Question

If $\alpha$ and $\beta$ are the roots of $x^2-2x\cos \varnothing +1=0$, then the equation, whose roots are ${\alpha }^n$and ${\beta }^n$

• A. $x^2–2x\cos n\varnothing –1=0$
• B. $x^2–2x\cos n\varnothing +1=0$
• C. $x^2–2x\sin n\varnothing +1=0$
• D. $x^2+2x\sin n\varnothing -1=0$

Answer 1

The correct option is B

$x^2–2x\cos n\varnothing +1=0$

Explanation for correct option:

Find the equation :

Given that $\alpha$ and $\beta$ are the roots of $x^2-2x\cos \varnothing +1=0$

$$\begin{gathered} \Rightarrow x=\frac{[2\cos \varnothing \pm \sqrt{4{\cos }^2\varnothing -4}]}{2} \\ \Rightarrow x=\frac{[2\cos \varnothing \pm 2\sqrt{{\cos }^2\varnothing -1}]}{2} \\ \Rightarrow x=\frac{[2\cos \varnothing \pm 2\sqrt{-{\sin }^2\varnothing }]}{2} \\ \Rightarrow x=\cos \varnothing \pm i\sin \varnothing \end{gathered}$$

Let $ɑ=\cos \varnothing +i\sin \varnothing ,$then $\beta =\cos \varnothing –i\sin \varnothing$

$$\begin{gathered} {\alpha }^n+{\beta }^n={(\cos \varnothing +i\sin \varnothing )}^n+{(\cos \varnothing –i\sin \varnothing )}^n=2\cos n\varnothing \\ and{\alpha }^n\times {\beta }^n={(\cos \varnothing +i\sin \varnothing )}^n{(\cos \varnothing –i\sin \varnothing )}^n={\cos }^{2}n\varnothing +{\sin }^{2}n\varnothing =1 \end{gathered}$$

The required equation is ${\mathit{x}}^{\mathbf{2}}\mathbf{–}\mathbf{}\mathbf{2}\mathit{x}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{n}\mathbf{\varnothing }\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$.

Hence, the correct option is (B).