Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\limits_{k=1}^{m}sin(kx)=\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}$
Induction step: $n=m+1$
$\sum\limits_{k=1}^{m+1}sin(kx)=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Prove: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}+sin(m+1)x=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Left side: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}+sin\frac{x}{2}sin(m+1)x}
{sin\frac{x}{2}}$
How to prove this equality? I used $sin(u)sin(v)$ identity but that didn't help.

Hint: If you want another approach you could probably do a series of deconvolutions in the Fourier domain. The euclidean algorithm should work.
