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

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(D) Given $x + \frac{1}{x} = 2\cos \alpha$.Multiplying by $x$, we get $x^2 - (2\cos \alpha)x + 1 = 0$.Using the quadratic formula, $x = \frac{2\cos \a

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$2\cos n\alpha$

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(D) Given $x + \frac{1}{x} = 2\cos \alpha$.Multiplying by $x$, we get $x^2 - (2\cos \alpha)x + 1 = 0$.Using the quadratic formula, $x = \frac{2\cos \alpha \pm \sqrt{4\cos^2 \alpha - 4}}{2} = \cos \alpha \pm i\sin \alpha$.Let $x = e^{i\alpha} = \cos \alpha + i\sin \alpha$. Then $\frac{1}{x} = e^{-i\alpha} = \cos \alpha - i\sin \alpha$.By De Moivre's Theorem, $x^n = e^{in\alpha} = \cos n\alpha + i\sin n\alpha$ and $\frac{1}{x^n} = e^{-in\alpha} = \cos n\alpha - i\sin n\alpha$.Adding these, we get ${x^n} + \frac{1}{{{x^n}}} = (\cos n\alpha + i\sin n\alpha) + (\cos n\alpha - i\sin n\alpha) = 2\cos n\alpha$.

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

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