<s>[INST] 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 
$$4 \sin\frac12(a+b) \cos\frac12a \cos\frac12b$$ 
 [/INST] Following Thomas Shelby and Dr. Sonnhard Graubner, I provide you a full proof:
\begin{align}\sin(a)+\sin(b)+\sin(a+b)&=2\cos\left(\frac{a-b}{2}\right)\sin\left(\frac{a+b}{2}\right)+2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a+b}{2}\right)\\&=2\sin\left(\frac{a+b}{2}\right)\left[\cos\left(\frac{a-b}{2}\right)+\cos\left(\frac{a+b}{2}\right)\right]\end{align}
Since 
\begin{align}\cos\left(a-b\right)+\cos\left(a+b\right)=2\cos(a)\cos(b)\end{align}
[see: last page of https://services.math.duke.edu/~leili/teaching/uwm/math222s11/problems/quizzes/trig.pdf or search for An elementary proof of two formulas in trigonometry on google] then,
\begin{align}\cos\left(\frac{a-b}{2}\right)+\cos\left(\frac{a+b}{2}\right)=2\cos\left(\frac{a}{2}\right)\cos\left(\frac{b}{2}\right)\end{align}
Hence, 
\begin{align}2\sin\left(\frac{a+b}{2}\right)\left[\cos\left(\frac{a-b}{2}\right)+\cos\left(\frac{a+b}{2}\right)\right]=&2\sin\left(\frac{a+b}{2}\right)\times 2\cos\left(\frac{a}{2}\right)\cos\left(\frac{b}{2}\right)\\=&4\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a}{2}\right)\cos\left(\frac{b}{2}\right)\end{align}
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