Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that 
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$ 
We also know that 
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have 
$$\sin a + \sin b = 2 \sin\frac12(a+b)\cos\frac12(a-b)$$
From there, we seem to be missing how to get to the right-hand side of the equation.
We first expand 
$$\sin a + \sin b = 2 \sin\frac12(a+b) \cos\frac12(a-b)$$
Then we add $\sin(a+b)$, which is $\sin a \cos b + \cos a \sin b$.
We now have:
$$2 \sin\frac12(a+b) \cos\frac12(a-b) + \frac12 \left(\sin(a-b) + \sin (a+b)\right) + \frac12 \left( \sin(b-a) + \sin (a+b)\right)$$
From there, we can't see how to obtain the right-hand side of the equation which is 
$$4 \sin\frac12(a+b) \cos\frac12a \cos\frac12b$$ 

It may help setting $A=a/2$ and $B=b/2$, so the left-hand side becomes
$$
\sin2A+\sin2B+\sin2(A+B)
$$
The right-hand side screams “sum-to-product”! OK, let's apply the formula
$$
\sin2A+\sin2B=2\sin(A+B)\cos(A-B)
$$
so the left-hand side becomes
\begin{align}
\sin2A+\sin2B+\sin2(A+B)
&=2\sin(A+B)\cos(A-B)+2\sin(A+B)\cos(A+B)\\
&=2\sin(A+B)\bigl(\cos(A-B)+\cos(A+B)\bigr) \\
&=2\sin(A+B)(\cos A\cos B+\sin A\sin B+\cos A\cos B-\sin A\sin B)\\
&=4\sin(A+B)\cos A\cos B
\end{align}
