Question

Find the value of $\cos \frac{\pi }{11}+\cos \frac{3\pi }{11}+\cos \frac{5\pi }{11}+\cos \frac{7\pi }{11}+\cos \frac{9\pi }{11}$.

Answer 1

Using $\cos \alpha +\cos (\alpha +\beta )+...+\cos (\alpha +\overline{n-1}\beta )=\frac{\sin \left(\frac{n\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}\cos (\alpha +\frac{n-1}{2}\beta )$
$\cos \frac{\pi }{11}+\cos \frac{3\pi }{11}+\cos \frac{5\pi }{11}+\cos \frac{7\pi }{11}+\cos \frac{9\pi }{11}$
$=\cos \frac{\pi }{11}+\cos (\frac{\pi }{11}+\frac{2\pi }{11})+...+\cos (\frac{\pi }{11}+4\times \frac{2\pi }{11})$
$=\frac{\sin (5\times \frac{\pi }{11})}{\sin \left(\frac{\pi }{11}\right)}\cos (\frac{\pi }{11}+2\times \frac{2\pi }{11})$
$=\frac{2\sin \frac{5\pi }{11}\cos \frac{5\pi }{11}}{2\sin \frac{2\pi }{11}}=\frac{\sin \frac{10\pi }{11}}{2\sin \frac{\pi }{11}}=\frac{1}{2}$
$(\because \sin (180-x)=\sin x⟹\sin \frac{10\pi }{11}=\sin \frac{\pi }{11})$