Question

$2^{n-1}(\cos \theta -\cos \frac{\pi }{n}\left)\right(\cos \theta -\cos \frac{2\pi }{n})...(\cos \theta -\cos \frac{n-1}{n}\pi )$ equals

• A. $\frac{\sin \theta }{\sin \left(n\theta \right)}$
• B. $\frac{\sin n\theta }{\sin \theta }$
• C. $1$
• D. $-n\cos \theta$

Answer 1

Answer: (B)

$\frac{\sin n\theta }{\sin \theta }$

From $(x-\alpha )(x-\overline{\alpha })=x^2-2x\cos \frac{\pi }{n}+1$
and $\frac{x^{2n}-1}{x^2-1}=x^{2n-2}+x^{2n+3}+...+x+1$
We get
$(x^2-2x\cos \frac{\pi }{n}+1)(x^2-2x\cos \frac{2\pi }{n}+1)...(x^2-2x\cos \frac{n-1}{n}\pi +1)=1+x+...+x^{2n-2}$
Divide both sides by $x^{n-1}$ to obtain
$(x+\frac{1}{x}-2\cos \frac{\pi }{n})(x+\frac{1}{x}-2\cos \frac{2\pi }{n})...(x+\frac{1}{x}-2\cos \frac{n-1}{n}\pi )=\frac{x^n-x^{-n}}{x-x^{-1}}$
Now put $x=\cos \theta +i\sin \theta$
we get
$2^{n-1}(\cos \theta -\cos \frac{\pi }{n}\left)\right(\cos \theta -\cos \frac{2\pi }{n})...(\cos \theta -\cos \frac{n-1}{n}\pi )=\frac{\sin n\theta }{\sin \theta }$