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 
$$4 \sin\frac12(a+b) \cos\frac12a \cos\frac12b$$ 

Use that $$\sin(A)+\sin(B)=2\cos\left(\frac{A-B}{2}\right)\sin\left(\frac{A+B}{2}\right)$$ and $$\sin(A+B)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$
