Deducing $\sum_{r=1}^{n}r$ from sine summation formula We know the famous formula
$$\sum_{r=1}^{n}\sin r\theta=\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\sin\frac{(n+1)\theta}{2}\ .$$
I have come across a question that use the above result to find $\sum_{r=1}^{n}r$.  I thought but could not deduce how this can be done.Is there a way to find $\sum_{r=1}^{n}r$ from the given sine summation series.  Thanks. 

$\sum_{r=1}^{n}\sin r\theta
=\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\sin\frac{(n+1)\theta}{2}
$
My first thought on seeing this
is to somehow use
$\lim_{x \to 0} \frac{\sin x}{x}
= 1
$.
Let's divide both sides by
$\theta$:
$\sum_{r=1}^{n}\frac{\sin r\theta}{\theta}
=\frac{\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\sin\frac{(n+1)\theta}{2}}{\theta}
=\frac{\sin \frac{n\theta}{2}\sin\frac{(n+1)\theta}{2}}{\theta\sin\frac{\theta}{2}}
$.
The left side,
as $\theta \to 0$,
becomes
$\sum_{r=1}^{n} r
$,
since
$\frac{\sin r\theta}{\theta}
=r\frac{\sin r\theta}{r\theta}
$.
This is a good sign.
The right side is
$\begin{array}\\
\frac{\sin \frac{n\theta}{2}\sin\frac{(n+1)\theta}{2}}{\theta\sin\frac{\theta}{2}}
&=\frac{n\theta}{2}\frac{\sin \frac{n\theta}{2}}{\frac{n}{2}\theta}
\frac{(n+1)\theta}{2}\frac{\sin \frac{(n+1)\theta}{2}}{\frac{n+1}{2}\theta}
\frac1{\theta\sin\frac{\theta}{2}}\\
&=\frac{n}{2}\frac{\sin \frac{n\theta}{2}}{\frac{n}{2}\theta}
\frac{(n+1)\theta}{2\sin\frac{\theta}{2}}\frac{\sin \frac{(n+1)\theta}{2}}{\frac{n+1}{2}\theta}\\
&\to \frac{n(n+1)}{2}
\quad\text{ as }\theta \to 0 \text{ (repeatedly using } \lim_{x \to 0} \frac{\sin x}{x}
= 1)\\
\end{array}
$
