# Simple trig problem

1. Sep 11, 2013

1. The problem statement, all variables and given/known data

We have cosx = -7/9 in the third quadrant, and my question is how to find cos(x/2) and sin(x/2).

I've tried using the sin^2x=1-cos2x/2 formula and its adjacent cosine formula. Is this the correct formula I should be using to get the answer for this question?

2. Sep 11, 2013

### MrAnchovy

Yes. You say you tried, what happened? Try rewriting the equality as $\sin^2\frac{x}{2} = \frac{1 - \cos x}{2}$; make sure your answer has the correct sign.

3. Sep 11, 2013

It worked for the sin(x/2), but when I tried for the cos(x/2)... I got 1+(-7/9) / 2 which becomes sqrt(2/18). Is this what you get ??

4. Sep 11, 2013

### eumyang

You can simplify $\sqrt{\frac{2}{18}}$, you know. :tongue: Also, there's one little thing you're forgetting. MrAnchovy mentions it in his post.