# How do you find the exact values of sin(u/2), cos(u/2), tan(u/2) using the half angle formulas given secu=-7/2, pi/2<u<pi?

Feb 25, 2018

See below.

#### Explanation:

Identities:

$\textcolor{red}{\boldsymbol{\sec x = \frac{1}{\cos} x}}$

$\textcolor{red}{\boldsymbol{\sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right) = \tan \left(\frac{x}{2}\right)}}$

$\textcolor{red}{\boldsymbol{\cos \left(\frac{x}{2}\right) = \sqrt{\frac{1}{2} \left(1 + \cos x\right)}}}$

$\textcolor{red}{\boldsymbol{\sin \left(\frac{x}{2}\right) = \sqrt{\frac{1}{2} \left(1 - \cos x\right)}}}$

$\sec \left(u\right) = - \frac{7}{2} \implies \frac{1}{\cos} \left(u\right) = - \frac{7}{2} \implies \cos \left(u\right) = - \frac{2}{7}$

$\sin \left(\frac{u}{2}\right) = \sqrt{\frac{1}{2} \left(1 - \left(- \frac{2}{7}\right)\right)} = \sqrt{\left(\frac{9}{14}\right)} = \frac{3}{\sqrt{14}} = \textcolor{b l u e}{\frac{3 \sqrt{14}}{14}}$

$\cos \left(\frac{u}{2}\right) = \sqrt{\frac{1}{2} \left(1 + \left(- \frac{2}{7}\right)\right)} = \sqrt{\frac{5}{14}} = \textcolor{b l u e}{\frac{\sqrt{70}}{14}}$

$\tan \left(\frac{u}{2}\right) = \frac{\frac{3 \sqrt{14}}{14}}{- \frac{\sqrt{70}}{14}} = \frac{3 \sqrt{14}}{\sqrt{70}} = \textcolor{b l u e}{\frac{3 \sqrt{5}}{5}}$

Note the signs:

If $u$ is in quadrant II

Then:

$\frac{u}{2}$ is in quadrant I