Class 11MathematicsTrigonometrical Ratios, Functions and IdentitiesTrigonometrical ratios of multiple and sub-multiple angles
MediumProve that $\sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x$.
(N/A) $L.H.S. = \sin 2x + 2 \sin 4x + \sin 6x$
$= (\sin 6x + \sin 2x) + 2 \sin 4x$
$= 2 \sin \left( \frac{6x + 2x}{2} \right) \cos \left( \frac{6x - 2x}{2} \right) + 2 \sin 4x$
$= 2 \sin 4x \cos 2x + 2 \sin 4x$
$= 2 \sin 4x (\cos 2x + 1)$
Using the identity $\cos 2x = 2 \cos^2 x - 1$,we get:
$= 2 \sin 4x (2 \cos^2 x - 1 + 1)$
$= 2 \sin 4x (2 \cos^2 x)$
$= 4 \cos^2 x \sin 4x = R.H.S.$
$= (\sin 6x + \sin 2x) + 2 \sin 4x$
$= 2 \sin \left( \frac{6x + 2x}{2} \right) \cos \left( \frac{6x - 2x}{2} \right) + 2 \sin 4x$
$= 2 \sin 4x \cos 2x + 2 \sin 4x$
$= 2 \sin 4x (\cos 2x + 1)$
Using the identity $\cos 2x = 2 \cos^2 x - 1$,we get:
$= 2 \sin 4x (2 \cos^2 x - 1 + 1)$
$= 2 \sin 4x (2 \cos^2 x)$
$= 4 \cos^2 x \sin 4x = R.H.S.$
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