This one really stumped me =[ Does anyone know how to verify it?
$\displaystyle \sin \frac{\alpha}{3} \cos \frac{\alpha}{3} = \frac{1}{2} \sin \frac{2\alpha}{3}$
$\displaystyle
\begin{aligned}
LHS &= \sin \frac{\alpha}{3} \cos \frac{\alpha}{3} \\
&= \frac{1}{2} \left(2\sin \frac{\alpha}{3} \cos \frac{\alpha}{3} \right) \\
\end{aligned}
$
Using $\displaystyle \sin 2x = 2 \sin x \cos x $
$\displaystyle = \frac{1}{2} \sin \frac{2\alpha}{3}$
$\displaystyle = RHS$