Trigonometric Identities Question 145
Question: If $ x+\frac{1}{x}=2,\cos \theta , $ then $ x^{3}+\frac{1}{x^{3}}= $
[MP PET 2004]
Options:
A) $ \cos 3\theta $
B) $ 2,\cos ,3\theta $
C) $ \frac{1}{2}\cos ,3\theta $
D) $ \frac{1}{3}\cos ,3\theta $
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ x+\frac{1}{x}=2\cos \theta $ , Now $ x^{3}+\frac{1}{x^{3}}={{( x+\frac{1}{x} )}^{3}}-3x\frac{1}{x}( x+\frac{1}{x} ) $ = $ {{(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 $ .
BETA
