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Prove the identity (cos 3x+cos x)/(sin3x-sinx)=cotx
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\[\cos x+\cos y=2\cos \frac{x+y}{2}\cos \frac{x-y}{2}\]
So cos 3x+cosx=2cos2xcosx Also: \[\sin x+\sin y=2\cos \frac{x+y}{2}\cos \frac{x-y}{2}\]
So sin3x-sinx = 2cos2xsinx
So \[\frac{\cos 3x+\cos x}{\sin 3x-\sin x}=\frac{2\cos 2x \cos x}{2\cos 2x \sin x}=\frac{\cos x}{\sin x}=\cot x\]
\[\sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{x-y}{2} ****\]
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