Chapter 10.7, Problem 15E

Solution

To find: The exact value of $\sin \frac{a}{2}$ , $\cos \frac{a}{2}$ and $\tan \frac{a}{2}$.

The exact value of $\sin \frac{a}{2}$ is $-\frac{3\sqrt{10}}{\sqrt{10}}$ , $\cos \frac{a}{2}$ is $\frac{2\sqrt{13}}{13}$ and $\tan \frac{a}{2}$ is $-\frac{\sqrt{10}}{10}$.

Given information: The given value is $\sin a=-\frac{3}{5}$ , $\pi <a<\frac{3\pi }{2}$.

Formula used: The half-angle formula for the sine function is given by $\sin \frac{a}{2}=\pm \sqrt{\frac{1-\cos a}{2}}$.

The half-angle formula for the cosine function is given by $\cos \frac{a}{2}=\pm \sqrt{\frac{1+\cos a}{2}}$.

The reciprocal identity of tangent function is given by $\tan a=\frac{\sin a}{\cos a}$.

Calculation:

Use Pythagorean identity to find $\sin a$.

  ${\sin }^{2}a+{\cos }^{2}a=1$

Substitute $-\frac{3}{5}$ for $\sin a$ in the above formula.

  $\begin{array}{c}{\left(-\frac{3}{5}\right)}^2+{\cos }^{2}a=1\\ {\cos }^{2}a=1-\frac{9}{25}\\ {\cos }^{2}a=\frac{16}{25}\\ \cos a=\pm \frac{4}{5}\end{array}$

Consider negative value of cosine function because $a$ lies in the third quadrant and the cosine function is negative in third quadrant.

  $\cos a=-\frac{4}{5}$

Apply the half-angle formula of sine function to find $\sin \frac{a}{2}$ . Consider negative value of sine as it is negative in the third quadrant.

  $\begin{array}{c}\sin \frac{a}{2}=-\sqrt{\frac{1-\left(-\frac{4}{5}\right)}{2}}\\ =-\sqrt{\frac{1+\frac{4}{5}}{2}}\\ =-\sqrt{\frac{5+4}{10}}\\ =-\sqrt{\frac{9}{10}}\end{array}$

Further simplify the above expression.

  $\begin{array}{c}\sin \frac{a}{2}=-\frac{3}{\sqrt{10}}\\ =-\frac{3\sqrt{10}}{\sqrt{10}}\end{array}$

Apply the half-angle formula of cosine function to find $\cos \frac{a}{2}$ . Consider negative value of cosine as it is negative in the third quadrant.

  $\begin{array}{c}\cos \frac{a}{2}=-\sqrt{\frac{1+\left(-\frac{4}{5}\right)}{2}}\\ =-\sqrt{\frac{5-4}{10}}\\ =-\sqrt{\frac{1}{10}}\\ =-\frac{\sqrt{10}}{10}\end{array}$

Further simplify the above expression.

  $\begin{array}{c}\cos \frac{a}{2}=\sqrt{\frac{4}{13}}\\ =\frac{2}{\sqrt{13}}\\ =\frac{2\sqrt{13}}{13}\end{array}$

Apply the reciprocal identity of tangent function.

  $\begin{array}{c}\tan \frac{a}{2}=\frac{\sin \frac{a}{2}}{\cos \frac{a}{2}}\\ =\frac{\frac{3\sqrt{13}}{13}}{\frac{2\sqrt{13}}{13}}\\ =\frac{3}{2}\end{array}$

Therefore, the exact value of $\sin \frac{a}{2}$ is $-\frac{3\sqrt{10}}{\sqrt{10}}$ , $\cos \frac{a}{2}$ is $\frac{2\sqrt{13}}{13}$ and $\tan \frac{a}{2}$ is $-\frac{\sqrt{10}}{10}$.