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

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}
Use that $$\sin(A)+\sin(B)=2\cos\left(\frac{A-B}{2}\right)\sin\left(\frac{A+B}{2}\right)$$ and $$\sin(A+B)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$
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}
\begin{align} \sin a + \sin b + \sin (a+b)&=2 \sin\left(\frac{a+b}2\right) \cos\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} Now use the identity $$\cos A +\cos B=2\cos \left(\frac{A+B}2\right)\cos \left(\frac{A-B}2\right)$$.
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$$\sin a+\sin b+\sin (a+b)=4\cos\tfrac{a}{2}\cos\tfrac{b}{2}\left(\sin\tfrac{a}{2}\cos\tfrac{b}{2}+\sin\tfrac{b}{2}\cos\tfrac{a}{2}\right)=4\cos\tfrac{a}{2} \cos\tfrac{b}{2}\sin\tfrac{a+b}{2}.$$