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  • Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 2 sub question (iii) maths textbook solution

Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 2 sub question (iii) maths textbook solution

Answers (1)

Answer:   \frac{2\pi }{3}

Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \cos ^{-1}\left ( \cos \frac{4\pi }{3} \right )

Explanation:

First we solve \cos \left ( \frac{4\pi }{3} \right )

            \begin{aligned} &\cos \frac{4 \pi}{3}=\cos \left(\pi+\frac{\pi}{3}\right) \\ &\therefore \cos [\pi+\theta]=-\cos \theta \end{aligned}

            \begin{aligned} &\cos \left(\pi+\frac{\pi}{3}\right)=-\frac{1}{2} \\ &\cos \left(\frac{4 \pi}{3}\right)=-\frac{1}{2} \end{aligned}

By substituting these value in \cos ^{-1}\left\{\cos \left(\frac{4 \pi}{3}\right)\right\}

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

Now,

Let        y=\cos ^{-1}\left(-\frac{1}{2}\right)

            \begin{aligned} &\cos y=-\frac{1}{2} \\ &-\cos \left(\frac{\pi}{3}\right)=\frac{1}{2} \\ &\cos \left(\pi-\frac{\pi}{3}\right)=-\frac{1}{2} \\ &\cos \left(\frac{2 \pi}{3}\right)=-\frac{1}{2} \end{aligned}

Hence, range of principal value of \cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{2 \pi}{3}\right)=-\frac{1}{2}

            \cos^{-1}(\cos\frac{2\pi}{3})=\frac{2\pi}{3}

 

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