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Math Help - Prove a mapping is 1 to 1 and onto

  1. #1
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    Prove a mapping is 1 to 1 and onto

    The mapping T:R^2 -> R^2 is (x,y)=T(u,v)=(sinu/cosv, sinv/cosu).

    I was able to prove that the mapping was 1 to 1 by setting T(u,v)=T(u',v') and showing that u=u' and v=v'.

    I'm not sure how to show its onto. I think since I have x and y in terms of u and v, if i write u and v in terms of x and y, this will prove onto. The only problem is I am having trouble doing this.
    I was given the hint to square both x(u,v) and y(u,v) and use sinx^2 +cosx^2=1 but i cant see how this helps.

    thanks
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  2. #2
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    Quote Originally Posted by BrianMath View Post
    The mapping T:R^2 -> R^2 is (x,y)=T(u,v)=(sinu/cosv, sinv/cosu).

    I was able to prove that the mapping was 1 to 1 by setting T(u,v)=T(u',v') and showing that u=u' and v=v'.
    Maybe I am misreading or there is something awfully wrong with it.
    Is this defined

    Is this true
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  3. #3
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    Sorry, forgot the domain. Its a projection from the triangle u>0, v>0, u+v=pi/2 to the unit square (0<x<1, 0<y<1). this eliminates your issue with it not being defined

    still need help with solving for u and v in terms of x and y
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  4. #4
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    Quote Originally Posted by BrianMath View Post
    Sorry, forgot the domain. Its a projection from the triangle u>0, v>0, u+v=pi/2 to the unit square (0<x<1, 0<y<1). this eliminates your issue with it not being defined

    still need help with solving for u and v in terms of x and y
    I'm not certain this will actually help you, but I can show you how to solve the problem. Here is a sketch:



    Then use



    Replace this with sin^2(v) = 1 - cos^2(v). Now solve for cos(v).

    -Dan
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  5. #5
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    ok thanks!. am i correct in assuming that if i can solve for u and v in terms of x and y, this will show that that mapping is onto? ie. for all values of x and y s.t. (xy)^2 does not equal one, we can find a value for u and v?
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