How do you find the exact value of sinx/2, cosx/2, & tanx/2 given that sinx= 5/13 is in Quadrant 2?

Apr 26, 2015

First find cos x by using trig identity: ${\cos}^{2} x + {\sin}^{2} x = 1.$
${\cos}^{2} x = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169}$
--> cos x = -12/13. (Quadrant II)
Next, use the trig identity: $\cos 2 a = 2. \left({\cos}^{2} a\right) - 1$.
$\cos x = 2 {\cos}^{2} \left(\frac{x}{2}\right) - 1$ -->
$2 {\cos}^{2} \left(\frac{x}{2}\right) = 1 - \frac{12}{13} = \frac{1}{13}$-->
${\cos}^{2} \left(\frac{x}{2}\right) = \frac{1}{26}$
--> $\cos \frac{x}{2} = - \frac{1}{5.1}$ (Quadrant II)
Next, use the trig identity: $\cos x = {\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right)$-->
${\sin}^{2} \left(\frac{x}{2}\right) = \frac{1}{26} + \frac{12}{13} = \frac{25}{26}$ -->
$\sin \left(\frac{x}{2}\right) = \frac{5}{5.1}$ (quadrant II)
Find $\tan \left(\frac{x}{2}\right) = \sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right) = \frac{\frac{5}{5.1}}{- \frac{1}{5.1}} = - \frac{5}{1} = - 5$