A) \[2\cos 12\theta \]
B) \[2\cos 6\,\theta \]
C) \[2\sin 3\,\theta \]
D) \[2\cos 3\,\theta \]
Correct Answer: A
Solution :
Let \[\sqrt{x}+\frac{1}{\sqrt{x}}=2\cos \theta \] \[\Rightarrow \] \[\frac{x+1}{\sqrt{x}}=2\cos \theta \] \[\Rightarrow \] \[x+1=2\sqrt{x}\cos \theta \] \[\Rightarrow \] \[x-2\sqrt{x}\cos \theta +1=0\] \[\Rightarrow \] \[(\sqrt{{{x}^{2}}})-2\cos \theta \sqrt{x}+1=0\] \[\therefore \] \[\sqrt{x}=\frac{2\cos \theta \pm \sqrt{4{{\cos }^{2}}\theta -4}}{2}\] \[=\frac{2\cos \theta \pm 2i\sin \theta }{2}\] \[[\sqrt{1-{{\cos }^{2}}\theta }=\sin \theta ]\] \[=\cos \theta \pm i\sin \theta \] \[\therefore \] \[x{{(\cos \theta \pm i\sin \theta )}^{2}}\] \[=(\cos 2\theta \pm i\sin 2\theta )\][by de-Moivres theorem \[{{(\cos \theta +i\sin \theta )}^{n}}=(\cos n\theta +i\sin n\theta )]\] \[\therefore \] \[{{x}^{6}}={{(\cos 2\theta \pm i\sin 2\theta )}^{6}}\] Now I Case : \[{{x}^{6}}={{(\cos 2\theta +i\sin 2\theta )}^{6}}\] \[=(\cos 12\theta +i\sin 12\theta )\] [De Moivres Theorem] and \[{{x}^{-6}}\frac{1}{{{(\cos 2\theta +i\sin 2\theta )}^{6}}}\] \[{{x}^{-6}}\frac{1}{\cos 12\theta +i\sin 12\theta }=\cos 12\theta -i\sin 12\theta \] \[\therefore \] \[{{x}^{6}}+{{x}^{-6}}=(\cos \,12\theta +i\sin 12\theta )\] \[+\frac{1}{\cos 12\theta +i\sin 12\theta }\] \[=\cos 12\theta +i\sin 12\theta +\cos 12\theta -i\sin 12\theta \] \[=2\cos 12\theta \] Now II Case : \[x=\cos 2\theta -i\sin 2\theta \] \[{{x}^{6}}=\cos 12\theta -i\sin 12\theta \] [By de-Moivres theorem] \[\therefore \] \[{{x}^{-6}}=\frac{1}{{{(\cos 2\theta -i\sin 2\theta )}^{6}}}\] \[=\frac{1}{\cos 12\theta -i\sin 12\theta }\] \[=\cos 12\theta +i\sin 12\theta \] \[\therefore \] \[{{x}^{6}}+{{x}^{-6}}=\cos 12\theta -i\sin 12\theta \] \[+\cos 12\theta +i\sin \theta \] \[=2\cos 12\theta \] So, in both case \[{{x}^{6}}+{{x}^{-6}}=2\cos 12\theta \]
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