# If Cos(t)=-7/8 with pi<=t<=3pi/2, how do you find the value of sin(t/2)?

Mar 24, 2016

$\frac{\sqrt{15}}{4}$

#### Explanation:

Apply the trig identity: $\cos 2 t = 1 - 2 {\sin}^{2} t$ -->
$\cos t = - \frac{7}{8} = 1 - 2 {\sin}^{2} \left(\frac{t}{2}\right)$
$2 {\sin}^{2} \left(\frac{t}{2}\right) = 1 + \frac{7}{8} = \frac{15}{8}$
${\sin}^{2} \left(\frac{t}{2}\right) = \frac{15}{16}$
$\sin \left(\frac{t}{2}\right) = \pm \frac{\sqrt{15}}{4}$
t is in Quadrant III, then, t/2 is in Quadrant II. Its sin is positive.
Therefor, $\sin \left(\frac{t}{2}\right) = \frac{\sqrt{15}}{4}$
Check by calculator
$\cos t = - \frac{7}{8} = - - 0.875$ --> $t = {151}^{\circ} 04$
$\frac{t}{2} = {75}^{\circ} 52$ --> $\sin \left(\frac{t}{2}\right) = \sin 75.52 = 0.968 = \frac{\sqrt{15}}{4}$. OK