A) \[\cos \,\,3\theta \]
B) \[2\,\cos \,3\theta \]
C) \[\frac{1}{2}\cos \,3\theta \]
D) \[\frac{1}{3}\cos \,3\theta \]
Correct Answer: B
Solution :
We have\[x+\frac{1}{x}=2\cos \theta \], Now \[{{x}^{3}}+\frac{1}{{{x}^{3}}}={{\left( x+\frac{1}{x} \right)}^{3}}-3x\frac{1}{x}\left( x+\frac{1}{x} \right)\] = \[{{(2\cos \theta )}^{3}}-3(2\cos \theta )=8{{\cos }^{3}}\theta -6\cos \theta \] = \[2(4{{\cos }^{3}}\theta -3\cos \theta )=2\cos 3\theta \]. Trick: Put \[x=1\] \[\Rightarrow \] \[\theta ={{0}^{{}^\circ }}\]. Then\[{{x}^{3}}+\frac{1}{{{x}^{3}}}=2=2\cos 3\theta \].
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