Question

If $x+\frac{1}{x}=2\cos \alpha$ then $x^n+\frac{1}{x^n}$ is equal to:

• A. $2^n\cos \alpha$
• B. $2^n\cos n\alpha$
• C. $2i\sin na$
• D. $2\cos n\alpha$

Answer 1

Answer: (D)

$2\cos n\alpha$

Put $x=\cos \alpha +i\sin \alpha =e^{i\alpha }$

So $\frac{1}{x}=x^{-1}=e^{-i\alpha }=\cos \alpha -i\sin \alpha$

Clearly $x+\frac{1}{x}=2\cos \alpha$

Hence $x^n+\frac{1}{x^n}=x^n+x^{-n}=(\cos \alpha +i\sin \alpha {)}^n+(\cos \alpha +i\sin \alpha {)}^{-n}$

$=\cos n\alpha +i\sin n\alpha +\sin n\alpha -i\sin n\alpha$
$=2\cos n\alpha$, Use Demoivre's theorem