If $\sec\theta=-\frac{13}{12}$, then find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$. The official answer differs from mine. 
Given $\sec\theta=-\frac{13}{12}$ find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$.

If the $\sec\theta$ is  $-\frac{13}{12}$ then, the $\cos \theta$ is $-\frac{12}{13}$, and the half angle formula tells us that $\cos{\frac{\theta}{2}}$ should be
$$\sqrt{\frac{1+\left(-\frac{12}{13}\right)}{2}}$$
which gives me $\sqrt{\dfrac{1}{26}}$ which rationalizes to $\dfrac{\sqrt{26}}{26}$.
The worksheet off which I'm working lists the answer as $\dfrac{5\sqrt{26}}{26}$.

Can someone explain what I've done wrong here?


The answer they gave $\left(\frac {5 \sqrt{26}}{26}\right)$ is the value for $$\sin \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1-\cos \theta}{2}}$$ however they're looking for $$\cos \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1+\cos \theta}{2}}$$
EDIT (thanks, DanielWainfleet!): For the range $\pi/2 \lt \theta \lt \pi$, $\pi/4 \lt \theta/2 \lt \pi/2$, so $\cos \dfrac {\theta}{2}$ will be positive. Thus, your answer will be $\left(\frac {\sqrt{26}}{26}\right).$
