A) \[2\cos n\theta \]
B) \[2\sin n\theta \]
C) \[\cos n\,\theta \]
D) \[\sin \,n\theta \]
Correct Answer: A
Solution :
\[x+\frac{1}{x}=2\cos \theta \] Þ \[{{x}^{2}}-2x\cos \theta +1=0\] Þ \[x=\cos \theta \pm i\sin \theta \] Þ \[{{x}^{n}}=\cos n\theta \pm i\sin n\theta \] Þ \[\frac{1}{x}=\frac{1}{\cos \theta \pm i\sin \theta }\] Þ \[\frac{1}{x}=\cos \theta \mp i\sin \theta \] Þ \[\frac{1}{{{x}^{n}}}=\cos n\theta \mp i\sin n\theta \] Thus, \[{{x}^{n}}+\frac{1}{{{x}^{n}}}=2\cos n\theta \].
You need to login to perform this action.
You will be redirected in
3 sec
