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= m

= n

or

150. Simultaneous Equations. Simultaneous trigonometric equations are solved by the same principles as simultaneous algebraic equations. 1. Required to solve for x and y the system

x sin a ty
sin b

(1) X cos a + y cos b

(2) From (1), x sin a cos a + y sin b cos a = m cos a.

(3) From (2), x sin a cos a + y cos b sin a = n sin a.

(4) From (3) and (4), y sin b cos a y cos b sin a = m cos a – n sin a,

y sin (b a) = m cos a – n sin a;

n sin a whence

sin (b a)

m cos b Similarly,

sin (b a) 2. Required to solve for x and

y

the system sin x + sin y = a

(1) Cos x + cos y = b

(2) By $ 103, 2 sin } (x + y) cos } (x y) = a,

(3) and

2 cos } (x + y) cos } (x y) = b. Dividing, tan } (x + y)

(4)

m cos a

=

n sin 6

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a

(5)

.:. sin } (x + y)

Va? + b2 Substituting the value of sin } (x + y) in (3),

cos } (x y) = } Val + 62. From (4),

x + y = 2 tan-1

20 From (6),

2 y = 2 cos-1} Va? + 62. From (6) and (7),

+ cos-1 } va? + b2 b

(6)

a

c = tan-1

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.: I = tan-1 Adding the squares of (1) and (2),

32 (sinax + cos x) = a 2 + 62. Therefore

y2 = a2 + 62, and

y=+ a + b2.

(1) (2)

= n.

4. Required to solve for x and y the system

y sin (v + a)=m

y cos (a + b)=n From (1), y sin x cos a + y cos sin a = m. From (2),

y cos x cos b y sin x sin b
We may now solve for y sin x and y cos x, and then solve for x and y.
5. Required to solve for r, x, and y the system

r cos x sin y = a
r cos x cos y = b

r sin x = 0

(1) (2) (3)

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sinx

Solve the following systems for x and y: 1. sin x + sin y = sin a

5. sino x + y = m cos x + cos y=1+ cos a

cosé x + y =n 2. sinox + sin’y = a

6. sin x + sin y=1 cos x cos” y = b

sin y=1 3. sin x sin

y
0.7038

7. COS X + cos y = a
0.7245

cos 2 x + cos 2 y b 4. x sin 21° + y cos 44° 179.70 8. sin x + sin y

= 2 m sin a X cos 21° + y sin 44°

COS X + cos y = 2 n cos a 9. Find two angles, x and y, knowing that the sum of their sines is a and the sum of their cosines is b.

COS X

COS Y

= 232.30

Solve the following systems for r and x : 10. r sin x = 92.344

11. r sin (x – 19° 18') = 59.4034 r cos x = 205.309

r cos (x - 30° 54') = 147.9347

151. Additional Symbols and Functions. It is the custom in advanced trigonometry and in higher mathematics to represent angles by the Greek letters, and this custom will be followed in the rest of this work where it seems desirable. The Greek letters most commonly used for this purpose are as follows: Q, alpha

0, theta

1, lambda
7, gamma
o, delta

b, phi
epsilon

W, omega Besides the six trigonometric functions already studied, there are, as mentioned on page 4, two others that were formerly used and that are still occasionally found in books on trigonometry. These two functions are as follows:

B, beta

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versed sine of a = 1 cos a, written versin a; coversed sine of a 1 - sin æ, written coversin a.

Exercise 79. Simultaneous Equations

=X

= a

=

=

=X

=a

1. Solve for $ and x:

4. Solve for 0 and $: versin •

sin 0 + cos 1 – sin $ 0.5

sin $ + cos O b 2. Solve for A and x :

5. Solve for 0 and $: 1 - sin 0

a sin4 0

b sin* $ 1+ sin 0 = a

a cos* 0 b cos* d = b 3. Solve for , and

M:

6. Solve for 0 :
sind =
V2 sin

sino0 + 2 cos 0 = 2
tan1 = V3 tan je

cos 0 - cos”A = 0 152. Eliminant. The equation resulting from the elimination of a certain letter, or of certain letters, between two or more given equations is called the eliminant of the given equations with respect to the letter or letters.

For example, if ox = b and a'ic B', it follows by division that a: a' = 6:6', or that ab' = a'b, and this equality, in which x does not appear, is the eliminant of the given equations with respect to x.

There is no definite rule for discovering the eliminant in trigonometric equations. The study of a few examples and the recalling of identities already considered will assist in the solutions of the problems that arise.

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153. Illustrative Examples. The following examples will serve to illustrate the method of finding the eliminant:

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Since sina $ + cos2 d = 1, we have a? + b2 = 1, the eliminant.

2. Find the eliminant, with respect to , of

secd= m

tan λ = η

Since seca, – tan2 X = 1, we have ma – na = 1, the eliminant.

n sin

COS

3. Find the eliminant, with respect to p, of

m sin u + COS je

1 M

M

1 Writing the equations m sinu =1- cos Hy n sin u =1+ cos y, and multiplying, we have

mn sin?u=1- cosa u Hence

mn = 1 is the eliminant.

sinau.

Exercise 80. Elimination

Find the eliminant with respect to a, e, i, j, or of the follow ing equations :

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CHAPTER XII

APPLICATIONS OF TRIGONOMETRY TO ALGEBRA

154. Extent of Applications. Trigonometry has numerous applications to algebra, particularly in the approximate solutions of equations and in the interpretation of imaginary roots.

These applications, however, are not essential to the study of spherical trigonometry, and hence this chapter may be omitted without interfering with the student's progress.

For example, if we had no better method of solving quadratic equations we could proceed by trigonometry, and in some cases it is even now advantageous to do so. Consider the equation x2 + px q = 0. Here the roots. are 2 tp + } Vp + 49, x=-{p - } Vp2 + 49.

2 Vq
If we let = tan •, or p= 2 Vq coto, we have

P
X
Vq cot $ + Vq Vcoto +1
√q

1
=- Vq cot • +

1

coto

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sin φ
sin •
cos P =Vq tano.
sin •

=VI

Similarly,
Hoy

Vq cot } $.
For example, if x2 + 1.1102 x – 3.3594 = 0 we have

2 V3.3594
tan o =

1.1102 whence

log tan p = 0.51876, and

$ = 73° 9' 2.6". Therefore

log tan} $ = 9.87041 – 10. and

log va = log V3.3594 = 0.26313. Hence

log xz = 0.13354, and

Xy = 1.360.

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