Simplify $2 \sin(x) \cos(7x) + \sin(6x)$ 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.
 A: There is a group of identities called the product-to-sum identities (Wikipedia has a very useful list of such things); the one that's relevant here is as follows:
$$\sin\alpha\cos\beta = \frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{2}$$
Applying the above, we have:
$$\begin{align}
2\sin x \cos 7x + \sin 6x &= 2\big( \frac{\sin 8x + \sin (-6x)}{2}\big) + \sin 6x\\
&=\sin 8x -\sin6x + \sin 6x\\
&=\sin 8x 
\end{align}$$
A: 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}
A: Another method. Take the derivative, which ends up being $8\cos(8x)$, and then integrate to get $\sin(8x)$
A: $$2\sin(x)\cos(7x)+\sin(6x)=2\sin(x)\cos(7x)+\sin(7x-x)$$
$$=2\sin(x)\cos(7x)+\sin(7x)\cos(x)-\sin(x)\cos(7x)$$
$$=\sin(x)\cos(7x)+\sin(7x)\cos(x)=\sin(8x)$$
