Prove the following:
$s i n 2 x + 2 s i n 4 x + s i n 6 x = 4 c o s^{2} x s i n 4 x$

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Solution

$L H S = s i n 2 x + 2 s i n 4 x + s i n 6 x = s i n 2 x + s i n 6 x + 2 s i n 4 x$
$= 2 sin (\frac{2 x + 6 x}{2}) cos (\frac{2 x - 6 x}{2}) + 2 s i n 4 x = 2 s i n 4 x [1 + cos 2 x] = 4 c o s^{2} x s i n 4 x = R H S$
Hence proved

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