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Trig problem, let sinA= -3/5 wiht A in Quad. 3 find Sin2A

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    SinA is = to -3/5 with A in Q3, find sinA?


    2. Relevant equations



    3. The attempt at a solution

    When I did this I set it up like
    First find cosA
    So to do that I used plus/minus √(1-sin^2A)
    The problem is that I really got confused when plugging in the values under the radical correctly, because i thought, -3/5 squared gives you positive 9/25 doesn't it and then the negative sign in the middle should make it 1-9/25, so you'd get 25/25-9/25 = 14/25
    and then the answer also says that since it is in quadrant 3 it is negative, so -14/25 ? right?
    Then plug that into the next step :
    Sin 2A = 2SinACosA
    =2(-3/5)(-√14/5)
    something is wrong with how I'm doing it and I'd really like to know what. :( :(
     
  2. jcsd
  3. Jul 17, 2011 #2

    eumyang

    User Avatar
    Homework Helper

    It's 16/25, not 14/25.

    No. Take the square root first, and then tack on the negative. So it's
    [itex]-\sqrt{\frac{16}{25}}=-\frac{4}{5}[/itex]
     
  4. Jul 17, 2011 #3
    omg thank u, some small silly mistake. ahhhh
    thanks for seeing that :)
     
  5. Jul 17, 2011 #4
    oh wow I thoguht it was sin(a) = -3/5 find a my bad
     
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