4
$\begingroup$

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?

$\endgroup$
  • 5
    $\begingroup$ 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 at 15:17
  • 1
    $\begingroup$ Is an intervall for $\theta$ given? $\endgroup$ – Dr. Sonnhard Graubner Feb 18 at 15:19
  • $\begingroup$ $\frac{\pi}{2} < \theta < \pi$ $\endgroup$ – dstarh Feb 18 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 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 at 21:24
3
$\begingroup$

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 $\left(\frac {\sqrt{26}}{26}\right).$

$\endgroup$
  • $\begingroup$ If $\pi/2<\theta<\pi$ then $\pi/4<\theta/2<\pi/2$ so $ \cos (\theta /2)>0.$ $\endgroup$ – DanielWainfleet Feb 18 at 21:31
  • $\begingroup$ Thanks! I've fixed my answer. $\endgroup$ – bjcolby15 Feb 18 at 22:51
0
$\begingroup$

Your answer is correct and the answer given in the working list is wrong because it's the value of $\sin \frac {\theta} {2}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.