Prove that (sin θ – 2 sin3 θ) / (2 cos3 θ – cos θ) = tan θ

Question

Prove that $\frac {\sin 	heta – 2 \sin^3 	heta}{2 \cos ^3 	heta – \cos 	heta} = 	an 	heta$

Sapling Academy · Accepted Answer

Let's start from LHS: LHS = $\frac {\sin 	heta – 2 \sin^3 	heta}{2 \cos ^3 	heta – \cos 	heta}$ = $\frac {\sin 	heta (1 – 2 \sin^2 	heta)}{\cos 	heta (2\cos ^2 	heta – 1)}$ We know that sin2 θ + cos2 θ = 1 By substituting this value in the above equation, we get: LHS = $\frac {\sin 	heta ((\sin^2 	heta + \cos^2 	heta) – 2 \sin^2 	heta)}{\cos 	heta (2\cos ^2 	heta – (\sin^2 	heta + \cos^2 	heta))}$ = $\frac {\sin 	heta (\sin^2 	heta + \cos^2 	heta – 2 \sin^2 	heta)}{\cos 	heta (2\cos ^2 	heta – \sin^2 	heta - \cos^2 	heta)}$ = $\frac {\sin 	heta (\cancel{\sin^2 	heta} + \cos^2 	heta – \cancel{2} \sin^2 	heta)}{\cos 	heta (\cancel{2}\cos ^2 	heta – \sin^2 	heta - \cancel{\cos^2 	heta)}}$ = $\frac {\sin 	heta (\cos^2 	heta – \sin^2 	heta)}{\cos 	heta (\cos ^2 	heta – \sin^2 	heta)}$ = $\frac {\sin 	heta \cancel{(\cos^2 	heta – \sin^2 	heta)}}{\cos 	heta \cancel{(\cos ^2 	heta – \sin^2 	heta)}}$ = $\frac {\sin 	heta}{\cos 	heta}$ = $	an 	heta$ Hence Proved ! Please do press “Heart” button if you liked the solution.

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