Class 12 Mathematics - Chapter Inverse Trigonometric Functions NCERT Solutions | Find the principal value of \begin{align

Welcome to the NCERT Solutions for Class 12th Mathematics - Chapter Inverse Trigonometric Functions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 1: find the principal value of begin align sin 1....
Question 1

Find the principal value of \begin{align} sin^{-1}\left(-\frac{1}{2}\right)\end{align}

Answer

\begin{align} sin^{-1}\left(-\frac{1}{2}\right)=y \;\;Then\;\; sin y = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align} 

We know that the range of the principal value branch of sin−1 is

\begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and \;\;sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\end{align} 

Therefore, the principal value of

\begin{align} sin^{-1}\left(-\frac{1}{2}\right) is -\frac{\pi}{6}\end{align} 

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