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Trigonometric Ratios of Multiples of an Angle

Prove that: ...

Question

Prove that:
$(s i n 3 x + s i n x) s i n x + (c o s 3 x - c o s x) c o s x = 0$

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Solution

$L H S = (s i n 3 x + s i n x) s i n x + (c o s 3 x - c o s x) c o s x$

$= [2 sin (\frac{3 x + x}{2}) cos (\frac{3 x - x}{2})] sin x + [- 2 sin (\frac{3 x + x}{2}) sin (\frac{3 x - x}{2})] cos x$

$\Rightarrow 2 [sin x cos x sin 2 x - sin x cos x sin 2 x] = 0 = R H S$

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