Find the principal value of egin{align | Class 12 Mathematics Chapter Inverse Trigonometric Functions, Inverse Trigonometric Functions NCERT Solutions

Question: Find the principal value of \begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right)\end{align}
Answer:

\begin{align} Let\;\; cos^{-1}\left(-\frac{1}{\sqrt2}\right)=y, \;\;Then,\;\; cos y = -\frac{1}{\sqrt2} = - cos\left(\frac{\pi}{4}\right)=cos\left(\pi - \frac{\pi}{4}\right) = cos\left(\frac{3\pi}{4}\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{3\pi}{4}\right) = -\frac{1}{\sqrt2}\end{align}

Therefore, the principal value of

 \begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right) is \frac{3\pi}{4}\end{align} 


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Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 9: Find the principal value of egin{align} cos^{-1}left(-frac{1}{sqrt2} ight)end{align}....