Inverse Trigonometric Functions Question 14
Question: If $ \cos (2{{\sin }^{-1}}x)=\frac{1}{9}, $ then $ x= $
[Roorkee 1975]
Options:
A) Only 2/3
B) Only -2/3
C) 2/3, -2/3
D) Neither 2/3 nor -2/3
Show Answer
Answer:
Correct Answer: C
Solution:
$ \cos (2{{\sin }^{-1}}x)=\frac{1}{9} $
$ \Rightarrow \cos ({{\sin }^{-1}}2x\sqrt{1-x^{2}})=\frac{1}{9} $
Therefore $ \cos ({{\cos }^{-1}}\sqrt{1-4x^{2}+4x^{4}})=\frac{1}{9} $
Therefore $ 1-2x^{2}=\frac{1}{9}\Rightarrow 2x^{2}=1-\frac{1}{9}=\frac{8}{9} $
Therefore $ x^{2}=\frac{4}{9}\Rightarrow x=\pm \frac{2}{3} $ .
BETA
