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161. It is important that the student should be thoroughly familiar with the second set of formula on p. 126.

Written as follows, they may be regarded as the inverse of the 'S, T'formulæ.

2 sin A.cos B = sin (A + B) + sin (A B),
2 cos A. sin B sin (A + B) – sin (4 B),

iv.
2 cos A.cos B = cos (A + B) +cos (4 B),
2 sin A. sin B = cos (A - B) - cos (A + B)..

EXAMPLES. XXXVIII,

Express as the sum or as the difference of two trigonometrical ratios the ten following expressions : (1) 2 sin 8. cos 0.

(2) 2 cos a. cos ß. (3) 2 sin 2a.cos 33. (4) 2 cos (a +B).cos(a - b).

30 A (5) 2 sin 38. cos 50. (6) 2 cos

2

COS

· sin

50 30 (7) sin 40. sin 0.

(8) cos

2 2 (9) 2 cos 100. sin 500. (10) cos 45o. sin 150. (11) Simplify 2 cos 20. cos 0 - 2 sin 40. sin 0. 50

30 (12) Simplify sin

sin COS 2

2 2

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COS

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MISCELLANEOUS EXAMPLES. XXXIX.

(1) If tan a=} and tan ß=}, prove that tan (a +B)=1.

(2) If tan a= and tan ß=ł, prove that one of the values of a +ß is +a

.

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COS a

cos 5a (4) Simplify

sin a+sin 5a

sin 5x – sin 3x (5) Simplify

cos 5x + cos 3.0
cos A +cos 3A

cos 2 A
(6) Prove that
cos3A+cos 5 A

cos 4A
sin 3x - sin a sin 3x +sin 2
(7) Simplify

+
cos 3x + cos cos 3x - cos x

(sin 4A -sin 2A) (cos A (8) Simplify

cos 3A) (cos 4A + cos 2A) (sin A + sin 34) (9) Prove that

2 sin 2a. cos a + 2 cos 4a.sin a=sin 5a + sin a. (10) Prove that

cos 2a . cos a -sin 4a.sin a=cos 3a . cos 2a. (11) tan 24. tan 3A . tan 5A=tan 54 - tan 3A – tan 24. (12) Solve

4 sin (0+0).cos (0-0)=31

4 cos (0+). sin (0-0)=1/ (13) Prove that sin A .sin 2A + sin 2A. sin 5 A+sin 3A.sin 10A

- tan 74. cos A. sin 2A +sin 2A.cos 5 A - cos 3A.sin 10A

A+B A-B 2 sin B (14) tan - tan 2

2 cos A +cos B

CHAPTER XII.

ON THE TRIGONOMETRICAL RATIOS OF MULTIPLE

ANGLES.

162. To express the Trigonometrical Ratios of the angle 2A in terms of those of the angle A. Since sin (A + B)=sin A.cos B + cos A. sin B; .. sin (A + A)=sin A.cos A + cos A. sin A ; .: sin 24 = 2 sin A.cos A

(1). Also, since cos (A + B)= cos A.cos B – sin A . sin B;

.. cos (A + A) = cos A.cos A – sin A. sin A ;
.. cos 2A = cos? A – sino A

(2). But

1=cos A + sin A ; .: 1 + cos 2A = 2 cos* A,

and 1- cos 2A = 2 sino A. The last two results are usually written

cos 2A = 2 cos* A -1 ........ (3), and cos 2A 1-2 sino A

(4).

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163.* * To prove the “2A' formulo geometrically.

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M'

M

R Let ROP be the angle 2A. With centre 0 and any radius describe the semicircle RPL. Draw PM perpendicular to OR. Join RP. PL.

Then the angle RPL in a semicircle is a right angle. The angle ROP=OLP+OPL=2 OLP [since OL=0P]. :: OLP= a half of ROP=A. Also MPR and OLP are each the complement of MPL. .: MPR=OLP=A. Hence

MP 2MP 2MP 2MP. PR
(1) sin 2A=

OP 20R LR PR. LR
=2 cos MPR, sin PLR=2 sin A, cos A.

OM LM-LO 2LM LO
(3) cos 2A =

OP OP 20P OP
2LM.LP OP

=2 cos MLP. cos PLR-1 LP. LR OP

=2 cosA -1. (2) Let OM'=OM. Then 20M=M'M=LM - LM'=LM - MR. Hence,

20M LM - MR LM MR
cos 2A=

20P LR LR LR'
LM.LP MR.PR

=cosA - sino A. LP. LR PR. LR

2MP 2MP (5) tan 2A=

20MLM - MR'

2MP LM

2 tan A 2 tan A LM MR MR.MP 1-tanA

1
LM LM MP.LM

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164. These five formule are very important,

sin 2A = 2 sin A. cos A
cos 2A = cos? A - sino A
cos 2A = 2 cos? A-1

1- 2 sino A

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cos 2 A

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(5),

165. The following result is important,
sin 2A 2 sin A. Cos A

=tan A.
1 + cos 2A 2 cos' A

166.

The student must notice that A is any angle, and therefore these formulæ will be true whatever we put for A.

А
Example. Write
A

..(1), 2

A А
COS A=cos2 siu?

(2),

4 instead of A, and we get sin A = 2 sin

COS

and so on.

EXAMPLES. XL.

Prove the following statements : (1) 2 cosec 2A=sec A. cosec A.

cosec? A cosec? A -2

=sec 2 A. 2-seca A (3)

=cos 2A. secA (4) cos? A (1 – tan? A)=cos 2A.

cot? A-1 (5) cot 2A=

2 cot A

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