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: 
$$2S=2\sin36^\circ\sin72^\circ=\cos36^\circ-\cos108^\circ=\cos36^\circ+\cos72^\circ\text{ as }\cos(180^\circ-y)=-\cos y$$

$$4S^2=(\cos36^\circ+\cos72^\circ)^2=(\cos36^\circ-\cos72^\circ)^2+4\cos36^\circ\cos72^\circ$$

Use [Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$](https://math.stackexchange.com/questions/827540/proving-trigonometric-equation-cos36-circ-cos72-circ-1-2) 

and for $\sin t\ne0$ and $\sin4t=\sin t$

$$4\cos t\cos2t=\dfrac{2\sin2t\cos2t}{\sin t}=\dfrac{\sin4t}{\sin t}=1$$

$$4S^2=\left(\dfrac12\right)^2+1$$
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:

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