Solving x/2 in 3rd Quadrant: Cosx=-7/9

[zaddyzad]

Homework Statement

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?

 

[pbuk]

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.

 

[zaddyzad]

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.

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 ??

 

[eumyang]

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