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Inverse trig functions with tan-1

  1. Jul 1, 2010 #1
    1. The problem statement, all variables and given/known data
    sin(tan-1(x))


    2. Relevant equations



    3. The attempt at a solution
    y=tan-1(x)
    tan(y)=x
    sec2(y)= 1+tan2(y)
    sec(y)=[tex]\sqrt{1+x^2}[/tex]
    This is where I'm getting stuck. I know that I have to say that the sin(y)= whatever, but I'm not sure how to tie the sin sec and tan together. Is there a trig identity or should I make tan= [tex]\frac{sin}{cos}[/tex]?
     
  2. jcsd
  3. Jul 1, 2010 #2

    Char. Limit

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    Gold Member

    What exactly are you trying to do here? Are you trying to differentiate the first expression? It's not clear.
     
  4. Jul 1, 2010 #3

    Mark44

    Staff: Mentor

    Draw a right triangle, labelled according to y = tan-1(x) or equivalently, tan(y) = x/1. One acute angle should be labelled y. The side opposite should be labelled x and the side adjacent should be labelled 1. From this you can figure out the hypotenuse.

    What then is sin(y)?

     
  5. Jul 1, 2010 #4
    Sorry, the problem says to simplify the expression. The final answer is supposed to be [tex]\frac{x}{\sqrt{1+x^2}}[/tex]
     
  6. Jul 1, 2010 #5

    Dick

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    Science Advisor
    Homework Helper

    sec(y)=1/cos(y). So you've got cos(y)=1/sqrt(1+x^2). To get sin(y) use sin^2(y)=1-cos^2(y).
     
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