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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?

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    $\begingroup$ 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. $\endgroup$ – B. Goddard Feb 18 '19 at 15:17
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    $\begingroup$ Is an intervall for $\theta$ given? $\endgroup$ – Dr. Sonnhard Graubner Feb 18 '19 at 15:19
  • $\begingroup$ $\frac{\pi}{2} < \theta < \pi$ $\endgroup$ – dstarh Feb 18 '19 at 15:21
  • $\begingroup$ @B.Goddard thats what I was thinking. Just wanted to make sure I wasn't missing something large $\endgroup$ – dstarh Feb 18 '19 at 15:22
  • $\begingroup$ 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}. $\endgroup$ – DanielWainfleet Feb 18 '19 at 21:24
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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 $\left(\frac {\sqrt{26}}{26}\right).$

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  • $\begingroup$ If $\pi/2<\theta<\pi$ then $\pi/4<\theta/2<\pi/2$ so $ \cos (\theta /2)>0.$ $\endgroup$ – DanielWainfleet Feb 18 '19 at 21:31
  • $\begingroup$ Thanks! I've fixed my answer. $\endgroup$ – bjcolby15 Feb 18 '19 at 22:51
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Your answer is correct and the answer given in the working list is wrong because it's the value of $\sin \frac {\theta} {2}.$

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