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. 
 A: Divide both sides of the given result by $\theta$ and then let $\theta\to0$.  We have
$$\sum_{r=1}^n \frac{\sin r\theta}{\theta}
  =\frac{\frac{\sin n\theta}{2}}{\theta}\frac{\theta}{\sin\frac\theta2}
    \frac{\sin\frac{(n+1)\theta}{2}}{\theta}$$
and
 so
$$\sum_{r=1}^n r=\frac n2\frac1{\frac12}\frac{n+1}2=\frac{n(n+1)}2\ .$$
A: $\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}
$
A: And yet another approach is to take a derivative of both sides with respect to $\theta$ and evaluate the results as $\theta \to 0$.  To that end, we have for the left-hand side
$$\begin{align}
\lim_{\theta \to 0}\left(\frac{d}{d\theta}\sum_{r=1}^n\sin r\theta\right)&=\lim_{\theta \to 0}\left(\sum_{r=1}^nr\cos r\theta\right) \\\\
&=\sum_{r=1}^nr \tag 1
\end{align}$$
and for the right-hand side
$$\begin{align}
\lim_{\theta \to 0}\left(\frac{d}{d\theta}\left[\sin\left(\frac{n\theta}{2}\right)\,\csc\left(\frac{\theta}{2}\right)\,\sin\left(\frac{(n+1)\theta}{2}\right)\right]\right)&=\lim_{\theta \to 0}\left(\frac{n}{2}\cos\left(\frac{n\theta}{2}\right)\,\csc\left(\frac{\theta}{2}\right)\,\sin\left(\frac{(n+1)\theta}{2}\right)\right)\\\\
&-\lim_{\theta \to 0}\left(\sin\left(\frac{n\theta}{2}\right)\,\frac12\csc\left(\frac{\theta}{2}\right)\cot\left(\frac{\theta}{2}\right)\,\sin\left(\frac{(n+1)\theta}{2}\right)\right)\\\\
&+\lim_{\theta \to 0}\left(\sin\left(\frac{n\theta}{2}\right)\,\csc\left(\frac{\theta}{2}\right)\,\frac{n+1}{2}\cos\left(\frac{(n+1)\theta}{2}\right)\right)\\\\
&=\frac{n(n+1)}{2}-\frac{n(n+1)}{2}+\frac{n(n+1)}{2}\\\\
&=\frac{n(n+1)}{2} \tag 2
\end{align} $$
Equating the right-hand sides of $(1)$ and $(2)$, we find 
$$\bbox[5px,border:2px solid #C0A000]{\sum_{r=1}^nr=\frac{n(n+1)}{2}}$$
as expected!
