# Use the information about the angle θ to find the exact value of?

## a. sin(θ/2) b. cos(θ/2) c. tan(θ/2) cot θ = 3/4, $\pi$ < θ < $\frac{3 \pi}{2}$

May 7, 2018

$\sin \left(\frac{\theta}{2}\right) = \frac{\sqrt{10}}{5}$

$\cos \left(\frac{\theta}{2}\right) = \frac{\sqrt{10}}{10}$

$\tan \left(\frac{\theta}{2}\right) = 2$

#### Explanation:

Given: $\cot \left(\theta\right) = \frac{3}{4} , \pi < \theta < \frac{3 \pi}{2}$

We need $\cos \left(\theta\right)$ for the formulas of the half-angles :

$\cos \left(\theta\right) = \cos \left({\cot}^{-} 1 \left(x\right)\right)$

$\cos \left(\theta\right) = \pm \frac{x}{\sqrt{{x}^{2} + 1}}$

Substitute $x = \frac{3}{4}$:

$\cos \left(\theta\right) = \pm \frac{\frac{3}{4}}{\sqrt{{\left(\frac{3}{4}\right)}^{2} + 1}}$

$\cos \left(\theta\right) = \pm \frac{\frac{3}{4}}{\sqrt{{\left(\frac{3}{4}\right)}^{2} + 1}}$

cos(theta) = +-3/sqrt((3^2+4^2)

$\cos \left(\theta\right) = \pm \frac{3}{5}$

The cosine function is negative in the third quadrant:

$\cos \left(\theta\right) = - \frac{3}{5} \text{ [1]}$

For the half-angle formulas, please observe that:

$\pi < \theta < \frac{3 \pi}{2} \to \frac{\pi}{2} < \frac{\theta}{2} < \frac{3 \pi}{4}$

This means that sine and cosine of the half-angles are positive.

$\sin \left(\frac{\theta}{2}\right) = \frac{\sqrt{1 - \cos \left(\theta\right)}}{2}$

Substitute equation [1]:

$\sin \left(\frac{\theta}{2}\right) = \frac{\sqrt{1 + \left(\frac{3}{5}\right)}}{2}$

$\sin \left(\frac{\theta}{2}\right) = \frac{\sqrt{10}}{5}$

$\cos \left(\frac{\theta}{2}\right) = \frac{\sqrt{1 + \cos \left(\theta\right)}}{2}$

Substitute equation [1]:

$\cos \left(\frac{\theta}{2}\right) = \frac{\sqrt{1 - \frac{3}{5}}}{2}$

$\cos \left(\frac{\theta}{2}\right) = \frac{\sqrt{10}}{10}$

$\tan \left(\frac{\theta}{2}\right) = \sin \frac{\frac{\theta}{2}}{\cos} \left(\frac{\theta}{2}\right)$

$\tan \left(\frac{\theta}{2}\right) = \frac{\frac{\sqrt{10}}{5}}{\frac{\sqrt{10}}{10}}$

$\tan \left(\frac{\theta}{2}\right) = 2$

May 7, 2018

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

#### Explanation:

$\cot t = \frac{3}{4}$ --> $\tan t = \frac{4}{3}$
First find cos t.
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{16}{9}} = \frac{9}{25}$
$\cos t = \pm \frac{3}{5}$.
Since t is in Quadrant 3, then, cos t is negative
$\cos t = - \frac{3}{5}$.
To find $\sin \left(\frac{t}{2}\right)$ and $\cos \left(\frac{t}{2}\right)$ use these identities:
$2 {\cos}^{2} \left(\frac{t}{2}\right) = 1 + \cos t$, and
$2 {\sin}^{2} \left(\frac{t}{2}\right) = 1 - \cos t$.
In this case:
a. $2 {\sin}^{2} \left(\frac{t}{2}\right) = 1 - \cos t = 1 + \frac{3}{5} = \frac{8}{5}$
${\sin}^{2} \left(\frac{t}{2}\right) = \frac{8}{10}$
$\sin \left(\frac{t}{2}\right) = \pm 2 \frac{\sqrt{2}}{\sqrt{10}}$
Because t lies in Q.3, then, $\left(\frac{t}{2}\right)$ lies in Q.2, and $\sin \left(\frac{t}{2}\right)$ is positive
b. $2 {\cos}^{2} \left(\frac{t}{2}\right) = 1 + \cos t = 1 - \frac{3}{5} = \frac{2}{5}$
${\cos}^{2} \left(\frac{t}{2}\right) = \frac{2}{10}$
$\cos \left(\frac{t}{2}\right) = \pm \frac{\sqrt{2}}{\sqrt{10}}$
Because $\left(\frac{t}{2}\right)$ lies in Q. 2, therefor, $\cos \frac{t}{2}$ is negative.
$\cos \left(\frac{t}{2}\right) = - \frac{\sqrt{2}}{\sqrt{10}}$
c. $\tan \left(\frac{t}{2}\right) = \frac{\sin}{\cos} = \left(\frac{2 \sqrt{2}}{\sqrt{10}}\right) \left(\frac{\sqrt{10}}{-} \sqrt{2}\right)$.
$\tan \left(\frac{t}{2}\right) = - 2$