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130. Given sin (A + B) = sin A cos B + cos A sin B......... (1), sin (A B) = sin A cos B.

cos A sin B

... (2) In (1) write A + B = D, so that A = D B. Then we have sin D= sin (D - B) cos B + cos (D- B) sin B.

But we have shown in the last article that from (1) it follows that

cos (D - B)= cos D cos B + sin D sin B. Substituting for cos (D B) in above, sin D=sin (D- B) cos B + cos D cos B sin B + sin D sino B, .. sin D (1 – sino B) = sin (D - B) cos B + cos D cos B sin B, :. sin D cosa B cos D cos B sin B = sin (D B) cos B, .. sin D cos B - cos D sin B = sin (D-B). Writing A for D we have (2). 131. Given

sin (A + B) = sin A cos B + cos A sin B............ .(1), and

sin (A B) = sin A cos B – cos A sin B.... :(2),

sin A +sin B = 2 sin | (A + B) cos } (A - B) ......(5), and sin A - sin B= 2 cos 1 (A + B) sin } (A B) ......(6).

(1) + (2) gives sin (A + B) + sin (A - B) = 2 sin A cos B,

(1)-(2) gives sin (A + B) – sin (4 B)= 2 cos A sin B.
Put

A + B=S and A B=T,
.: (adding) 2A = S+T, (subtracting) 2B=S-T,

.: sin $ + sin T = 2 sin } (S + T') cos } (S T'), and sin S-sin T' = 2 cos } (S + T') sin } (S - T').

Writing A for S and B for T we have (5) and (6).

Similarly Given cos (A + B) = cos A cos B sin A sin B...... (3), and cos (A B) = cos A cos B + sin A sin B.... ...(4), we can prove

cos B + cos A = 2 cos 1 (A B) cos 1 (A B) ..........(7), and cos B - cos A = 2 sin } (A + B) sin } (A B) .........(8). J. T.

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to prove

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132. The signs in the formulæ (1)–(8) may be connected with the fact that the sine increases but the cosine decreases as the angle increases.

133. In sin (A + B) = sin A cos B+cos A sin B ......... (1), and cos (A + B)=cos A cos B – sin A sin B ...(3),

tan A + tan B and tan (A + B)

.(21). 1-tan A tan B

Put B= A.

(9),

.(10),

Thus sin (A + A) = sin A cos A + cos A sin A,

i.e. sin 2A = 2 sin A cos A
cos (A + A) = cos A cos A – sin A sin A,
i.e. cos 2A = cosA – sin’A

tan A + tan A
tan (A +A)

1-tan A tan A'

2 tan A i.e. tan 2A

1- tan’A These results are very important.

... (17).

134. To express the sum or difference of two sines or cosines as a product.

This is done in the general case in Arts. 124 and 125. But the following method is recommended as an exercise in the four important fundamental formulæ of Arts. 120, 121, 122, 123.

Express the two given angles as the sum and difference, respectively, of two new angles ; either by inspection or by solving an equation. Then apply the formula of Arts. 120-123. Thus

Example 1. sin 74 + sin 34=sin (5A+2A)+sin (54 – 24) =(sin 5A cos 2A +cos 5 A sin 2A)+(sin 5 A cos 2A - cos 5 A sin 2A)

=2 sin 5 A cos 2A. Example 2. cos 5A – cos 11A=cos (8A – 34)- cos (8A +34) =(cos 8A cos 3A +sin 8A sin 3A)-(cos 8A cos 3A - sin 8A sin 3A)

=2 sin 8 A sin 3A.

Example 3. sin A -sin B: assuming A to be the greater. Let

x+y=A and x-y=B. Then

sin (x+y) -- sin (x - y)=2 cos x sin y. But solving for x and y, 2x=A+B and 2y=A B,

.. sin A -sin B=2 cos (A + B) sin (A-B). Example 4. cos A - cos B: assuming B to be the greater. Let

x+y=B and 2-y=A. Then

cos (x -- y) - cos (x+y)=2 sin x sin y. But solving for x and y, 2x=B+A and 2y=B-A.

.. cos A - cos B=2 sin } (B+ A) sin }(B-A).

135. To express the product of sines or cosines as a sum or difference.

This is the converse problem to that of the last article. The method is simply to introduce a product containing the ratios complementary to those in the given product. Thus Example 1.

2 cos 4A sin 3A =(sin 4A cos 3A +cos 4A sin 3A) – (sin 4A cos 3A - cos 4A sin 3A)

=sin 7A - sin A.

Example 2.

2 cos 7 A cos 2 A =(cos 7A cos 2A +sin 7A sin 2A)+(cos 7A cos 2A – sin 7A sin 2A)

=cos 54 +cos 9 A.

EXAMPLES V.

1. Show that cos (A - B) + sin (A + B) = (cos A + sin A) (cos B + sin B) cos (A + B) + sin (A – B) = (cos A + sin A) (cos B - sin B) cos (A B) – sin (A + B) = (cos A – sin A) (cos B - sin B) cos (A + B) – sin (4 B) = (cos A - sin A) (cos B+ sin B). 2. Find the sine and cosine of 15° from those of 60° and 45°.

3.

Find the sine and cosine of 75° from those of 45° and 30°.

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5. If sin A = cos B = , find sin (A + B) and cos (A + B).
6. If
sin A = 45 and sin B=33, find sin (A B) and cos (A – B).
7. Find tan (A + 45°) and cot (45o – A).
8. Show that

sin A + cos A= 12. sin (A + 45°)= /2 cos (45° – A).
9. Show that

cos A – sin A = 1/2. sin (45o – A) = 12 cos (45° + A). 10. Find sin 30° from the identity cos 30o = sin 60°.

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Prove the following identities :
11. (sin A + cos A)? = 1 + sin 2A.
12. (cos A + sin A) (cos A – sin A) = cos 2A.
13.
cos 2 A=2 cos' A -1=1 – 2 sin’A.
1- cos 2 A

1 + cos 2A
14. tan’A

15.

cot A = 1 + cos 2 A

sin 2A

16. 1 + sec 2 A = 2 cos”A . sec 2A.

Put the following expressions (17—28) in the form of twice the product of two ratios :

sin 13 A - sin 9 A.

17. sin 11 A + sin 7A. 18.
19.
cos 3A + cos 5A.

20.
21. sin (n + 1) A + sin(n-1) A.
23. sin 3A - sin 2A.

24. 25. sin 0 + cos p.

26. 27. cos (n − 3) A - cos (n + 3) A.

cos 7A

- cos 17A.

22. cos 2 A + sin 2 A. cOS 2A

cos 3A. COS $ - sin 0.

28. sin 2A - Cos 2B.

29.

33.

50 2 cos

COS A

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Put the following expressions (29—36) in the form of the sum or difference of two ratios : 2 sin A cos B.

30. 2 sin A sin B.
31. 2 sin 10° cos 50°. 32. 2 sin 50° cos 10°.
30

7B 5B
sin

34. 2 sin sin
2 2

2 2
35. 2 sin (a + b). cos (a - b). 36. 2 cos 7° sin 11°.
Prove the following identities (37—58) :

sin A - sin B A - B
37.

tan cos A + cos B

2 sin A + sin B A - B 38.

cot

2 sin A + sin B A + B 39.

tan cOS A + cos B

2 sin A sin B A + B 40.

cot

2
sin A + sin B tan! (A + B)
41.

sin A - sin B tan 1 (A B)

cos A - cos B tan (A B) 42.

COS A + cos B cot } (A + B) 43. sin 43° + sin 17° 44. cos 81° + cos 39° 45. sin 73° – sin 47° - sin 13°. 46. cos 11° – cos 49° = sin 19°. 47. cos 25° – sin 5° = cos 35o.

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COS A

cos B

= cos 13°.
= cos 21°.

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tan sin 20 - sin 0 2 sin 0 + sin Ø

= tan; (+) cot }(0-0). sin 0 – sind

51.

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