Question

Prove that sin 2x+2sin 4x+sin6x=4cos2 x sin4x.

Answer 1

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= sin 2x+2sin 4x+sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

$=[2\,sin(\frac{2x+6x}{2})cos(\frac{2x-6x}{2})]+2sin\,4x$

$[∵sin\,A+sin\,B=2\,sin(\frac{A+B}{2})cos(\frac{A-B}{2})]$

= 2 sin 4x cos (-2x) + 2 sin 4x

= 2 sin 4x cos 2x+2 sin 4x

= 2 sin 4x (cos 2x+1)

= 2 sin 4x (2 cos2 x-1+1)

= 2 sin 4x (2 cos2 x)

= 4cos2 x sin 4x

= R.H.S.