# How do you find the exact values of sin(u/2), cos(u/2), tan(u/2) using the half angle formulas given cscu=-5/3, pi<u<(3pi)/2?

May 24, 2017

$\sin \left(\frac{u}{2}\right) = 3 \frac{\sqrt{10}}{10}$
$\cos \left(\frac{u}{2}\right) = - \frac{\sqrt{10}}{10}$
$\tan \left(\frac{u}{2}\right) = - 3$

#### Explanation:

$\frac{1}{\sin u} = - \frac{5}{3}$ --> $\sin u = - \frac{3}{5}$
$\cos u = 1 - {\sin}^{2} u = 1 - \frac{9}{25} = \frac{16}{25}$ --> $\cos u = \pm \frac{4}{5}$
$\cos u = - \frac{4}{5}$ because u is in Quadrant 3.
Use trig identity:
$2 {\sin}^{2} a = 1 - \sin 2 a$.
In this case:
$2 {\sin}^{2} \left(\frac{u}{2}\right) = 1 - \cos u = 1 + \frac{4}{5} = \frac{9}{5}$
${\sin}^{2} \left(\frac{u}{2}\right) = \frac{9}{10}$
$\sin \left(\frac{u}{2}\right) = \pm \frac{3}{\sqrt{10}}$
$\sin \left(\frac{u}{2}\right) = \frac{3}{\sqrt{10}}$ --> because $\frac{u}{2}$ is in Quadrant 2
The same way -->
2cos^2 (u/2) = 1 + cos u = 1 - 4/5 = 1/5
$\cos \left(\frac{u}{2}\right) = \pm \frac{1}{\sqrt{10}}$
$\cos \left(\frac{u}{2}\right) = - \frac{1}{\sqrt{10}}$ --> because $\frac{u}{2}$ is in Quadrant 2
$\tan \left(\frac{u}{2}\right) = \frac{\sin}{\cos} = \frac{3}{-} 1 = - 3$