Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Prove the following\[3\sin^{-1}x=\sin^{-1}(3x-4x^3)\;x\in\bigg[-\frac{1}{2},\frac{1}{2}\bigg]\]

Can you answer this question?

1 Answer

0 votes
  • \( sin\: 3A=3\: sinA-4sin^3A\)
  • \( sin^{-1}sinx=x\: \: \: \: if\: x \: \in\: \bigg[ \large\frac{-\pi}{2} \large\frac{\pi}{2} \bigg] \)
Let $sin^{-1}x = A, \rightarrow x = sin A$.
R.H.S.: $sin^{-1}(3x - 4x^3)$
Substituting for $x = sinA$, R.H.S. = \( sin^{-1} \bigg[ 3sinA-4sin^3A \bigg] \)
Substituting the value of \( sin\: 3A=3\: sinA-4sin^3A\), we get
R.H.S.:$ sin^{-1}(sin 3A) \rightarrow 3A$.
L.H.S. $3 sin^{-1} x$
Substituting the value of $sin^{-1}x = A$, L.H.S.= $3A$.
Therefore, L.H.S. = R.H.S.


answered Feb 22, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App