As resolved the identity of this series?

Question

Use the identity $\cos((k-\frac{1}{2})x) - \cos((k+\frac{1}{2})x) = 2\sin kx \sin \frac{x}{2}$ to show that

$S_n:=\sum_{k=1}^{n}\sin kx=\frac{1}{2\sin \frac{x}{2}}(\cos\frac{x}{2}-\cos((nx+\frac{x}{2}))$

for $\sin \frac{x}{2}\neq0$, and conclude that $(S_n)$ is bounded. Get a similar result for $$T_n:=\sum_{k=1}^{n}\cos kx $$

Answer 1

Hint You have a telescopic sum in $k$.$$\cos\left(k-\frac{1}{2}\right) x- \cos\left(k+\frac{1}{2}\right) x=a_k-a_{k+1} $$