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.
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