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??

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- June 27th 2010, 03:28 PMninobrn99having 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 - June 27th 2010, 04:25 PMskeeter
- June 27th 2010, 04:28 PMninobrn99
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. - June 27th 2010, 04:57 PMskeeter
- June 28th 2010, 03:20 AMninobrn99
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.

- June 28th 2010, 07:20 AMskeeter
if , then ... so ?

- June 28th 2010, 01:36 PMninobrn99
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!