# Find the average of the number $n \sin n^\circ$ for $n=2,4,6\cdots,180$ [duplicate]

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?

• So $n$ is to be interpreted in degrees, right? Dec 1, 2020 at 8:19
• @Matti P. yes, can you suggest how to write degree with math jax Dec 1, 2020 at 8:20
• like this: 180^{\circ} Dec 1, 2020 at 8:23
• thanks @Matti P. Dec 1, 2020 at 8:24

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!