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  • Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 15 math

Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 15 math

Answers (1)

Answer: \frac{7}{9}

Given:

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

Hint: Try to solve the bracket portion

i.e. \sin ^{-1}  function.

Solution:

Let\ \ , y=\sin ^{-1} \frac{1}{3} \\\\ Then\ \ , \sin y=\frac{1}{3}

Now,              

\begin{array}{l} \cos y=\sqrt{1-\sin ^{2} y}, \cos y=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3} \\\\ \cos \left(2 \sin ^{-1} \frac{1}{3}\right)=\cos 2 y \\\\ \end{array}

                                   \begin{array}{l} =\cos ^{2} y-\sin ^{2} y \quad\left[\therefore \cos 2 x=\cos ^{2} x-\sin ^{2} x\right] \\\\ =\left(\frac{2 \sqrt{2}}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2} \\\\ =\frac{8}{9}-\frac{1}{9} \\\\ =\frac{7}{9} \end{array}

Posted by

infoexpert22

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