(B) Given that $\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta$. Squaring both sides,we get $x+\frac{1}{x}+2=4 \cos^2 \theta$. Thus,$x+\frac{1}{x}=2(2 \cos

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(B) Given that $\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta$. Squaring both sides,we get $x+\frac{1}{x}+2=4 \cos^2 \theta$. Thus,$x+\frac{1}{x}=2(2 \cos^2 \theta - 1) = 2 \cos 2 \theta$. Let $x = e^{i2\theta}$,then $x^n + x^{-n} = 2 \cos(n \cdot 2\theta) = 2 \cos(2n\theta)$. For $n=6$,$x^6 + x^{-6} = 2 \cos(2 \times 6 \theta) = 2 \cos 12 \theta$.

Class 11 Mathematics 4-1.Complex numbers De Moivre's theorem and Roots of unity

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If $\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta$,then $x^6+x^{-6}=$

TS EAMCET 2020 All TS EAMCET

A
$2 \cos 6 \theta$
B
$2 \cos 12 \theta$
C
$2 \cos 3 \theta$
D
$2 \sin 3 \theta$

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4-1.Complex numbers

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De Moivre's theorem and Roots of unity

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