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

May 23, 2017

$\sin \left(\frac{u}{2}\right) = \sqrt{\frac{1 + 3 \sqrt{10}}{2}}$, $\cos \left(\frac{u}{2}\right) = - \sqrt{\frac{1 - 3 \sqrt{10}}{2}}$ and $\tan \left(\frac{u}{2}\right) = - \sqrt{\frac{1 + 3 \sqrt{10}}{1 - 3 \sqrt{10}}}$

#### Explanation:

As $\cot u = 3$ and we have $\pi < u < \frac{3 \pi}{2}$ i.e. $u$ is in $Q 3$

for $\frac{\pi}{2}$ we have $\frac{\pi}{2} < \frac{u}{2} < \frac{3 \pi}{4}$

i.e. $\frac{u}{2}$ is in $Q 2$ and while $\sin \left(\frac{u}{2}\right)$ is positive, $\cos \left(\frac{u}{2}\right)$ and $\tan \left(\frac{u}{2}\right)$ are negative.

As $\cot u = 3$, we have $\tan u = \frac{1}{3}$ and $\csc u = - \sqrt{1 + {\cot}^{2} u} = - \sqrt{10}$ i.e. $\sin u = - \frac{1}{\sqrt{10}}$ and $\cos u = \sin \frac{u}{\tan} u = \frac{- \frac{1}{\sqrt{10}}}{\frac{1}{3}} = - \frac{3}{\sqrt{10}}$.

Now $\cos u = 2 {\cos}^{2} \left(\frac{u}{2}\right) - 1$ gives us $2 {\cos}^{2} \left(\frac{u}{2}\right) = 1 + \left(- 3 \sqrt{10}\right)$

and $\cos \left(\frac{u}{2}\right) = - \sqrt{\frac{1 - 3 \sqrt{10}}{2}}$

Similarly from $\cos u = 1 - 2 {\sin}^{2} \left(\frac{u}{2}\right)$ we get

$\sin \left(\frac{u}{2}\right) = \sqrt{\frac{1 - \cos u}{2}} = \sqrt{\frac{1 + 3 \sqrt{10}}{2}}$

and $\tan \left(\frac{u}{2}\right) = - \sqrt{\frac{1 + 3 \sqrt{10}}{1 - 3 \sqrt{10}}}$