Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

Asked by Pragya Singh | 1 year ago |  101

##### Solution :-

We know that, sinA+sinB=2sin$$( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})$$

L.H.S.=sin2x+2sin4x+sin6x

=[sin2x+sin6x]+2sin4x

$$[2sin ( \dfrac{2x+6x}{2})cos( \dfrac{2x-6x}{2})]+2sin4x$$

By further simplification

= 2 sin 4x cos (– 2x) + 2 sin 4x

It can be written as

= 2 sin 4x cos 2x + 2 sin 4x

Taking common terms

= 2 sin 4x (cos 2x + 1)

Using the formula

= 2 sin 4x (2 cos2 x – 1 + 1)

We get

= 2 sin 4x (2 cos2 x)

= 4cos2 x sin 4x

= R.H.S.

Answered by Abhisek | 1 year ago

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