Question: Find the value of $$\sum\_{r=1}^4 \log\_2 (\sin(\frac{r\pi}{5}))$$

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**My apporach:-**

$$\sum\_{r=1}^4 \log\_2 (\sin(\frac{r\pi}{5}))$$
$$=\log\_2 (\sin(36^{\circ}))+\log\_2 (\sin(2\*36^{\circ}))+\log\_2 (\sin(3\*36^{\circ}))+\log\_2 (\sin(4\*36^{\circ}))$$
$$=\log\_2 (\sin(36^{\circ})\*\sin(2\*36^{\circ})\sin(3\*36^{\circ})\sin(4\*36^{\circ}))$$

After this i was unable to solve this question ?
And also i want to ask one question more is there any general formula for this type of series $$ \sin(36^{\circ})\*\sin(2\*36^{\circ})\sin(3\*36^{\circ})\sin(4\*36^{\circ})$$

Answer: 
Observe that $\sin5x=0$ for $x=0^\circ,36^\circ,72^\circ,108^\circ,144^\circ$

As $\sin5x=5\sin x-20\sin^3x+16\sin^5x$

The roots of $16s^5-20s^3+5s=0$ are $s\_r=\sin(r36^\circ)$ where $r\equiv0,\pm1,\pm2\pmod5$

So, the roots of $16s^4-20s^2+5=0$ are $s\_r=\sin(r36^\circ)$ where $r\equiv\pm1,\pm2\pmod5$

Using [Vieta's formulas](https://en.wikipedia.org/wiki/Vieta%27s_formulas)
$$\prod\_{r=1}^4\sin(r36^\circ)=\prod\_{r=-2,\ne0}^2\sin(r36^\circ)=\dfrac5{16}$$

as $s\_{-r}=-s\_r$
More in general let's call $\Sigma\_n$ the n-th partial sum. We can notice that the function $f(r)$ in the sum is periodic with period $T=10$. These means that we need to calculate only the first ten values of the sum $\{\Sigma\_1,...,\Sigma\_{10}\}$ and then we'll have:

$$\Sigma\_{n}=\lfloor \frac{n}{10} \rfloor \Sigma\_{10}+\Sigma\_{n\text { mod } 10} $$