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Trigonometric Ratios of Multiples of an Angle

Prove that: ...

Question

Prove that:
$s i n x + s i n 3 x + s i n 5 x + s i n 7 x = 4 c o s x c o s 2 x s i n 4 x$

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Solution

$L H S = s i n x + s i n 3 x + s i n 5 x + s i n 7 x = s i n x + s i n 7 x + s i n 3 x + s i n 5 x$
$= 2 sin 4 x cos 3 x + 2 sin 4 x cos x = 2 sin 4 x (cos 3 x + cos x)$
$= 4 s i n 4 x c o s x c o s 2 x$
$= R H S$

Hence proved

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