Question

If α satisfies the equation $x^2-2xcos\theta +1=0$. Find the value of ${\alpha }^n$ + $\frac{1}{{\alpha }^n}$ n $ϵ$ Z

• A. 2sin.
• B. 2cos.
• C. sin.
• D. cos.

Answer 1

The correct option is B

2cos.

Solution:

From quadratic formulae,we know roots of the quadratic equation

α = $\frac{-b\underset{–}{+}\sqrt{b^2-4ac}}{2a}$

= $\frac{2cos\theta \underset{–}{+}\sqrt{4co{s}^2\theta -4.1.1}}{2}$

= $cos\theta \underset{–}{+}isin\theta$ , $\alpha$ = cos$\theta$ + i sin$\theta$, cos$\theta$ - i sin$\theta$

Let $\alpha$ = cos$\theta$ + i sin$\theta$

Subsititute the value of $\alpha$ in ${\alpha }^n$ + $\frac{1}{{\alpha }^n}$

= $(cos\theta +isin\theta {)}^n$ + $\frac{1}{(cos\theta +isin\theta {)}^n}$

{Use de moivre's theorem $(cos\theta +isin\theta {)}^n$ = cos$n\theta$ + i sin$n\theta$ }

$(cos\theta +isin\theta {)}^n$ + $(cos\theta +isin\theta {)}^{-n}$

we know, cos$(-\theta )$ = cos$\theta$ and sin$(-\theta )$ = - sin$\theta$

= [(cos$n\theta$ + i sin$n\theta )$ + (cos$\left(n\theta \right)$ + i sin$(-n\theta ))$]

= cos$n\theta$ + i sin$n\theta$ + cos$n\theta$ - i sin$n\theta )$

= 2 cos$n\theta$.

OR,

Let $\alpha$ = cos$\theta$ - i sin$\theta$

Subsititute the value of $\alpha$ in ${\alpha }^n$ + $\frac{1}{{\alpha }^n}$

= $(cos\theta -isin\theta {)}^n$ + $\frac{1}{(cos\theta -isin\theta {)}^n}$

{Use de moivre's theorem $(cos\theta +isin\theta {)}^n$ = cos$n\theta$ + i sin$n\theta$ }

we know, cos$(-\theta )$ = cos$\theta$ and sin$(-\theta )$ = - sin$\theta$

= (cos$(-\theta$) + i sin$(-\theta ){)}^n$ + (cos$(-\theta )$ + i sin$(-\theta ){)}^{-n}$

= (cos$(-n\theta$) + i sin$(-n\theta ))$ + (cos$\left(n\theta \right)$ + i sin$\left(n\theta \right))$

= cos$n\theta$ - i sin$n\theta )$ + cos$n\theta$ + i sin$n\theta )$

= 2 cos$n\theta$.