1981 AHSME Problems/Problem 24
Problem
If 
is a constant such that 
and 
, then for each positive integer 
, 
equals
Solution
Multiply both sides by 
and rearrange to 
. Using the quadratic equation, we can solve for . After some simplifying:
![\[x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}\]](http://latex.artofproblemsolving.com/4/d/f/4dfaa96d4f48bb666eb8c48146e1492987ef6c2f.png)
![\[x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}\]](http://latex.artofproblemsolving.com/6/9/c/69c6dddc307c908580702605d7feffffbf3cbc9c.png)
![\[x=\cos(\theta) + i\sin(\theta)\]](http://latex.artofproblemsolving.com/e/9/5/e952a9c3624de3f2c1b1dab34054d82ca63ee950.png)
Substituting this expression in to the desired gives:
![\[(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}\]](http://latex.artofproblemsolving.com/1/4/7/147b08fd15b530e18f6e3c39644bc552a3e5dcf4.png)
Using DeMoivre's Theorem:
![\[=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)\]](http://latex.artofproblemsolving.com/1/0/a/10a4879a58d6eb32e0da94dfaf3552f7f04b2d80.png)
Because 
is even and 
is odd:
![\[=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)\]](http://latex.artofproblemsolving.com/9/c/4/9c4ab1d8bcdb50a156082686378683e6c4732c1c.png)
![\[=\boxed{2\cos(n\theta)},\]](http://latex.artofproblemsolving.com/5/0/b/50bdee4e39e193a9e8010c2118f0531fea2deb50.png)
which gives the answer 
