Question

If $x^2-2x\cos \theta +1=0$, then $x^{2n}-2x^n\cos n\theta +1$ is equal to

• A. $\cos 2n\theta$
• B. $\sin 2n\theta$
• C. $0$
• D. $R-\left\{0\right\}$

Answer 1

Answer: (C)

$0$

Given relation is

$x^2-2x\cos \theta +1=0$

$\Rightarrow x^2-2x\cos \theta +{\cos }^2\theta -{\cos }^2\theta +1=0$

$\Rightarrow (x-\cos \theta {)}^2+{\sin }^2\theta =0$
$\Rightarrow x=\cos \theta \pm i\sin \theta$
Take $x=\cos \theta +i\sin \theta$
$x^n=\cos n\theta +i\sin n\theta$
and $x^{-n}=\frac{1}{x^n}=\cos n\theta -i\sin n\theta$
$\therefore x^n+\left(x^{-n}\right)=2\cos n\theta$
$\therefore x^{2n}-2x^n\cos n\theta +1=0$.