1981 AHSME Problems/Problem 24

Problem

If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$ , then for each positive integer $n$ , $x^n + \dfrac{1}{x^n}$ equals

$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta$

Solution

Multiply both sides by $x$ and rearrange to $x^2-2x\cos(\theta)+1=0$ . Using the quadratic equation, we can solve for $x$ . After some simplifying:

$x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}$ $x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}$ $x=\cos(\theta) + i\sin(\theta)$

Substituting this expression in to the desired $x^n + \dfrac{1}{x^n}$ gives:

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

Using DeMoivre's Theorem:

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

Because $\cos$ is even and $\sin$ is odd:

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

$=\boxed{2\cos(n\theta)},$

which gives the answer $\boxed{\textbf{D}}.$

Solution 2

Because we have $x + \frac{1}{x} = 2\cos \theta$, squaring both sides gives $x^2 + 2 + \frac{1}{x^2} = 4\cos^2 \theta.$ Subtracting 2 from both sides, we get $x^2 + \frac{1}{x^2} = 4\cos^2 \theta - 2.$ Rewriting, $x^2 - \frac{1}{x^2} = 2(2\cos^2 \theta - 1) = 2\cos(2\theta).$ Now, looking at the answer choices, the only one matching this form for $n = 2$ is $\boxed{\textbf{D}}.$

~Voidling

1981 AHSME (Problems • Answer Key • Resources)
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1981 AHSME Problems/Problem 24

Contents

Problem

Solution

Solution 2

See also