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Simple trig problem

  1. Sep 11, 2013 #1
    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. jcsd
  3. Sep 11, 2013 #2
    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.
  4. Sep 11, 2013 #3
    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 ??
  5. Sep 11, 2013 #4


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    Homework Helper

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