Problem:
If θ is a constant such that 0<θ<π and x+x1=2cosθ, then for each positive integer n,xn+xn1 equals
Answer Choices:
A. 2cosθ
B. 2ncosθ
C. 2cosnθ
D. 2cosnθ
E. 2ncosnθ
Solution:
Write x+x1=2cosθ as
x2−2xcosθ+1=0
Then x=cosθ±cos2θ−1=cosθ±isinθ(=e±iθ). By De Moivre's theorem
xn=cosnθ±isinnθ(=e±inθ)xn1=cosnθ±isinnθ1=cosnθ∓isinnθ(=e∓iθ).
Thus
xn+xn1=2cosnθ.