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: 
$$\sin36^{\circ}\sin72^{\circ}\sin108^{\circ}\sin144^{\circ}=\sin^236^{\circ}\sin^272^{\circ}=\frac{5}{16}.$$
For the proof use $$\cos36^{\circ}=\frac{1+\sqrt5}{4}$$ and
$$ \sin18^{\circ}=\frac{\sqrt5-1}{4}.$$
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:

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