# Thread: having trouble with a 1/2 angle

1. ## having trouble with a 1/2 angle

Ive run into a point I cant seem to figure out. Where do I go from where I stopped at cos^2(x/2)= 6/14??

2. Originally Posted by ninobrn99

Ive run into a point I cant seem to figure out. Where do I go from where I stopped at cos^2(x/2)= 6/14??
$\displaystyle \displaystyle \cos^2\left(\frac{x}{2}\right) = \frac{3}{7}$

$\displaystyle \displaystyle \cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{3}{7}}$

since x is in quad II , which value works?

3. okay, Im not crazy then. I came up with that b4, but it wasn't one of my answer choices. Since x is in quad 2, then cos is neg and sin is posi.

Thank you.

4. Originally Posted by ninobrn99
okay, Im not crazy then. I came up with that b4, but it wasn't one of my answer choices. Since x is in quad 2, then cos is neg and sin is posi.

Thank you.
for x it is ... you need to remember that the question is asking about $\displaystyle \cos\left(\frac{x}{2}\right)$ ... care to rethink that conclusion?

5. Ive been staring at this problem and cant figure out what to do from there. I would think to set cos(1/2x)=-(sqrt3/7) but that doesnt work out either.

6. if $\displaystyle 90 < x < 180$ , then $\displaystyle 45 < \frac{x}{2} < 90$ ... so $\displaystyle \cos\left(\frac{x}{2}\right) =$ ?

7. Got it! Ugh, I've been going at this solo this past weekend trying to get ahead and was finally shown today what you meant. Thank you again for the help!