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

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== Solution ==
 
== Solution ==
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Multiply both sides by <math>x</math> and rearrange to <math>x^2-2\sin(\theta)+1=0</math>. Using the quadratic equation, we can solve for <math>x</math>. After some simplifying:
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Multiply both sides by <math>x</math> and rearrange to <math>x^2-2x\cos(\theta)+1=0</math>. Using the quadratic equation, we can solve for <math>x</math>. After some simplifying:
  
 
<cmath>x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}</cmath>
 
<cmath>x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}</cmath>
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<cmath>x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta)}</cmath>
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<cmath>x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}</cmath>
 
<cmath>x=\cos(\theta) + i\sin(\theta)</cmath>
 
<cmath>x=\cos(\theta) + i\sin(\theta)</cmath>
  
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<cmath>=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)</cmath>
 
<cmath>=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)</cmath>
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<cmath>=2\cos(n\theta)</cmath>
 
  
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Which gives the answer <math>\boxed{\textbf{D}}</math>
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<cmath>=\boxed{2\cos(n\theta)},</cmath>
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which gives the answer <math>\boxed{\textbf{D}}.</math>
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==See also==
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{{AHSME box|year=1981|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 14:29, 28 June 2025

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 $\boxed{\textbf{D}}.$

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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