Prove this double angle identity?

Question

How would I prove the following double angle identity?

$$\frac{\sin2A - \sin A}{\cos2A + \sin A}=\tan \frac{3A}{2} . \cot\frac{A}{2}$$

Sadly I am stuck.

This is similar to one of the tangent half-angle formulas: $\dfrac{\sin A+\sin B}{\cos A+\cos B}=\tan\dfrac{A+B}{2}$.
@Mob The half-angle identity $\tan(\frac{\theta}{2}) = \frac{\sin(\theta)}{1+\cos(\theta)}$ and the similar identity for $\cot(\frac{\theta}{2})$ might be useful on the right-hand side.

Ma Ming · Answer 1 · 2013-05-05 23:53:07Z

up vote 0 down vote

This is simply wrong, if there is no typo. Take $A=\frac{\pi}{3}$, $LHS=0\neq RHS=\infty$.

answered May 5 at 23:53

Ma Ming
2,631315

chenbai · Answer 2 · 2013-05-06 01:15:47Z

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RHS=$\dfrac{sin2A+sinA}{sin2A-sinA}$, so only some $A$ will let LHS=RHS. but it is 6 degree equation.are you sure you have right information?

answered May 6 at 1:15

chenbai
86325

2 Answers