Prove that sin x - sin 3x-sin2x - cos2x - 2sin x

It is known that sin A sin B = 2 cos(A + B2) sin(A B2) , cos^2 A sin^2 A = cos 2A ∴ L.H.S. = sin x sin 3xsin^2 x cos^2 x = 2 cos(x + 3x ...

Prove that ...

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Prove that $\frac{sin x - sin 3 x}{{sin}^{2} x - {cos}^{2} x} = 2 sin x$

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\frac{sin x - sin 3 x}{{sin}^{2} x - {cos}^{2} x} = 2 sin x

Prove that $\frac{sin 3 x - sin x}{cos 2 x} = 2 sin x$

Prove the following:
$\frac{sin x - sin 3x}{{sin}^{2} x - {cos}^{2} x} = 2 sin$ x.

\cos x + \cos 2 x + \cos 3 x = 0

\cos x + \cos 3 x - \cos 2 x = 0

\sin x + \sin 5 x = \sin 3 x

\cos x \cos 2 x \cos 3 x = \frac{1}{4}

\cos x + \sin x = \cos 2 x + \sin 2 x

\sin x + \sin 2 x + \sin 3 = 0

\sin x + \sin 2 x + \sin 3 x + \sin 4 x = 0

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\sin 2 x - \sin 4 x + \sin 6 x = 0

\sin 3 x + \sin 2 x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}