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?
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 $$ \frac{\cos{1^\circ} - \cos{89^\circ}}{\sin{1^\circ}} + 1 = \frac{\cos{1^\circ} - \sin{1^\circ}}{\sin{1^\circ}} + 1 = \color{red}{\cot{1^\circ}}\text. $$
\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!
