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Question: If \(\frac{1}{x} + x = 2\cos\theta,\) then \(x^{n} + \frac{1}{x^{n}}\) is equal to...

If 1x+x=2cos⁡θ,\frac{1}{x} + x = 2\cos\theta, then xn+1xnx^{n} + \frac{1}{x^{n}} is equal to

A

2cos⁡nθ2\cos n\theta

B

2sin⁡nθ2\sin n\theta

C

cos⁡nθ\cos n\theta

D

sin⁡nθ\sin n\theta

Answer

2cos⁡nθ2\cos n\theta

Explanation

Solution

Sol. Let x=cos⁡θ+isin⁡θ=eiθx = \cos\theta + i\sin\theta = e^{i\theta} then xn+1xn=einθ+1einθx^{n} + \frac{1}{x^{n}} = e^{in\theta} + \frac{1}{e^{in\theta}}

=einθ+e−inθ=cos⁡nθ+isin⁡nθ+cos⁡nθ−isin⁡nθ=2cos⁡nθ.= e^{in\theta} + e^{- in\theta} = \cos n\theta + i\sin n\theta + \cos n\theta - i\sin n\theta = 2\cos n\theta.