Art of Problem Solving

AoPS Online

1981 AHSME Problems/Problem 24

Revision as of 18:36, 2 February 2019 by Qcumber (talk | contribs) (created page + solution)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $x^2-2\sin(\theta)+1=0$ . 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 $\boxed{\textbf{D}}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_24&oldid=101158"

© 2024 AoPS Incorporated