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Simple Trig Question

  1. Jun 5, 2006 #1
    If cos x = 0.6, and 3Pie/2 < x < 2Pie, then find

    a. sin (x) and B. sin (-x)

    I figured out x to be 5.357 by trial and error but Is there another way to do this?

    Is there some sort of rule that can help me out?

    I tried using Cos(x)^2 + Sin(x)^2 = 1, but I ended up getting a different answer.
     
  2. jcsd
  3. Jun 5, 2006 #2
    x is given in radians.
     
  4. Jun 5, 2006 #3
    I'm sorry, but could you be more specific. I'm not sure what that means
     
    Last edited: Jun 5, 2006
  5. Jun 5, 2006 #4
    What you found was in degrees, i.e cos(53.13º) ~ 0.6. But the x in the problem is given in terms of radian - http://mathworld.wolfram.com/Radian.html
     
  6. Jun 5, 2006 #5
    I converted .6 radians to 34.3774677 degree.

    I don't really see the connection.
     
  7. Jun 5, 2006 #6

    TD

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    This is a good idea, since you are given cos(x) and wish to find sin(x).

    [tex]
    \cos ^2 x + \sin ^2 x = 1 \Rightarrow \sin x = \pm \sqrt {1 - \cos ^2 x}
    [/tex]

    And: sin(-x) = -sin(x).
     
  8. Jun 5, 2006 #7
    Thanks,

    I get .8, but I think the answer is -0.8 for a.

    How do I know to use the - instead of +?
     
  9. Jun 5, 2006 #8
    Actually I'm wondering why you found the angle in the first place. Just do what TD said.
     
  10. Jun 5, 2006 #9
    Oh, nevermind. I think I see it. It's because of the 3Pie/2 < x < 2Pie right?
     
  11. Jun 5, 2006 #10
    :rofl: , ok.
     
  12. Jun 5, 2006 #11

    TD

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    Indeed, sin(x) is negative in the fourth quadrant.
     
  13. Jun 5, 2006 #12
    Thanks:smile:
     
  14. Jun 5, 2006 #13

    TD

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    You're welcome :smile:
     
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