Difference between revisions of "1981 AHSME Problems/Problem 24"

Revision as of 14:20, 6 July 2021

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: \begin{align*} &=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta) \\ &=\boxed{\textbf{2\cos(n\theta)}}, \end{align*}

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

@@ Line 20: / Line 20: @@
 Because <math>\cos</math> is even and <math>\sin</math> is odd:
+\begin{align*}
-<cmath>=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)</cmath>
+&=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta) \\
-<cmath>=\boxed{\textbf{2\cos(n\theta)}},</cmath>
+&=\boxed{\textbf{2\cos(n\theta)}},
+\end{align*}
 which gives the answer <math>\boxed{\textbf{D}}.</math>

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Difference between revisions of "1981 AHSME Problems/Problem 24"

Revision as of 14:20, 6 July 2021

Problem

Solution