Question

If the value of $\cos \frac{3\pi }{7}-\cos \frac{2\pi }{7}+\cos \frac{\pi }{7}=\frac{1}{m}$.Find $m$

Answer 1

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

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

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

=Real part of ($m+m^5+m^3$)

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

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

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

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

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

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

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

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

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

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

$\therefore m=2$