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 $$ \sin(36^{\circ})*\sin(2*36^{\circ})\sin(3*36^{\circ})\sin(4*36^{\circ})$$

  • $\begingroup$ Note that $\sin(36^\circ)\sin(4*36^\circ)=\frac{-1}{2}(\cos(5*36^\circ)-\cos(3*36^\circ)$ and ... $\endgroup$ – Qurultay Apr 14 at 15:13
  • $\begingroup$ @–Qurultay explain me more $\endgroup$ – Abhishek Kumar Apr 14 at 15:21
  • 1
    $\begingroup$ $$\prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n},$$ then you may apply $\log$ to both sides. $\endgroup$ – Jack D'Aurizio Apr 14 at 16:10
  • $\begingroup$ @jack is this valid for $$\prod_{r=1}^4\sin(rA)$$ $\endgroup$ – Abhishek Kumar Apr 14 at 17:32

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


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


Use Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$

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

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



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 $s_{-r}=-s_r$


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:

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


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