A) \[2\cos 6\theta \]
B) \[2\cos 12\theta \]
C) \[2\cos 3\theta \]
D) \[2\sin 3\theta \]
Correct Answer: B
Solution :
Given, \[\sqrt{x}+\frac{1}{\sqrt{x}}=2\cos \theta \] ?..(i) On squaring both sides, we get \[x+\frac{1}{x}+2=4\,{{\cos }^{2}}\theta \] Þ \[x+\frac{1}{x}=4{{\cos }^{2}}\theta -2\] Þ \[x+\frac{1}{x}=\]\[2(2{{\cos }^{2}}\theta -1)\] \[=2\cos 2\theta \] ?..(ii) Again squaring both sides, \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=4{{\cos }^{2}}2\theta \] Þ \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=4{{\cos }^{2}}2\theta -2\]\[=2(2{{\cos }^{2}}2\theta -1)\] Þ \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=2\cos 4\theta \] ?..(iii) Now take cube of both sides, \[{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{3}}={{(2\cos 4\theta )}^{3}}\] Þ \[{{x}^{6}}+\frac{1}{{{x}^{6}}}+3{{x}^{2}}\times \frac{1}{{{x}^{2}}}\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=8{{\cos }^{3}}4\theta \] Þ \[{{x}^{6}}+\frac{1}{{{x}^{6}}}+3\,(2\cos 4\theta )=8{{\cos }^{3}}4\theta \] \[\Rightarrow {{x}^{6}}+\frac{1}{{{x}^{6}}}=8{{\cos }^{3}}4\theta -6\cos 4\theta \] = \[2\,(4{{\cos }^{3}}4\theta -3\cos 4\theta )\]= \[2\cos 3(4\theta )\] = \[2\cos 12\theta \].
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