Prove by induction $ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $ 
Prove by induction 
  $$ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $$

What I have for now:
$$  \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} + \sin(n + 1)x =  \frac {\sin (\frac {n + 2} {2} x)} {\sin \frac{x}{2}} \sin \frac{(n + 1)x}{2}$$
Letting $y = \frac {(n + 1)x} {2} $ 
$$\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + \sin2y = 
\frac {\sin (y + \frac {x} {2} )} {\sin \frac{x}{2}} \sin y $$
Then I used $\sin (\alpha + \beta)$ and $\sin (\alpha - \beta)$ formulas, bit they didn't help.

You are almost there, but use $\sin 2y = 2\sin y\cos y$, considering your left hand side we get:
$$\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + \sin2y  = 
\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + 2\sin y\cos y = $$
$$\left(\frac {\sin (y-\frac{x}{2})} {\sin \frac{x}{2}} +2\cos y\right)\sin y = \left(\frac {\sin (y-\frac{x}{2}) +2\cos y\sin\frac{x}{2}} {\sin \frac{x}{2}}\right)\sin y =$$
Now use the subtraction of angles formula $\sin (\alpha-\beta)$:
$$=\left(\frac {\sin y\cos \frac{x}{2} - \cos y \sin \frac{x}{2} +2\cos y\sin\frac{x}{2}} {\sin \frac{x}{2}}\right)\sin y = \left(\frac {\sin y\cos \frac{x}{2} +\cos y\sin\frac{x}{2}} {\sin \frac{x}{2}}\right)\sin y =$$
And lastly use the addition of angles formula $\sin(\alpha+\beta)$:
$$=\left(\frac {\sin (y+ \frac{x}{2}) } {\sin \frac{x}{2}}\right)\sin y $$
Which is what you had on your right hand side, and thus the induction is done.
