If x + frac{1}{x} = 2cos alpha , then {x^n} + | vedclass

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(d) We have, $x + \frac{1}{x} = 2\cos \alpha $

${x^2} + \frac{1}{{{x^2}}} + 2 = 4{\cos ^2}\alpha $.

${x^2} + \frac{1}{{{x^2}}} = 4{\cos ^2}\alph

Vedclass · Accepted Answer

$2\cos \,n\alpha $

Vedclass · Answer

(d) We have, $x + \frac{1}{x} = 2\cos \alpha $

${x^2} + \frac{1}{{{x^2}}} + 2 = 4{\cos ^2}\alpha $.

${x^2} + \frac{1}{{{x^2}}} = 4{\cos ^2}\alpha  - 2$ 

${x^2} + \frac{1}{{{x^2}}} = 2(2{\cos ^2}\alpha  - 1) = 2\cos 2\alpha $ 

Similarly ${x^n} + \frac{1}{{{x^n}}} = 2\cos \,n\alpha $.

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

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