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

Question

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.

Answer 1

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.

Answer 2

You can also see that \begin{align} \sum_{k=0}^{n}\sin kx & =\Im \sum_{k=0}^{n}\mathrm{e}^{\mathrm{i}kx}\\ & = \Im \sum_{k=0}^{n}\left(\mathrm{e}^{\mathrm{i}x}\right)^k\\ & = \Im\frac{\left(\mathrm{e}^{\mathrm{i}x}\right)^{n+1}-1}{\mathrm{e}^{\mathrm{i}x}-1}\\ & = \Im\frac{\mathrm{e}^{\mathrm{i}(n+1)x}-1}{\mathrm{e}^{\mathrm{i}x}-1}\\ & = \Im\frac{\mathrm{e}^{\mathrm{i}\frac{n+1}{2}x}}{\mathrm{e}^{\mathrm{i}\frac{x}{2}}}\frac{\mathrm{e}^{\mathrm{i}\frac{n+1}{2}x}-\mathrm{e}^{-\mathrm{i}\frac{n+1}{2}x}}{\mathrm{e}^{\mathrm{i}\frac{x}{2}}-\mathrm{e}^{-\mathrm{i}\frac{x}{2}}}\\ & = \Im\mathrm{e}^{\mathrm{i}\frac{nx}{2}}\frac{\sin\left(\frac{n+1}{2}x\right)}{\sin\left(\frac{x}{2}\right)}\\ & = \sin\left(\frac{nx}{2}\right)\frac{\sin\left(\frac{n+1}{2}x\right)}{\sin\left(\frac{x}{2}\right)}. \end{align}