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

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)

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

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

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