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  • Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 3 Maths Textbbok Solution.

Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 3 Maths Textbbok Solution.

Answers (1)

Answer: \frac{1}{\sqrt{10}}

Hint: First we will convert \cos ^{-1} \frac{4}{5} into \sin ^{-1}

Given: \sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)

Explanation:

Let \cos ^{-1} \frac{4}{5}=\theta

\cos \theta=\frac{4}{5}

we know that

\begin{aligned} &\cos \theta=1-2 \sin ^{2} \frac{\theta}{2} \\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \\ &\sin ^{2} \frac{\theta}{2}=\frac{1-\cos \theta}{2} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}} \end{aligned}

\begin{aligned} &\sin \frac{\theta}{2}=\sqrt{\frac{1-\frac{4}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{\frac{1}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1}{10}} \\ &\sin \frac{\theta}{2}=\frac{1}{\sqrt{10}} \\ &\frac{\theta}{2}=\sin ^{-1} \frac{1}{\sqrt{10}} \end{aligned}                                           from equation (1)

\frac{1}{2} \cos ^{-1} \frac{4}{5}=\sin ^{-1} \frac{1}{\sqrt{10}}     ......(2)

Now

\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)                                           from equation (2)

=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)

=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)                                   \left[\because \sin \left[\sin ^{-1}(\theta)\right]=\theta\right]

=\frac{1}{\sqrt{10}}

 

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