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 Excercise ".$ex_no." , Question 12: find the principal value of begin align cos 1....
Question 12

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

Answer

\begin{align} Let \;\;cos^{-1}\left(\frac{1}{2}\right) =x. \;\;Then,\;\; cos x = \frac{1}{2} = cos\left(\frac{\pi}{3}\right).\end{align}

\begin{align}  \therefore cos^{-1}\left(\frac{1}{2}\right)  = \frac{\pi}{3}\end{align}

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

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

\begin{align} \therefore cos^{-1}\left(\frac{1}{2}\right) + 2sin^{-1}\left(\frac{1}{2}\right)\end{align}

\begin{align} =\frac{\pi}{3} + \frac{2\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}\end{align}

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