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  • Please solve rd sharma class 12 chapter inverse trigonometric functions exercise 3.2 question 4 sub question (iii) maths textbook solution

Please solve rd sharma class 12 chapter inverse trigonometric functions exercise 3.2 question 4 sub question (iii) maths textbook solution

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Answer: \frac{5\pi }{6}

Hint: The range of the principal value branch of  \cos ^{-1} is \left [ 0,\pi \right ]

Given:    \cos ^{-1}\left ( \sin \left ( \frac{4\pi }{3} \right ) \right )

Solution:

\cos ^{-1}\left ( \sin \left ( \frac{4\pi }{3} \right ) \right )=\cos ^{-1}\left ( \sin \left ( \pi +\frac{\pi }{3} \right ) \right )

                                       =\cos^{-1} \left ( \frac{-\sqrt{3} }{2} \right )

For any  x\in \left [ -1,1 \right ],\cos ^{-1}x  represents an angle in \left ( 0,\pi \right )                

           \cos^{-1} \left ( \frac{-\sqrt{3} }{2} \right )=\pi -\frac{\pi }{6}=\frac{5\pi }{6}                              

Therefore, the principal value of  \cos ^{-1}\left ( \sin \left ( \frac{4\pi }{3} \right ) \right ) is \frac{5\pi }{6}.

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