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1981 AHSME Problems/Problem 24

Revision as of 21:20, 1 May 2020 by Alexwin0806 (talk | contribs) (→‎Solution)

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:

\[=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)\] \[=2\cos(n\theta)\]

Which gives the answer $\boxed{\textbf{D}}$

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