A) \[\cos 2n\theta \]
B) \[\sin 2n\theta \]
C) 0
D) some real number greater than 0
Correct Answer: C
Solution :
\[{{x}^{2}}-2x\cos \theta +1=0\] \[\therefore \]\[x=\frac{2\cos \theta \pm \sqrt{4{{\cos }^{2}}\theta -4}}{2}=\cos \theta \pm i\sin \theta \] Let \[x=\cos \theta +i\sin \theta \] \[\therefore \]\[{{x}^{2n}}-2{{x}^{n}}\cos n\theta +1\] \[=\cos 2n\theta +i\sin 2n\theta -2(\cos n\theta +i\sin n\theta )\]\[\cos n\theta +1\] \[=\cos 2n\theta +1-2{{\cos }^{2}}n\theta +i\]\[(\sin 2n\theta -2\sin n\theta \cos n\theta )\] \[=0+i0=0\]
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