Question

Find a if $4\cos \frac{2\pi }{7}\cos \frac{\pi }{7}-1=a\cos \frac{a\pi }{7}$

Answer 1

$e^{i\left(\frac{\pi }{7}\right)}=i\sin \left(\frac{\pi }{7}\right)+\cos \left(\frac{\pi }{7}\right)$

Let $m=e^{i\left(\frac{\pi }{7}\right)}$

$m^7=-1\Rightarrow m^6+m^4+m^2+1=m^5+m^3+m$

Let $a=m^5+m^3+m$

$ma=m^6+m^4+m^2$

$ma=a-1\Rightarrow a=\frac{1}{1-m}$

$a=\frac{1}{1-\cos \frac{\pi }{7}-i\sin \frac{\pi }{7}}$

$=\frac{1-\cos \frac{\pi }{7}+i\sin \frac{\pi }{7}}{(1-\cos \frac{\pi }{7}{)}^2+{\sin }^2\frac{\pi }{7}}$

$=\frac{1-\cos \frac{\pi }{7}+i\sin \frac{\pi }{7}}{2(1-\cos \frac{\pi }{7})}$

$=\frac{1}{2}+i\left(\frac{\sin \frac{\pi }{7}}{(1-\cos \frac{\pi }{7})}\right)$

$m+m^3+m^5=\frac{1}{2}+i\left(\sin \frac{\pi }{7}2(1-\cos \frac{\pi }{7})\right)$

Real part(a)=Real part $(m+m^3+m^5)=\frac{1}{2}$

Given , $\cos \frac{3\pi }{7}-\cos \frac{2\pi }{7}+\cos \frac{\pi }{7}=\frac{1}{m}$

$2\cos \frac{\pi }{7}+2\cos \frac{3\pi }{7}=1-2\cos \frac{5\pi }{7}$

$4\cos \frac{\pi }{7}\cos \frac{2\pi }{7}-1=-2\cos \frac{5\pi }{7}$

$\therefore 4\cos \frac{\pi }{7}\cos \frac{2\pi }{7}-1=-2\cos \frac{2\pi }{7}$

Given, $4\cos \frac{\pi }{7}\cos \frac{2\pi }{7}-1=-a\cos \frac{a\pi }{7}$

$\therefore a=2$