# How do you find the exact values of sin(theta/2), cos(theta) where cos theta=4/5 0<=theta<=pi/2?

Sep 23, 2016

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

#### Explanation:

$\cos t = \frac{4}{5}$. Find $\sin \left(\frac{t}{2}\right) \mathmr{and} \cos \left(\frac{t}{2}\right)$.
Use the 2 trig identities:
$2 {\cos}^{2} \left(\frac{t}{2}\right) = 1 + \cos t$
$2 {\sin}^{2} \left(\frac{t}{2}\right) = 1 - \cos t$
We have:
$2 {\cos}^{2} \left(\frac{t}{2}\right) = 1 + \cos t = 1 + \frac{4}{5} = \frac{9}{5}$
${\cos}^{2} \left(\frac{t}{2}\right) = \frac{9}{10}$
$\cos \left(\frac{t}{2}\right) = \pm \frac{3}{\sqrt{10}} = \pm \frac{3 \sqrt{10}}{10}$
Since t is in Quadrant I, $\cos \left(\frac{t}{2}\right)$ is positive --> $\cos \frac{t}{2} = \frac{3 \sqrt{10}}{10}$
$2 {\sin}^{2} \left(\frac{t}{2}\right) = 1 - \cos t = 1 - \frac{4}{5} = \frac{1}{5}$
${\sin}^{2} \left(\frac{t}{2}\right) = \frac{1}{10}$
$\sin \left(\frac{t}{2}\right) = \pm \frac{1}{\sqrt{10}} = \pm \frac{\sqrt{10}}{10}$
Since t is in quadrant I, $\sin \left(\frac{t}{2}\right)$ is positive -->
$\sin \left(\frac{t}{2}\right) = \frac{\sqrt{10}}{10}$