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2 tan B (0)

=sin 2B. 1 + tano B (7) tan B+cot B=2 cosec 2B.

1- tan’ P (8) 1 + tan2B

=cos 2B. (9) cot B-tan B=2 cot 2B.

cotB+1 (10) cot? B-1

=sec 2B.

0 A\2 (11) sin + cos 2

=l+sin 0.

2

A 012 (12) sin - COS =l-sina.

2

A )

1 + tan =l+sin 0. 2

A (14) sin

cot =1-sin 0. 2

cos 09

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sin ß (16)

tan
1+cos B 2

sin ß (17)

Ecot 1- cos ß

2• 1- cos

B (18)

=tan 1 + cos ß

2 1 + sec ß (19) = 2 cog2B sec ß

2“

B (20) cosec ß – cot B=tan.

1- tan (21) 1 + sin 2x 1+tan

C

cos 2.x

1 + tan COS (22) 1-sins

1-tan

Olol

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sin 3B

(34)

cos 3B (32)

+ =2 cot 23. sin ß

cos B
sin 43
(33)

2 cos 2ß.
sin 2B
sin 5B
cos 58

4 cos 2B.
sin ß

cos ß
57 57
sin COS
12 12

=23.
TT

TT
sin COS
12

12 (36) tan (45° + A) – tan (450 – A)=2 tan 2A. (37) tan (45o – A) +cot (45o – A)=2 sec 2A.

(35)

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NOTE. The similarity of these two results is apt to cause confusion. This may be avoided by observing that the second formula must be true when A=0°; and then cos 3A=cos 0o=1. In which case the formula gives cos 0o=4 cos 0° – 3 cos Oo, or l=4-3, which is true.

The first formula may be proved thus :
sin 34=sin (2A + A)=sin 2A.cos A + cos 2A. sin A

=(2 sin A.cos A) cos A+(1 – 2 sin? A) sin A

= 2 sin A .cos? A +sin A - 2 sind A
=2 sin A (1 – sin? A)+sin A - 2 sin3 A
=2 sin A - 2 sin3 A+sin 4 - 2 sin3 A

=3 sin A - 4 sin3 A.

The second formula may be proved in a similar manner.

3 tan A - tan3 A Example. Prove that tan 3A=

1-3 tan’ A
sin 34 3 sin A-4 sinA
tan 3A

cos 3A 4 cos3 A - 3 cos A
3 sin A (sino A + cos2 A) – 4 sin3 A
4 cos3 A-3 cos A (sin? A + cos2 A)
3 sin A.cos? A -sin3 A
cos3 A - 3 cos A. sinA
3 sin A. cos? A sinA
coS3 A

cos? A

3 tan A - tanA cOS A

3 cos A. sino A 1-3 tan’ A cos3 A cos3 A

EXAMPLES. XLI.

(3)

Prove the following statements :

sin 3A (1) =2 cos 2A +1.

sin A

cos 3A (2)

2 cos 2A - 1.
COS A
3 sin A - sin 3A

=tan? A.
cos 34+3 cos A

cot3 A - 3 cot A (4) cot 3A

3 cota A-1 sin 3A - sin A (5)

tan A. cos 3A +cos

sin 3A -cos 3A (6)

2 sin 2A -1. sin A +cos A

sin 3A +cos 3A (7)

=2 sin 2A +1.
cos A - sin A
1

1 (8)

+

=cot 2A.
tan 3A –tan A cot A -cot 3A
(3 sin A -sin 3A\2

sec 2A-113 (9)

3 cos A +cos 3A

1- cos 3A (10) 1-cos A

=(1+2 cos A).

** MISCELLANEOUS EXAMPLES. XLII. Prove the following statements :

sin A +cos A (1)

=tan 2A +sec 2A. cos A -sin A

A tan

2 (2)

=tan A + sec A.

A 1-tan

2 (3) sin (n+1)a.cos (n-1)a- sin 2a=sin(n-1)a.cos (n + l)a

sin a+sin B a+ß (4)

=tan
cos a + cosß 2
cos 2a +cos 12a cos 7a - cos 3a

sin 4a (5)

+

+ 2

-0. cos 6a + cos 8a cos a - cos 3a

sin 2a (6) If A=18', prove that sin 2A=cos 34 ; hence prove that

15-1 sin 180

4 sin (7)

cot

+)
(8) sin 2A.sin 2B=sin? (A + B) - sin(4 B).
(9) COS 4A=8 cos4 A - 8 cos? A +1.
(10) tan 500 + cot 50o=2 sec 100.
(11) sin 32=4 sin A. sin (60° + A) sin (60° – A).

А
cot - tan

2
cos3a - sinß.sin 5a - cos 7a
(13)
sin 3a + sinß.cos 5a - sin 7a

is independent of B. (14) (cos x+cos y)2 + (sin x+sin y)2=4 cos?

2 (15) 2 cosA. cosa B+2 sino A. sino B=l+cos 2A.cos 2B. (16) cot 5 – tan = 2.

4 tan 0 (1 - tana 0) (17) tan 40=

1-6 tano 0 +tan0'

+ )

a

A) *cot A – 2 cot 24)= 4 cot 4.

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