prove $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4 \cdot \sin(\frac{B - C}{2}) \cdot \sin(\frac{C - A}{2}) \cdot \sin(\frac{A - B}{2})$ If $A + B + C = \pi$,
then prove $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4 \cdot \sin(\frac{B - C}{2}) \cdot \sin(\frac{C - A}{2}) \cdot \sin(\frac{A - B}{2})$
My try:

$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = \sin(A - C) + \sin(B - A) + \sin(C -B)$
$\sin(A - C) + \sin(B - A) + \sin(C -B) = 2 \cdot \sin(\frac{B -C}{2}) \cdot \cos(A - \frac{C + B}{2}) + \sin(C - B)$
$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 2 \cdot \sin(\frac{B -C}{2}) \cdot \sin(\frac{A}{2}) + \sin(C - B)$

I am confused what steps should follow?
 A: Use the product-to-sum identities on the RHS a couple of times:
\begin{align}
  4\sin\frac{B-C}{2}\sin\frac{C-A}{2}\sin\frac{A-B}{2}
      &= 4\cdot\frac{1}{2}\left(\cos\frac{B+A-2C}{2} - \cos\frac{B-A}{2}\right)\sin\frac{A-B}{2} \\
      &= 2\left(\cos\frac{B+A-2C}{2}\sin\frac{A-B}{2} - \cos\frac{B-A}{2}\sin\frac{A-B}{2}\right) \\
      &= \sin(A-C) - \sin(B-C) - \sin 0 + \sin(B-A) \\
      &= \sin(A-C) + \sin(C-B) + \sin(B-A).
\end{align}
Then, as you observed in the original post, since $A+B+C=\pi$, this is equal to
$$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B).$$
A: First let $C=\pi -A -B$, and then rewrite both $LHS$ and $RHS$:


*

*$$\color{green}{LHS}=\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B)=\\\sin(B + 2\pi-2A-2B) + \sin(\pi-A-B + 2A) + \sin(A + 2B)=\\\sin(-B-2A) + \sin(\pi+A-B) + \sin(A+2B)=\\\sin(-B-2A)+\sin(A+2B)-\sin(A-B)=\\2\sin\left(\frac{A+2B-B-2A}{2}\right)\cos\left(\frac{A+2B+B+2A}{2}\right)+\sin(B-A)=\\\color{green}{2\sin\left(\frac{B-A}{2}\right)\cos\left(\frac{3A+3B}{2}\right)+\sin(B-A)}$$

*$$\color{blue}{RHS}=4 \cdot \sin\left(\frac{B - C}{2}\right) \cdot \sin\left(\frac{C - A}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\4 \cdot \sin\left(\frac{B - \pi+A+B}{2}\right) \cdot \sin\left(\frac{\pi-A-B - A}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\4 \cdot \sin\left(-\left(\frac{\pi}{2}-\frac{A+2B}{2}\right)\right) \cdot \sin\left(\frac{\pi}{2}-\frac{2A+B}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\-4 \cdot \sin\left(\frac{\pi}{2}-\frac{A+2B}{2}\right) \cdot \cos\left(\frac{2A+B}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\-4 \cdot \cos\left(\frac{A+2B}{2}\right) \cdot \cos\left(\frac{2A+B}{2}\right) \cdot \sin\left(\frac{A - B}{2}\right)=\\-4\cdot \frac12\sin\left(\frac{A - B}{2}\right)\left[\cos\left(\frac{A+2B}{2}-\frac{2A+B}{2}\right)+\cos\left(\frac{A+2B}{2}+\frac{2A+B}{2}\right)\right]=-2\sin\left(\frac{A - B}{2}\right)\left[\cos\left(\frac{B-A}{2}\right)+\cos\left(\frac{3A+3B}{2}\right)\right]=\\2\sin\left(\frac{B-A}{2}\right)\cos\left(\frac{B-A}{2}\right)+2\sin\left(\frac{B-A}{2}\right)\cos\left(\frac{3A+3B}{2}\right)=\\\color{blue}{\sin(B-A)+2\sin\left(\frac{B-A}{2}\right)\cos\left(\frac{3A+3B}{2}\right)}$$
So if $C=\pi -A -B$,$$LHS=RHS\\\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B)=4\sin\left(\frac{B - C}{2}\right)\sin\left(\frac{C - A}{2}\right)\sin\left(\frac{A - B}{2}\right)\\\color{yellow}{QED}$$

A: Another way to do it, is to use exponential notation.
(1) First, as $A+B+C=\pi$ you can simplify your LHS to
$$\sin(B+2C)+\sin(C+2A)+\sin(A+2B)\\=\sin(\pi-A+C)+\sin(\pi-B+A)+\sin(\pi-C+B)\\
=-\sin(C-A)-\sin(A-B)-\sin(B-C)$$
Now our problem only depends on the distances of the three values, so we can reduce its dimension by substituting
$$A-B=x,\quad B-C=y, \quad\Rightarrow C-A=-x-y$$ and obtain an identity of only two variables $x,y\in\mathbb{R}$ which is
$$\sin(x+y)-\sin(x)-\sin(y)=-4\sin\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\sin\left(\frac{x+y}{2}\right).$$
(2) Now we can proof this equation using exponential identity $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$ and simple exponential algebra. The RHS is
$$=\underbrace{\frac{-4}{(2i)^3}}_{\displaystyle{\frac{1}{2i}}}\underbrace{\underbrace{\left(e^{ix/2}-e^{-ix/2}\right)
\left(e^{iy/2}-e^{-iy/2}\right)}_{\displaystyle{\left(e^{i(x+y)/2}-e^{i(x-y)/2}-e^{i(y-x)/2}+e^{-i(x+y)/2}\right)}}
\left(e^{i(x+y)/2}-e^{-i(x+y)/2}\right)}_{\displaystyle{\left(e^{i(x+y)}-e^0-e^{ix}+e^{-iy}-e^{iy}+e^{-ix}+e^0-e^{-i(x+y)}\right)}}$$
which can obviously grouped back to form
$$=\sin(x+y)-\sin(x)-\sin(y).$$
