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Double Angle Trig

  1. Feb 23, 2007 #1
    1. The problem statement, all variables and given/known data

    (x=theta for typing here)
    Find exact values of sin2x, cos2x, and tan2x when sec(x)=-6; 90degrees<x<180degrees

    2. Relevant equations

    sin(2x)=2sinxcosx
    etc..

    3. The attempt at a solution

    If I can figure out what I am doing wrong for just sin(x) I should be good to go.

    sec(x)=-6 therefore cos(x)=-1/6

    sin(2x)=(2)((sqr37)/36)(-1/6)

    sin2x=(-2sqr37)/36

    I have done this problem EXACTLY how one like it is done in the notes, and keep coming up with the incorrect answer. To check myself sin2x=-.338 and what I got for sin2x=-.328
     
  2. jcsd
  3. Feb 23, 2007 #2

    cristo

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    where does sqrt(37/36) come from? I presume you used sin2x+cos2x=1, and used cosx=-1/6. This would give sin2x=1-(-1/6)2=35/36.
     
  4. Feb 23, 2007 #3
    I used the pythagereon therom.

    a^2+b^2=c^2
    b^2=c^2-a^2
    b^2=6^2--1^2
    b=sqr37

    wouldn't I have to take the square root of 35/36 using pythag id to get sin(x)?


    Edit:Ok, I got the correct answer, but I don't undestand why the pythag therom did not work...
     
    Last edited: Feb 23, 2007
  5. Feb 23, 2007 #4
    what is the value of a^2 when a = -1?

    Remember you square first then subtract!!
     
  6. Feb 23, 2007 #5
    Careless mistakes can be lethal :frown:
     
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