Class 12 Mathematics - Chapter Inverse Trigonometric Functions NCERT Solutions | Find the principal value of \begin{align
Find the principal value of \begin{align} cos^{-1}\left(\frac{1}{2}\right) + 2sin^{-1}\left(\frac{1}{2}\right)\end{align}
\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}
More Questions From Class 12 Mathematics - Chapter Inverse Trigonometric Functions
- Q:- Find the principal value of \begin{align} tan^{-1}\left(-\sqrt3\right)\end{align}
- Q:- Find the principal value of \begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right)\end{align}
- Q:- Find the value of if \begin{align} tan^{-1}\sqrt3 - sec^{-1}(-2)\end{align}
is equal to
\begin{align} (A) \pi \;\;(B) -\frac{\pi}{3}\;\; (C) \frac{\pi}{3} \;\;(D) \frac{2\pi}{3}\end{align} - Q:- Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right)\end{align}
- Q:- Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{2}\right)\end{align}
- Q:- Find the principal value of \begin{align} sin^{-1}\left(-\frac{1}{2}\right)\end{align}
- Q:- Find the principal value of \begin{align} cos^{-1}\left(\frac{\sqrt3}{2}\right)\end{align}
- Q:- Find the principal value of \begin{align} tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}
- Q:- Find the principal value of \begin{align} cot^{-1}\left(\sqrt3\right)\end{align}
- Q:- Find the principal value of \begin{align} cosec^{-1}\left({-\sqrt2}\right)\end{align}
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