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:
$$f_n(x) = \sum_{n=1}^\infty \frac{\sin x \sin nx}{\sqrt {n+x^2}}$$ 
 A: 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}
A: In this answer, I showed that 
$$\begin{align}
\left|\sum_{n=1}^N \sin(nx)\right| \le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\csc\left(\frac{x}{2}\right)\right|
\end{align}$$
Thus, 
$$\begin{align}
\left|\sum_{n=1}^N \sin x \sin(nx)\right| &\le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\sin x\csc\left(\frac{x}{2}\right)\right|\\\\
&=\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\cos\left(\frac{x}{2}\right)\right|\\\\
&\le2 \tag 1
\end{align}$$
For the Dirichlet test of $f_n(x)=\sum_{n=1}^{\infty}\frac{\sin x\sin nx}{\sqrt{n+x^2}}$, we only require the following two conditions:
Condition $(1)$ 
The sequence $\frac{1}{\sqrt{n+x^2}}$ decreases monotonically to zero.
Condition $(2)$ 
The partial sums $\sum_{n=1}^N \sin x \sin(nx)$ be bounded by a constant.
Condition $(1)$ is trivially confirmed while equation $(1)$ confirms Condition $(2)$.
