1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    eumyang

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simple trig problem
  1. Simple Trig problem (Replies: 3)

  2. Simple trig problem (Replies: 3)

  3. Trig problems (Replies: 7)

  4. Simple trig problem (Replies: 4)

Loading...