Use the identity $\cos(A-B) -\cos(A+B) = 2\sin(A)\sin(B)$ to prove another identity and evaluate a sum Use the identity $$\cos(A-B) -\cos(A+B) = 2\sin(A)\sin(B)$$ to prove that:
$$2\sin(j\theta)\sin(\frac{1}{2}\theta)=\cos((j-\frac{1}{2})(\theta))-\cos((j+\frac{1}{2})\theta).$$
This seemed almost too easy, so I am wondering if I am missing something? 
I just let $A = j\theta$ and $B = \frac{1}{2}\theta$, and substituted it directly into the identity. Since the question has asked me to use the identity, I am assuming that we can use it directly.
Have I missed anything?
The next part of the question asks:
Deduce that
$$\sum^n_{j=1}\sin(j\theta) = \frac{\cos(\frac{1}{2}\theta)-\cos((n+\frac{1}{2})\theta)}{2\sin(\frac{1}{2}\theta)}, \text{ if $\theta$ is not a multiple of $2\pi$}$$
Writing this out in general terms I see that the middle terms all cancel out (i have omitted the denominator here):
$$\cos((j-\frac{1}{2})\theta)-\cos((j+\frac{1}{2})\theta) + \cos(((j+1)-\frac{1}{2})\theta)-\cos(((j+1)+\frac{1}{2})\theta) + \cos(((j+2)-\frac{1}{2})\theta)-\cos(((j+2)+\frac{1}{2})\theta)+\dotsb+ \cos((n-\frac{1}{2})\theta)-\cos((n+\frac{1}{2})\theta)$$
I notice that if $\theta$ is a multiple of $2\pi$, then I would lose the 1st term, thus $\theta$ cannot be a multiple of $2\pi$.
Can anyone give me some direction to make this more mathematically rigorous? I feel like I am just a few steps away from answering this correctly.
Thanks in advance!
 A: Your proof of the second identity is correct. If $\sin \left( \frac{1}{2}\theta \right) \neq 0$ (what does this mean in terms of $\theta$?) this identity is equivalent to
$$\begin{equation*}
\sin \left( j\theta \right) =\frac{\cos \left( \left( j-\frac{1}{2}\right)
\theta \right) -\cos \left( \left( j+\frac{1}{2}\right) \theta \right) }{
2\sin \left( \frac{1}{2}\theta \right) }.
\end{equation*}$$
The first terms of the numerators of the sum, which you didn't wrote in your proof, are $$\cos \left(  \frac{1}{2} \theta \right) -\cos \left(  
\frac{3}{2} \theta \right) +\cos \left(  \frac{3}{2}
\theta \right) -\cos \left(  \frac{5}{2} \theta\right)  +\cdots 
$$
Evaluation of the sum, which is a  telescoping sum, as noticed by André Nicolas
\begin{eqnarray*}
\sum_{j=1}^{n}\sin \left( j\theta \right)  &=&\sum_{j=1}^{n}\frac{\cos
\left( \left( j-\frac{1}{2}\right) \theta \right) -\cos \left( \left( j+
\frac{1}{2}\right) \theta \right) }{2\sin \left( \frac{1}{2}\theta \right) }
\\
&=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) }\sum_{j=1}^{n}\cos
\left( \left( j-\frac{1}{2}\right) \theta \right) -\cos \left( \left( j+
\frac{1}{2}\right) \theta \right)  \\
&=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) }
\sum_{j=1}^{n}a_{j}-a_{j+1},\qquad a_{j}=\cos \left( \left( j-\frac{1}{2}
\right) \theta \right)  \\
&=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) }\left(
a_{1}-a_{n+1}\right) ,\qquad \text{See below} \\
&=&\cdots 
\end{eqnarray*}
For the telescoping sum $\sum_{j=1}^{n}a_{j}-a_{j+1}$ we have
\begin{eqnarray*}
\sum_{j=1}^{n}a_{j}-a_{j+1} &=&\left( a_{1}-a_{2}\right) +\left(
a_{2}-a_{3}\right) +\cdots +\left( a_{n-1}-a_{n}\right) +\left( a_{n}-a_{n+1}\right)  \\
&=&a_{1}-a_{n+1}.
\end{eqnarray*}
