Question

The equation whose roots are $n^{th}$ power of the roots of the equation, $x^2-2x\cos \theta +1=0$ is given by

• A. $(x+\cos n\theta {)}^2+{\sin }^{2}n\theta =0$
• B. $(x-\cos n\theta {)}^2+{\sin }^{2}n\theta =0$
• C. $x^2+2x\cos n\theta +1=0$
• D. $x^2-2x\cos n\theta +1=0$

Answer 1

The correct option is A $(x+\cos n\theta {)}^2+{\sin }^{2}n\theta =0$

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

$x=\frac{2\cos \theta \pm \sqrt{(2\cos \theta {)}^2-4(1\left)\right(1)}}{2\left(1\right)}=\frac{2\cos \theta \pm \sqrt{4({\cos }^2\theta -1)}}{2}$

$[\because {\sin }^2\theta +{\cos }^2\theta =1;1-{\cos }^2\theta =-{\sin }^2\theta ]$

$x=\frac{\cos \theta \pm i\sin \theta }{1}$ $=e^{i\theta }$ & $e^{-i\theta }$

$\alpha$ & $\beta$

By demovries theorem.

${\alpha }^n$ & ${\beta }^n$ will be $e^{in\theta },e^{-in\theta }$.

$(x-e^{in\theta })(x-e^{-in\theta })$

$[\frac{x-\cos \left(n\theta \right)}{a}-\frac{i\sin \left(n\theta \right)}{b}][\frac{x-\cos \left(n\theta \right)}{a}+\frac{i\sin n\theta }{b}]$

$\left[\right(a-b\left)\right(a+b)=a^2-b^2)]$

$(x-\cos n\theta {)}^2+{\sin }^2(n\theta )=0$ is required equation.