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Trig Help

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data

    [itex]4 \ sin \ \theta \ = \ 3 \ csc\ \theta[/itex]

    3. The attempt at a solution

    [itex]sin\ \theta \ = \ \frac {3}{4} \ csc \ \theta[/itex]

    [itex]sin^2 \ \theta \ = \ \frac {3}{4} [/itex]

    [itex]sin \ \theta \ = \ \pm \ \frac {\sqrt{3}}{2}[/itex]

    [itex]30 \ \deg \ in \ QI, \ 150 \ \deg \ in \ QII, \ 210 \ \deg \ in \ QIII, \ 330 \ \deg \ in \ QIV[/itex]

    Am I doing this correctly?
     
  2. jcsd
  3. Oct 30, 2009 #2

    lurflurf

    User Avatar
    Homework Helper

    Almost, but I think you have confused sine and cosine
    sin(30 degrees)=1/2
    sin(60 degrees)=sqrt(3)/2
    cos(30 degrees)=sqrt(3)/2
    cos(60 degrees)=1/2
     
  4. Oct 30, 2009 #3
    I understand the difference between the two, I must have hit the wrong button in my calculator. Thanks brotha.
     
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