Find the average of the number $n \sin n^\circ$ for $n=2,4,6\cdots,180$ I have been asked in a exam to find the average of the number: $$n \sin n^\circ$$ for $n$=$2,4,6,\cdots,180$
I have tried a lot basically with sum product, or pairing the inputs...but at the end don't able to find any way to solve it, can someone help me with the approach?
 A: Since $\sin(180^\circ - \theta) = \sin(\theta)$, $\sin{90^\circ} = 1$, and $\sin{180^\circ} = 0$, we can write the sum as
$$
(2 \sin{2^\circ} + 178 \sin{2^\circ}) + (4 \sin{4^\circ} + 176 \sin{4^\circ}) + \ldots + (88 \sin{88^\circ} + 92 \sin{88^\circ}) + 90\text.
$$
To get the average, divide by the number of terms, $90$, and get
$$
2 \sin{2^\circ} + 2 \sin{4^\circ} + \ldots + 2 \sin{88^\circ} + 1\text.\tag{*}
$$
Now, $\cos(\theta - 1^\circ) - \cos(\theta + 1^\circ) = 2 \sin\theta \sin 1^\circ$.  Therefore,
$$
2\sin\theta = \frac{\cos(\theta - 1^\circ) - \cos(\theta + 1^\circ)}{\sin{1^\circ}}\text.\tag{**}
$$
When you plug $\text{(**)}$ into $\text{(*)}$, most of the $\cos$ terms cancel out and you are left with
$$
\frac{\cos{1^\circ} - \cos{89^\circ}}{\sin{1^\circ}} + 1 = \frac{\cos{1^\circ} - \sin{1^\circ}}{\sin{1^\circ}} + 1 = \color{red}{\cot{1^\circ}}\text.
$$
A: \begin{align}
\sum_{r=1}^{90}2r\sin\left(\dfrac{2r\pi}{180}\right)&=2\sum_{r=1}^{45}r\sin\left(\dfrac{r\pi}{90}\right)+2\sum_{r=1}^{45}(90-r)\sin\left(\pi-\dfrac{r\pi}{90}\right)\\
&=2\sum_{r=1}^{45}r\sin\left(\dfrac{r\pi}{90}\right)+180\sum_{r=1}^{45}\sin\left(\dfrac{r\pi}{90}\right)-2\sum_{r=1}^{45}r\sin\left(\dfrac{r\pi}{90}\right)\\
&=180\times\sum_{r=1}^{45}\sin\left(\dfrac{r\pi}{90}\right)
\end{align}
Now apply sum of sine of AP formula and you're done!
