1. Trig Identity

Verify:
Sin(α/3)Cos(α/3)=.5Sin(2α/3)

2. Use the sine double-angle identity.
\begin{aligned}
\tfrac{1}{2}\sin \left(\tfrac{2\alpha}{3} \right) &= \tfrac{1}{2}\sin \left(2 \cdot \tfrac{\alpha}{3} \right) \\
&= \tfrac{1}{2} \cdot 2 \sin \left(\tfrac{\alpha}{3} \right) \cos \left(\tfrac{\alpha}{3} \right) \\
&= \sin \left(\tfrac{\alpha}{3} \right) \cos \left(\tfrac{\alpha}{3} \right)
\end{aligned}

3. Originally Posted by eumyang
Use the sine double-angle identity.
\begin{aligned}
\tfrac{1}{2}\sin \left(\tfrac{2\alpha}{3} \right) &= \tfrac{1}{2}\sin \left(2 \cdot \tfrac{\alpha}{3} \right) \\
&= \tfrac{1}{2} \cdot 2 \sin \left(\tfrac{\alpha}{3} \right) \cos \left(\tfrac{\alpha}{3} \right) \\
&= \sin \left(\tfrac{\alpha}{3} \right) \cos \left(\tfrac{\alpha}{3} \right)
\end{aligned}
In the second step how did you get Cos to appear?

4. Like I said, I used the sine double angle identity:
$\sin\,2\theta = 2\sin\,\theta\,\cos\,\theta$.
You should have learned this. If not, you could still verify the problem using the sine of a sum identity instead:
$\sin\,(u + v) = \sin\,u\,\cos\,v + \cos\,u\,\sin\,v$.