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

An alternative strategy uses $1+\cos x=2\cos^2\tfrac{x}{2}$ to obtain $$\sin a+\sin b+\sin (a+b)=2\left(\sin a\cos^2\tfrac{b}{2}+\sin b\cos^2\tfrac{a}{2}\right).$$To finish, use $\sin x=2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$ to obtain$$\sin a+\sin b+\sin (a+b)=4\cos\tfrac{a}{2}\cos\tfrac{b}{2}\left(\sin\tfrac{a}{2}\cos\tfrac{b}{2}+\sin\tfrac{b}{2}\cos\tfrac{a}{2}\right)=4\cos\tfrac{a}{2} \cos\tfrac{b}{2}\sin\tfrac{a+b}{2}.$$
