Prove the following identities
1) (sin A + sin 3A + sin 5A) / cos 3A = tan3A (2 cos 2A + 1)
2) sin² 5X - sin² 3X = sin 8X sin 2X
3) cos 8A sin 2A + sin 5A cos A = sin 7A cos 3A
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Answer:
1)
LHS = ( sin A + sin 3A + sin 5A ) / cos 3A
= (sin 3A + 2 sin 3A cos 2A ) / cos 3A => sin A + sin 5A = 2 sin 3A cos 2A
= sin 3A (1 + 2 cos 2A) / cos 3A
= tan3A(2cos2A + 1)
= RHS (shown)
2)
LHS = sin² 5x - sin² 3x
= ½ (1 - cos 10x) - ½(1 - cos 6x) => because cos 10x = 1 - 2 sin² 5x and cos 6x = 1 - 2 sin² 3x
= ½(cos 6x - cos 10x)
= ½( -2 sin (8x) sin (-2x) )
= sin 8x sin 2x
= RHS (shown)
3)
LHS = cos8A sin2A + sin5A cosA
= ½ (2 cos 8A sin 2A) + ½ (2 sin 5A cos A)
= ½ (sin 10A - sin 6A) + ½(sin 6A + sin 4A)
= ½ (sin 10A + sin 4A)
= ½ (2 sin 7A cos 3A)
= sin 7A cos 3A
= RHS (shown)
