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Sum and difference of sines and cosines.

sin. Asin. B= 2 sin. 3 (A + B) cos. (A — B) (43)

sin. A sin. B=2 cos. (A+B) sin. (AB) (44) 1 1⁄2

cos. A cos. B = 2 cos. 1⁄2 (A + B) cos. ¿ (A — B) (45) cos. B―cos. A = 2 sin. § (A+B) sin. § (A — B). (46) 47. Corollary. The quotient, obtained by dividing (43) by (44), is

sin. Asin. B

sin. (A+B) cos. (AB) sin. A - sin. B cos. (A+B) sin. § (A — B)

=

Reducing the second member by means of equations (6), (7), (8), we have

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48. Corollary. The quotient of (46) divided by (43) is,

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49. Corollary. Putting in (33) and (35), M and N each equal to A, we obtain

sin. 2 Asin. A cos. A+sin. A cos. A-2 sin. A cos. A (49)

cos. 2 A cos. A cos. A sin. A sin. A

=

(cos. A)2- (sin. A)2.

(50)

Sine &c. of double, and half of an angle.

50. Corollary. The sum of (50), and of the following equation, which is the same as (9),

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52. Problem. To find the tangent of the sum and of the difference of two angles.

Solution. First. To find the tangent of the sum of two angles, which we will suppose to be M and N, we have from

(7),

sin. (MN)

tang. (M+N) =

cos. (MN)

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Divide every term of both numerator and denominator of the second member by cos. M cos. N;

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which, reduced by means of (7), becomes

tang. Mtang. N

tang. (MN) =

(60)

1-tang. Mtang. N

Secondly. To find the tangent of the difference of M and

N. Since by (7)

sin. (M-N)

tang. (MN) =

cos. (M― N)'

a bare inspection of (37) and (38) shows that we have only to change the signs, which connect the terms in the value of tang. (MN) to obtain that of tang. (MN). This change, being made in (60), produces

tang. (MN) =

tang. Mtang. N
1+tang. Mtang. N

(61)

Tangent and cotangent of double an angle.

53. Corollary. As the contangent is merely the reciprocal of the tangent, we have, by inverting the fractions, from (60) and (61),

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54. Corollary. Make MN= A, in (60) and (62).

They become

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Sine, &c. of 0° and 90°.

CHAPTER V.

VALUES OF THE SINES, COSINES, TANGENTS, COTANGENTS, SECANTS, AND COSECANTS OF CERTAIN ANGLES.

55. Problem. To find the sine, &c. of 0° and 90°.

Solution. It is evident, from § 22, that the sine and tangent of zero are, each of them, zero, while its cosine is unity. Hence, and from the consideration that 0° and 90° are complements of each other, we have

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