hyperbola, 196
Unit
Triangle: area of, 62, 96, 99 of reference, 38 oblique, 51
Variation of the trigonometric func-
right,
29
solution of, 29, 51 Trigonometric functions, 21, 38
circle, 43,
tions,
44
42
Vector, radius, 37 Vertex: of ellipse, 181 of hyperbola, 194 of parabola, 172
