Prove that : $\sin 2 x + 2 \sin 4 x + \sin 6 x = 4 \cos^{2} x \sin 4 x$

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Solution

As we have $\sin 2 x + 2 \sin 4 x + \sin 6 x = 4 \cos^{2} x \sin 4 x$
$L H S = \sin 2 x + 2 \sin 4 x + \sin 6 x = (\sin 6 x + \sin 2 x) + 2 \sin 4 x \dots (i) W e k n o w \sin A + \sin B = 2 \sin \frac{(A + B)}{2} \cos \frac{(A - B)}{2} (i) b e c o m e s 2 \sin 4 x \cos 2 x + 2 \sin 4 x = 2 \sin 4 x (\cos 2 x + 1) = 2 \sin 4 x (2 \cos^{2} x - 1 + 1) = 2 \sin 4 x (2 \cos^{2} x) = 4 \cos^{2} x \sin 4 x = R H S$
Hence proved.

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