Question

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

• A. $\cos 2n\theta$
• B. $\sin 2n\theta$
• C. $0$
• D. some real number greater than $0$

Answer 1

Answer: (C)

$0$

$x^2-2x\cos \theta +1=0$
$\therefore x=\frac{2cos\theta \pm \sqrt{4{\cos }^2\theta -4}}{2}=cos\theta \pm i\sin \theta$
Let us consider,
$x=\cos \theta +i\sin \theta$
Now,
$x^{2n}-2x^n\cos \theta +1$
$=\cos 2n\theta +i\sin \left(2n\theta \right)-2(cosn\theta +i\sin (n\theta \left)\right)\cos n\theta +1$
$=\cos 2n\theta -2{\cos }^2\theta +1+i\sin 2n\theta -2i\sin \left(n\theta \right)cosn\theta$
$=0$
If we replace $\theta$ by $-\theta$, we will obtain the same result for the other value of $x$.
Hence, option C is the correct answer