if:
$S=\sin x+2\sin (2x)+\cdots+n\sin nx$,
$C=\cos x +2\cos (2x)+\cdots+n\cos (nx).$
prove that $4\sin^2 (x/2).S=(n+1)\sin (nx)-n\sin(nx+x)$
I can solve this easily using complex numbers(ie taking $C+iS$,which becomes an AGP) but was wondering if it can be solved using basic trig identities.
Source S.L.Loney plane trigonometry
Thank you!
Easy to see that for $\sin\frac{x}{2}=0$ it's true.
But for $\sin\frac{x}{2}\neq0$ by the telescopic summation we obtain: $$S=-\left(\sum_{k=1}^n\cos{kx}\right)'=-\left(\frac{\sum\limits_{k=1}^n2\sin\frac{x}{2}\cos{kx}}{2\sin\frac{x}{2}}\right)'=$$ $$=-\left(\frac{\sum\limits_{k=1}^n\left(\sin\left(k+\frac{1}{2}\right)x-\sin\left(k-\frac{1}{2}\right)x\right)}{2\sin\frac{x}{2}}\right)'=-\left(\frac{\sin\left(n+\frac{1}{2}\right)x-\sin\frac{x}{2}}{2\sin\frac{x}{2}}\right)'=$$ $$=-\frac{\left(n+\frac{1}{2}\right)\cos\left(n+\frac{1}{2}\right)x\sin\frac{x}{2}-\frac{1}{2}\sin\left(n+\frac{1}{2}\right)x\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}=$$ $$=\frac{\frac{1}{2}\left(\sin(n+1)x+\sin{nx}\right)-\left(n+\frac{1}{2}\right)(\sin(n+1)x-\sin{nx})}{4\sin^2\frac{x}{2}}=$$ $$=\frac{(n+1)\sin{nx}-n\sin(n+1)x}{4\sin^2\frac{x}{2}}.$$