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Math Help - simplying trig expression

  1. #1
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    simplying trig expression

    i have to simply the expression which is:

    sin(tan-1(x))

    now i'm not exactly where to start

    what i've thought of so far is that since tan t = y/x which would be tan theta = sin/cos, tan-1 would be cos/sin but that is what cot theta is equal to.

    i'm quite stumped as it is and don't exactly understand the problem.

    thanks in advance
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by deltemis View Post
    i have to simply the expression which is:

    sin(tan-1(x))

    now i'm not exactly where to start

    what i've thought of so far is that since tan t = y/x which would be tan theta = sin/cos, tan-1 would be cos/sin but that is what cot theta is equal to.

    i'm quite stumped as it is and don't exactly understand the problem.

    thanks in advance
    Let \theta=\tan^{-1}x.

    Then if you construct a [right] triangle containing \theta, the side opposite of \theta will have a length of x, the side adjacent to \theta will have a length of 1, and the hypotenuse of the triangle will have a length of \sqrt{x^2+1}.

    Therefore, \sin\left(\tan^{-1}x\right)=\sin\theta=\frac{\text{opp}}{\text{hyp}  }=\frac{x}{\sqrt{1+x^2}}

    Does this make sense?
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  3. #3
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    thanks chris.

    i keep forgetting the basics of trig and that is to work w/ triangles
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