# 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?

• I'm pretty sure there's a typo. I think your answer is correct. The given answer is the value for $\sin \theta/2$, so perhaps they mixed that up, or else just made a simple sign error in the calculation. Feb 18, 2019 at 15:17
• Is an intervall for $\theta$ given? Feb 18, 2019 at 15:19
• $\frac{\pi}{2} < \theta < \pi$ Feb 18, 2019 at 15:21
• @B.Goddard thats what I was thinking. Just wanted to make sure I wasn't missing something large Feb 18, 2019 at 15:22
• My edit was to put brackets in the half-angle formula so you do not have "$+-$". An alternative would be to type \frac {-12}{13} instead of -\frac {12}{13}. Feb 18, 2019 at 21:24

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 $$\left(\frac {\sqrt{26}}{26}\right).$$
• If $\pi/2<\theta<\pi$ then $\pi/4<\theta/2<\pi/2$ so $\cos (\theta /2)>0.$ Feb 18, 2019 at 21:31
Your answer is correct and the answer given in the working list is wrong because it's the value of $$\sin \frac {\theta} {2}.$$