# having trouble with a 1/2 angle

• Jun 27th 2010, 04:28 PM
ninobrn99
having trouble with a 1/2 angle
Attachment 18028

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??
http://www.mathhelpforum.com/math-he...isc/pencil.png
• Jun 27th 2010, 05:25 PM
skeeter
Quote:

Originally Posted by ninobrn99
Attachment 18028

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??
http://www.mathhelpforum.com/math-he...isc/pencil.png

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

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

since x is in quad II , which value works?
• Jun 27th 2010, 05:28 PM
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.
• Jun 27th 2010, 05:57 PM
skeeter
Quote:

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 $\cos\left(\frac{x}{2}\right)$ ... care to rethink that conclusion?
• Jun 28th 2010, 04:20 AM
ninobrn99
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.
• Jun 28th 2010, 08:20 AM
skeeter
if $90 < x < 180$ , then $45 < \frac{x}{2} < 90$ ... so $\cos\left(\frac{x}{2}\right) =$ ?
• Jun 28th 2010, 02:36 PM
ninobrn99
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!