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The given expression will then become, by substitution,

k cos ε sin + k sin a cos 0 = k sin (0 + 8). We might equally have supposed k sin ε = a and k cos ε = :b, when the given expression would have become k cos (0 £). Example. Reduce the expression

1239.3 sin a - 724.6 cos a to a monomial. k sin E = 724.6

log

2.860 10 k cos E = 1239.3 log = 3.093 18

log tan a, 9.766 92

?, – 30° 18'.8 log cosa,

9.936 15 log k, 3.157 03

k, 1435.6 We therefore have

1239.3 sin x — 724.6 cos x= 1435.6 sin (x – 30° 18'.8).

EXERCISES.

Reduce to monomials the expressions :
1. 27.615 cos u — 23.208 sin u.
2. 3.600 3 sin (0 7° 52'.6) + 5.9070 sin (0 + 53° 57'.6).
3. Reduce to a monomial the expression

92.65 sin 0 cos a 196.23 cos O sin a,
when a = 162° 48'.7.
4. Reduce to a monomial the expression

cos a cos +sin a sin cos H, when a = 62° 39'.5 and H= 22° 36'.8.

52. PROBLEM V. To reduce an expression of the form
a sin x cos 0 + b cos x sin O

(1) to a form which shall not contain the product of any two trigonometric functions. Solution. We have, from $ 43,

sin x cos 6 = 4 sin (x + 8) + sin(x – 0);
cos x sin 0 = 4 sin (x + 0) – $sin (x – 6).

Making these substitutions in the expression (1), it becomes

fla + b) sin (2+0) + a b) sin (x - 0)

EXERCISES.

Clear from products of sines and cosines :

1. m cos a sin ß n sin a cos ß.
2. a cos O cos u + b sin o sin M.
3. a cos O cos u b sin sin u.

=

= C,

a

COS λ

53. PROBLEM VI. From the equations

r cos ß cos a,
r cos ß sin =b,

(1) sin ß to find the values of r, B, and 1, the values of a, b, and c being given.

Method of solution. Dividing the second equation by the first we obtain tan λ=

(2) from this equation we find 1, and then sin 1 or cos 1 from the tables. Then

b re cos B

(3)

sin 1 can be computed. The value of r sin ß being given by the third equation (1), the values of r and ß are found by Problem I. Example. Find r, B, and 1 from the equations

r cos ß cos =- .53.953;
rcos B sin 1= + 197.207;
r sin ß

39.062. Work

log cos B, 9.992 21 log r cos ß cos 2, - 1.732 01 log r cos , 2.310 60 log r cos ß sin 1, 2.294 93

log r sin B, 1.591 75 sin , 9.984 33

log tan ß, -9.281 15 log tan 1, – 0.562 92

2.318 39 1, 105° 18'.0

1,

208.16 log r cos ß, 2.310 60

B, 10° 49'

log r,

EXERCISES.

Find the values of r, B, and 1 from: 1. r cos ß cos 1 =

1.271 83; r cos ß sin i = – 0.981 52; r sin ß

0.890 02. 2. p sin ß sin 1 = 19.765 3; η sin β coς λ=

7.1928; po cos ß

12.124 2.

MISCELLANEOUS EXERCISES. 1. Compute w and r from the equations

1.268 22 sin w = 0.948 30 + r sin (25° 27'.2);

1.268 22 cos w = 0.281 16+r cos (25° 27'.2). First eliminate r.

Ans. w= 60° 53'. 8;

p= 0.371 7. 2. Find w and x from the equations

3 sin w+ cos w = 2.c. Ans. W = 71° 34';
sin w+ 2 cos w = *.

V.
After finding sin w and cos w in terms of x employ the equation

sin? w + cos' w = 1. Find « from the following quadratic equations, and express the results without surds:

x =

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2x 3. X* +1

Ans. x=cot ja or tan } a. sin a

1 sin a 4. **+1= 22 sec a.

Ans. 2 =

COS a. 5. 1-3 = 2xcot a. Ans. x = tan fa or - cot fa.

sin a +1
6. 2 – 1= 2 x tan a. Ans. x =

COS a.
Find 0 from the equations:
7. 27.615 cos 0 - 23.208 sin 0 = 19.094.
8. 3.6003 sin (0 – 8°) + 5.907 sin (0 + 54°) = 2.6253.
Reduce the first members by Prob. IV.
9. a sin 0 +b cos 0+c=0.

From this last equation find sin 0, cos 0, and tan O by separate quadratic equations.

CHAPTER VI.

SOLUTION OF TRIANGLES IN GENERAL.

sin a =

54. A plane triangle has six parts, three sides and three angles. Of these parts the three angles are not independent of each other, because when two angles are given the third may be found from the condition that the sum of the three angles is 180°. Hence, if two angles are given, the case is the same as if all three were given.

When any three independent parts are given the remaining three may be found, but in order to be independent one of the three given parts must be a side.

55. The fact that the sum of the three angles of a plane triangle is 180° enables us to express a trigonometric function of any one angle as a similar function of the sum of the other two angles. It has been shown that

sin (180° — x);

cos (180° — x); tan X =- tan (180° — 2); cot x = – cot (180° — x);

sec (180° — x);

cosec (180°— x).
If a, b, and y are the three angles of a triangle, we have

a= 180o (B +v);
8 = 180° - (x + a);

(a) y = 180° — (a + B). Therefore

sin (B + v);
sin ß =
sin (y + a);

(1) sin (a + b);

cos (B+y); etc.

COS X =

sec =

cosec x =

sin a =

sin y =

etc.

COS A =

(6)

By dividing the equations (a) by 2 we find

fa = 90° – }(+»);)

B = 90° – 4 (v + a);

y = 90° – +(a+b). Therefore

sin f a = cos } (B +v);
cos fa = sin 3 (B + r);
tan fa = cot} (B +v);
cot fa = tan (B + y);

etc. etc.

(2)

From what has been said the given parts may be

One side and the angles ;
Two sides and one angle;

The three sides.
Also, when the sides and one angle are given, this angle may be
either that between the given sides or 'opposite one of them.
Hence there are four cases in all to be considered.

a

56. CASE I. Given the angles and one side.

THEOREM I. The sides of a plane triangle are proportional to the sines of their opposite angles. Proof. Put

a, b, c, the three sides;

a, b, y, their opposite angles. From one angle, as y, drop a perpendicular y D upon the opposite side, c. Then

yD= b sin a; y D= a sin ß. ($35) Therefore

a sin B = b sin a. Dividing by sin a sin ß,

7 sin

8 By dropping a perpendicular from a upon a we should find, in the same way,

7

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D

a

sin a

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