Prove that sin x - sin 3xsin2 x - cos2 x - 2 sin x

sin x sin 3x sin^ 2 x cos^ 2 x ⟹sin x 3sin x+4sin^ 3 x 2sin^ 2 x 1 2sin x(2sin^ 2 x 1) 2sin #160; #160;^ 2 x 1 =2sin x (Proved) ...

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Basic Trigonometric Identities

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

\frac{{cos}^{3} x - {sin}^{3} x}{cos x - sin x} = \frac{1}{2} (2 + sin 2 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

\sin 3 x - \sin x = 4 \cos^{2} x - 2

\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}

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