calculus - Showing $sum_{n=1}^infty sin x sin nx$ is uniformly bounded

I need to show that for every $x$: $$\sum_{n=1}^\infty \sin x \sin nx \lt M$$

So the first thing came into my mind is applying a well-known trigonometric identity:

$$\sum_{n=1}^\infty \sin x \sin nx = \frac{1}{2} \sum_{n=1}^\infty \cos (x-nx) - \cos(x+nx)$$

For a second I thought I'd get a telescoping series but it isn't.

What should I do next?

EDIT
Basically I'm trying to use here Dirichlet's test to show uniform converges for the functions series:

$$f_n(x) = \sum_{n=1}^\infty \frac{\sin x \sin nx}{\sqrt {n+x^2}}$$

Answer

Since $\cos$ is an even function, you have in fact a telescoping series:

\begin{align}
\sum_{n = 1}^N \sin x\sin (nx) &= \frac{1}{2}\sum_{n = 1}^N \bigl(\cos\bigl((n-1)x\bigr) - \cos \bigl((n+1)x\bigr)\bigr)\\
&= \frac{1}{2}\bigl( 1 + \cos x - \cos (Nx) - \cos \bigl((N+1)x\bigr)\bigr).
\end{align}