Question

The value of $2\cos \frac{\mathrm{\pi }}{13}\cos \frac{9\mathrm{\pi }}{13}+\cos \frac{3\mathrm{\pi }}{13}+\cos \frac{5\mathrm{\pi }}{13}$ is ________________.

Answer 1

​$2\cos \frac{\mathrm{\pi }}{13}\cos \frac{9\mathrm{\pi }}{13}+\cos \frac{3\mathrm{\pi }}{13}+\cos \frac{5\mathrm{\pi }}{13}$
Using identity:- 2 cos x cos y = cos (x + y) + cos (x – y)
$$\begin{gathered} =\cos \left(\frac{9\pi }{13}+\frac{\pi }{13}\right)+\cos \left(\frac{9\pi }{13}-\frac{\pi }{13}\right)+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13} \\ =\cos \left(\frac{10\pi }{13}\right)+\cos \left(\frac{8\pi }{13}\right)+\cos \left(\frac{3\pi }{13}\right)+\cos \left(\frac{5\pi }{13}\right) \\ =\cos \left(\frac{10\pi }{13}\right)+\cos \left(\frac{3\pi }{13}\right)+\cos \left(\frac{8\pi }{13}\right)+\cos \left(\frac{5\pi }{13}\right) \\ \left[\mathrm{using}\mathrm{identity},\cos x+\cos y=2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)\right] \\ =2\cos \left(\frac{{\displaystyle \frac{10\pi }{13}}+{\displaystyle \frac{3\pi }{13}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{10\pi }{13}}-{\displaystyle \frac{3\pi }{13}}}{2}\right)+2\cos \left(\frac{{\displaystyle \frac{8\pi }{13}}+{\displaystyle \frac{5\pi }{13}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{8\pi }{13}}-{\displaystyle \frac{5\pi }{13}}}{2}\right) \\ =2\cos \left(\frac{{\displaystyle \frac{13\pi }{13}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{7\pi }{13}}}{2}\right)+2\cos \left(\frac{{\displaystyle \frac{13\pi }{13}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{4\pi }{13}}}{2}\right) \\ =2\cos \frac{\pi }{2}\cos \frac{7\pi }{26}+2\cos \frac{\pi }{2}\cos \frac{4\pi }{26} \\ =2\cos \frac{\pi }{2}\left(\cos \frac{7\pi }{26}+\frac{4\pi }{26}\right) \\ =2\times 0\left(\cos \frac{7\pi }{26}+\cos \frac{4\pi }{26}\right)=0 \end{gathered}$$
Hence, the answer is 0.