• # question_answer If $\sqrt{x}+\frac{1}{\sqrt{x}}=2\cos \theta ,$then ${{x}^{6}}+{{x}^{-6}}=$ [Karnataka CET 2003] A) $2\cos 6\theta$ B) $2\cos 12\theta$ C) $2\cos 3\theta$ D) $2\sin 3\theta$

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 ${{x}^{6}}+\frac{1}{{{x}^{6}}}+3\,(2\cos 4\theta )=8{{\cos }^{3}}4\theta$ $\Rightarrow {{x}^{6}}+\frac{1}{{{x}^{6}}}=8{{\cos }^{3}}4\theta -6\cos 4\theta$ = $2\,(4{{\cos }^{3}}4\theta -3\cos 4\theta )$= $2\cos 3(4\theta )$ = $2\cos 12\theta$.