Chapter 10.7, Problem 14E

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 $\frac{3\sqrt{13}}{13}$ , $\cos \frac{a}{2}$ is $\frac{2\sqrt{13}}{13}$ and $\tan \frac{a}{2}$ is $\frac{3}{2}$.

Given information: The given value is $\sin a=\frac{12}{13}$ , $\frac{\pi }{2}<a<\pi$.

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 $\frac{12}{13}$ for $\sin a$ in the above formula.

  $\begin{array}{c}{\left(\frac{12}{13}\right)}^2+{\cos }^{2}a=1\\ {\cos }^{2}a=1-\frac{144}{169}\\ {\cos }^{2}a=\frac{25}{169}\\ \cos a=\pm \frac{5}{13}\end{array}$

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

  $\cos a=-\frac{5}{13}$

Divide the inequality by $2$.

  $\begin{array}{c}\frac{\pi }{2\cdot 2}<\frac{a}{2}<\frac{\pi }{2}\\ \frac{\pi }{4}<\frac{a}{2}<\frac{\pi }{2}\end{array}$

The angle $\frac{a}{2}$ lies in the first quadrant. Sine and cosine are positive in the first quadrant.

Apply the half-angle formula of sine function to find $\sin \frac{a}{2}$.

  $\begin{array}{c}\sin \frac{a}{2}=\sqrt{\frac{1-\left(-\frac{5}{13}\right)}{2}}\\ =\sqrt{\frac{1+\frac{5}{13}}{2}}\\ =\sqrt{\frac{13+5}{26}}\\ =\sqrt{\frac{18}{26}}\end{array}$

Further simplify the above expression.

  $\begin{array}{c}\sin \frac{a}{2}=\sqrt{\frac{9}{13}}\\ =\frac{3}{\sqrt{13}}\\ =\frac{3\sqrt{13}}{13}\end{array}$

Apply the half-angle formula of cosine function to find $\cos \frac{a}{2}$.

  $\begin{array}{c}\cos \frac{a}{2}=\sqrt{\frac{1+\left(-\frac{5}{13}\right)}{2}}\\ =\sqrt{\frac{1-\frac{5}{13}}{2}}\\ =\sqrt{\frac{13-5}{26}}\\ =\sqrt{\frac{8}{26}}\end{array}$

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 $\frac{3\sqrt{13}}{13}$ , $\cos \frac{a}{2}$ is $\frac{2\sqrt{13}}{13}$ and $\tan \frac{a}{2}$ is $\frac{3}{2}$.