If θ is a constant such that 0<θ<π and x+1x=2cosθ, then for each positive integer n, x^n+1x^n equals

1981 AMC 12 Problem 24

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24.

\theta

is a constant such that

0\lt\theta\lt\pi

and

x+\frac1x=2\cos\theta,

then for each positive integer

n,

x^n+\frac1{x^n}

equals

2\cos\theta

2^n\cos\theta

2\cos^n\theta

2\cos n\theta

2^n\cos^n\theta

Rewrite the condition as a quadratic equation in

x

Its two roots are

\cos\theta\pm i\sin\theta

The equation is

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

whose roots are

e^{i\theta}

and

e^{-i\theta}.

Thus

x=e^{\pm i\theta}

and

x^{-1}=e^{\mp i\theta}.

By De Moivre’s theorem,

\begin{aligned} x^n+x^{-n} &=e^{in\theta}+e^{-in\theta}\\ &=2\cos n\theta. \end{aligned}

Therefore, the correct answer is D.