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 2: find the principal value of begin align cos 1....
Question 2

Find the principal value of \begin{align} cos^{-1}\left(\frac{\sqrt3}{2}\right)\end{align}

Answer

\begin{align} Let\;\; cos^{-1}\left(\frac{\sqrt3}{2}\right)=y, \;\;Then,\;\; cos y = \frac{\sqrt3}{2} = cos\left(\frac{\pi}{6}\right)\end{align} 

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

 \begin{align} \left[0,\pi\right] and \;\;cos\left(\frac{\pi}{6}\right) = \frac{\sqrt3}{2}\end{align} 

Therefore, the principal value of

 \begin{align} cos^{-1}\left(\frac{\sqrt3}{2}\right) is \frac{\pi}{6}\end{align} 

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