Question

If $ x + \frac{1}{x} = 2 \cos \theta $, find the value of \[ x^5 + \frac{1}{x^5} \] in terms of $\theta$.

• A. $2 \cos 5\theta$
• B. $5 \cos \theta$
• C. $2 \cos 4\theta$
• D. $5 \cos 5\theta$

Hint:

Use the recurrence relation $a_n = (x + \frac{1}{x}) a_{n-1} - a_{n-2}$ and trigonometric identities to express powers in terms of cosines.

Answer 1

The Correct Option is A

Given: \[ x + \frac{1}{x} = 2 \cos \theta \] Define: \[ a_n = x^n + \frac{1}{x^n} \] Then, \[ a_1 = 2 \cos \theta, \quad a_0 = 2 \] Using recurrence relation: \[ a_n = (x + \frac{1}{x}) a_{n-1} - a_{n-2} = 2 \cos \theta \cdot a_{n-1} - a_{n-2} \] Calculate terms stepwise: \[ a_2 = 2 \cos \theta \cdot a_1 - a_0 = 2 \cos \theta \cdot 2 \cos \theta - 2 = 4 \cos^2 \theta - 2 = 2 \cos 2\theta \] \[ a_3 = 2 \cos \theta \cdot a_2 - a_1 = 2 \cos \theta \cdot 2 \cos 2\theta - 2 \cos \theta = 2 \cos 3\theta \] \[ a_4 = 2 \cos \theta \cdot a_3 - a_2 = 2 \cos \theta \cdot 2 \cos 3\theta - 2 \cos 2\theta = 2 \cos 4\theta \] \[ a_5 = 2 \cos \theta \cdot a_4 - a_3 = 2 \cos \theta \cdot 2 \cos 4\theta - 2 \cos 3\theta = 2 \cos 5\theta \] Therefore, \[ x^5 + \frac{1}{x^5} = a_5 = 2 \cos 5\theta \]