TITLE: Simplify $2 \sin(x) \cos(7x) + \sin(6x)$
QUESTION [5 upvotes]: I was doing a problem and in my chain of computations I arrived at a seemingly complicated function 
$$2 \sin(x) \cos(7x) + \sin(6x)$$ 
However, I typed it into Wolfram and was surprised to find
$$2 \sin(x) \cos(7x) + \sin(6x) = \sin(8x)$$   
Is there a way I could have seen this beforehand? Even now I'm unsure how I would prove this identity.

REPLY [4 votes]: If you don't want to remember or look up any trig identities at all, you can turn everything into exponentials:
\begin{eqnarray}
2 \sin x \cos (7x) + \sin (6x) &=& 2\left(\frac{e^{ix}-e^{-ix}}{2i}\right)\left(\frac{e^{7ix}+e^{-7ix}}{2}\right) + \frac{e^{6ix}-e^{-6ix}}{2i}\\
&=&\frac{1}{2i}\left(e^{8ix}-e^{6ix}+e^{-6ix}-e^{-8ix} + e^{6ix}-e^{-6ix}\right)\\
&=&\frac{e^{8ix}-e^{-8ix}}{2i}\\
&=&\sin(8x)
\end{eqnarray}