Trigonometric Identities Question 322
Question: If $ \sqrt{x}+\frac{1}{\sqrt{x}}=2\cos \theta , $ then $ x^{6}+{x^{-6}}= $
[Karnataka CET 2003]
Options:
A) $ 2\cos 6\theta $
B) $ 2\cos 12\theta $
C) $ 2\cos 3\theta $
D) $ 2\sin 3\theta $
Show Answer
Answer:
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, $ {{( x^{2}+\frac{1}{x^{2}} )}^{3}}={{(2\cos 4\theta )}^{3}} $
Þ $ x^{6}+\frac{1}{x^{6}}+3x^{2}\times \frac{1}{x^{2}}( x^{2}+\frac{1}{x^{2}} )=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 $ .
BETA
