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The given expression will then become, by substitution,

k cos e sin 0+k sine cos 0k sin (0+ ε).

=

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

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

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2. 3.600 3 sin ( — 7° 52′.6) + 5.907 0 sin (0 + 53° 57′.6). 3. Reduce to a monomial the expression

92.65 sin cos a

196.23 cos sin a,

when a 162° 48'.7.

4. Reduce to a monomial the expression

cos a cos + sin a sin O 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 +b cos x sin

(1)

to a form which shall not contain the product of any two trigonometric functions.

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Making these substitutions in the expression (1), it becomes (a+b) sin (x + 0) + (a − b) sin (x — 0)

EXERCISES.

Clear from products of sines and cosines:

1. m cos a sin ẞ n sin a cos B.

2. a cos e cos μμ

3. a cos e cos μ

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to find the values of r, ß, and A, the values of a, b, and c being

given.

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

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from this equation we find λ, and then sin λ or cos λ from the tables.

Then

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cos λ sin

=

(3)

can be computed. The value of r sin 6 being given by the third equation (1), the values of r and B are found by Problem I. Example. Find r, B, and λ from the equations

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MISCELLANEOUS EXERCISES.

1. Compute ∞ and r from the equations

1.268 22 sin ∞ = 0.948 30+ r sin (25° 27′.2);
1.268 22 cos ∞ = 0.281 16+ cos (25° 27′.2).

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After finding sin ∞ and cos ∞ in terms of a employ the equation

sin2 + cos2 ∞ = 1.

Find a from the following quadratic equations, and express the results without surds:

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7. 27.615 cos — 23.208 sin = 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 b cos 0+c=0.

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

CHAPTER VI.

SOLUTION OF TRIANGLES IN GENERAL.

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

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If a, B, and y are the three angles of a triangle, we have

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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.

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=bsin a; yD= a sin ß. (835) Therefore

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By dropping a perpendicular from a upon a we should find, in

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