Inverse Trigonometric Functions Question 112
Question: What is the value of x that satisfies the equation $ {{\cos }^{-1}}x=2{{\sin }^{-1}}x $ ?
Options:
A) $ \frac{1}{2} $
B) $ -1 $
C) $ 1 $
D) $ -\frac{1}{2} $
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Answer:
Correct Answer: A
Given that $ {{\cos }^{-1}}x=2{{\sin }^{-1}}x $
$ \Rightarrow \frac{\pi }{2}-{{\sin }^{-1}}x=2{{\sin }^{-1}}x $
$ \Rightarrow \frac{\pi }{2}=3{{\sin }^{-1}}x\Rightarrow {{\sin }^{-1}}x=\frac{\pi }{6} $ So, $ x=\sin \frac{\pi }{6}=\frac{1}{2} $
BETA
