Find the value of $\sum_{r=1}^4 \log_2 (\sin(\frac{r\pi}{5}))$ Find the value of $$\sum_{r=1}^4 \log_2 (\sin(\frac{r\pi}{5}))$$

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 $$ \sin(36^{\circ})*\sin(2*36^{\circ})\sin(3*36^{\circ})\sin(4*36^{\circ})$$
 A: $$\sin36^{\circ}\sin72^{\circ}\sin108^{\circ}\sin144^{\circ}=\sin^236^{\circ}\sin^272^{\circ}=\frac{5}{16}.$$
For the proof use $$\cos36^{\circ}=\frac{1+\sqrt5}{4}$$ and
$$ \sin18^{\circ}=\frac{\sqrt5-1}{4}.$$
A: 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
$$\prod_{r=1}^4\sin(r36^\circ)=\prod_{r=-2,\ne0}^2\sin(r36^\circ)=\dfrac5{16}$$
as $s_{-r}=-s_r$
A: $$2S=2\sin36^\circ\sin72^\circ=\cos36^\circ-\cos108^\circ=\cos36^\circ+\cos72^\circ\text{ as  }\cos(180^\circ-y)=-\cos y$$
$$4S^2=(\cos36^\circ+\cos72^\circ)^2=(\cos36^\circ-\cos72^\circ)^2+4\cos36^\circ\cos72^\circ$$
Use Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$ 
and  for $\sin t\ne0$ and $\sin4t=\sin t$
$$4\cos t\cos2t=\dfrac{2\sin2t\cos2t}{\sin t}=\dfrac{\sin4t}{\sin t}=1$$
$$4S^2=\left(\dfrac12\right)^2+1$$
A: 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:
$$\Sigma_{n}=\lfloor \frac{n}{10} \rfloor \Sigma_{10}+\Sigma_{n\text  { mod }  10} $$ 
