# How do you find the exact values of sin(u/2), cos(u/2), tan(u/2) using the half angle formulas given tanu=-5/8, (3pi)/2<u<2pi?

Feb 5, 2017

$\sin \left(\frac{u}{2}\right) = \frac{\sqrt{89} - 8}{\sqrt{178 - 16 \sqrt{89}}}$
cos(u/2)=-5/sqrt((178-16sqrt89)
$\tan \left(\frac{u}{2}\right) = - \frac{\sqrt{89} - 8}{5}$

#### Explanation:

As tanu=(2tan(u/2))/(1-tan^2(u/2)

$\frac{2 \tan \left(\frac{u}{2}\right)}{\left(1 - {\tan}^{2} \left(\frac{u}{2}\right)\right)} = - \frac{5}{8}$

or 16tan(u/2))=-5+5tan^2(u/2)

or $5 {\tan}^{2} \left(\frac{u}{2}\right) - 16 \tan \left(\frac{u}{2}\right) - 5 = 0$

and using quadratic formula

$\tan \left(\frac{u}{2}\right) = \frac{- \left(- 16\right) \pm \sqrt{{\left(- 16\right)}^{2} - 4 \times 5 \times \left(- 5\right)}}{10}$

= $\frac{16 \pm \sqrt{256 + 100}}{10} = \frac{16 \pm \sqrt{356}}{10} = \frac{8 \pm \sqrt{89}}{5}$

As $\frac{3 \pi}{2} < u < 2 \pi$, we have $\frac{3 \pi}{4} < \frac{u}{2} < \pi$ i.e. $\frac{u}{2}$ lies in Q2

and hence $\tan \left(\frac{u}{2}\right)$ is negative i.e. $\tan \left(\frac{u}{2}\right) = - \frac{\sqrt{89} - 8}{5}$

and $\sec \left(\frac{u}{2}\right) = - \sqrt{1 + {\left(\sqrt{89} - 8\right)}^{2} / 25}$

= $- \sqrt{1 + \frac{89 + 64 - 16 \sqrt{89}}{25}}$

= $- \sqrt{\frac{178 - 16 \sqrt{89}}{25}}$

and cos(u/2)=-5/sqrt((178-16sqrt89)

and $\sin \left(\frac{u}{2}\right) = \cos \left(\frac{u}{2}\right) \tan \left(\frac{u}{2}\right) = - \frac{5}{\sqrt{\left(178 - 16 \sqrt{89}\right)}} \times \left(- \frac{\sqrt{89} - 8}{5}\right)$

= $\frac{\sqrt{89} - 8}{\sqrt{178 - 16 \sqrt{89}}}$