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Question

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

If x+1x=2cosαx + \frac{1}{x} = 2\cos\alpha, then xn+1xn=x^{n} + \frac{1}{x^{n}} =

A

2ncosα2^{n}\cos\alpha

B

2ncosnα2^{n}\cos n\alpha

C

2isinnα2i\sin n\alpha

D

2cosnα2\cos n\alpha

Answer

2cosnα2\cos n\alpha

Explanation

Solution

We have, x+1x=2cosαx + \frac{1}{x} = 2\cos\alpha

x2+1x2+2=4cos2αx^{2} + \frac{1}{x^{2}} + 2 = 4\cos^{2}\alpha.

x2+1x2=4cos2α2x^{2} + \frac{1}{x^{2}} = 4\cos^{2}\alpha - 2,

x2+1x2=2(2cos2α1)=2cos2αx^{2} + \frac{1}{x^{2}} = 2(2\cos^{2}\alpha - 1) = 2\cos 2\alpha

Similarly xn+1xn=2cosnαx^{n} + \frac{1}{x^{n}} = 2\cos n\alpha.