Question

If ${\log }_{2}5-\sum _{k=1}^4{\log }_2\left(\sin \frac{k\pi }{5}\right)=\frac{p}{q},$ where $p$ and $q$ are co-prime, then the value of $p+q$ is

Answer 1

$E={\log }_{2}5-\sum _{k=1}^4{\log }_2\left(\sin \frac{k\pi }{5}\right)={\log }_{2}5-({\log }_2\sin \frac{\pi }{5}+{\log }_2\sin \frac{2\pi }{5}+{\log }_2\sin \frac{3\pi }{5}+{\log }_2\sin \frac{4\pi }{5})={\log }_{2}5-\left({\log }_2\left({\sin }^2\frac{\pi }{5}{\sin }^2\frac{2\pi }{5}\right)\right)={\log }_{2}5-{\log }_2\left(\frac{1}{4}(2{\sin }^2\frac{\pi }{5}\cdot 2{\sin }^2\frac{2\pi }{5})\right)={\log }_{2}5-{\log }_2\left(\frac{1}{4}\right(1-\cos {72}^{\circ }\left)\right(1-\cos {144}^{\circ }\left)\right)={\log }_{2}5-{\log }_2\left(\frac{1}{4}(1-\frac{\sqrt{5}-1}{4})(1+\frac{\sqrt{5}+1}{4})\right)={\log }_{2}5-{\log }_2\left(\frac{1}{4}\left(\frac{5}{4}\right)\right)={\log }_{2}5-{\log }_2\left(\frac{5}{16}\right)={\log }_{2}5-{\log }_{2}5+4\therefore E=\frac{4}{1}=\frac{p}{q}$