Chapter 7, Problem 19CT

To determine

To calculate: The exact value of $\sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)$ using sum, difference, product, or half-angle formulas.

Answer to Problem 19CT

Solution:

The exact value of $\sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)$ is $\sqrt{\frac{1}{5}}$.

Explanation of Solution

Given information:

The expression is $\sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)$.

Formula used:

Half angle formula of sine function: $\sin \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{2}}$

Calculation:

Consider the expression $\sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)$.

Compare the expression $\sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)$ and $\sin \frac{\alpha }{2}$.

$\Rightarrow \alpha ={\cos }^{-1}\frac{3}{5}$

By using the formula $\sin \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{2}}$,

$\sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)=\pm \sqrt{\frac{1-\cos \left({\cos }^{-1}\frac{3}{5}\right)}{2}}$

By using $\cos \left({\cos }^{-1}x\right)=x$ where $-1\le x\le 1$

Since $-1\le x=\frac{3}{5}\le 1$, $\cos \left({\cos }^{-1}\frac{3}{5}\right)=\frac{3}{5}$

$\Rightarrow \sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)=\pm \sqrt{\frac{1-\frac{3}{5}}{2}}$

$\Rightarrow \sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)=\pm \sqrt{\frac{\frac{2}{5}}{2}}$

$\Rightarrow \sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)=\pm \sqrt{\frac{1}{5}}$

Let $\alpha ={\cos }^{-1}\frac{3}{5}$. Then, $\cos \alpha =\frac{3}{5}$, where $0\le \alpha \le \pi$

Since $\cos \alpha >0$, it follows that $0\le \alpha \le \frac{\pi }{2}$.

So $\alpha$ lies in $I$ quadrant.

This implies that $\frac{\alpha }{2}=\frac{1}{2}{\cos }^{-1}\frac{3}{5}$ also lies in the first quadrant.

Since sine is positive in the first quadrant,

$\Rightarrow \sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)=\sqrt{\frac{1}{5}}$

Therefore, the exact value of $\sin \left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)$ is $\sqrt{\frac{1}{5}}$.