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Show that arcsin is the inverse of sin

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    I have a problem, I don't know to substantiate, why arcsin(sin(x)) = sin(arcsin(x)) = x ?
    Thank you very much for each advice.



    2. Relevant equations
    3. The attempt at a solution
     
  2. jcsd
  3. Sep 19, 2009 #2
    Re: arcsin(sin(x))

    i didnt exactly get your question. Are you trying to find the range of values of x for which the above eq is valid?

    what are u asking? can u be clearer.
     
  4. Sep 19, 2009 #3
    Re: arcsin(sin(x))

    My question is: why arcsin(sin(x)) is possible to regulate for x. Why arcsin(sin(x)) = x?

    Why graph for y = arcsin(sin(x)) is y = x?

    Thank you.
     
  5. Sep 19, 2009 #4
    Re: arcsin(sin(x))

    Do you know about the composition of inverse functions?
    Also, arcsin(sin(x)) = sin(arcsin(x)) = x only for -1≤x≤1 for all three.
     
  6. Sep 19, 2009 #5
    Re: arcsin(sin(x))

    [itex]\sin(\arcsin(x)) = x[/itex] for all real [itex]x[/itex] but [itex]\arcsin(\sin(x))[/itex] only for [itex]-\pi/2 \le x \le \pi/2[/itex] . That's with one standard way of defining [itex]\arcsin[/itex]
     
  7. Sep 19, 2009 #6
    Re: arcsin(sin(x))

    well look at it this way :

    sin( asin(x ) )

    let x = 1;

    the asin(1) = P;
    then sin(P) = 1;
     
  8. Sep 19, 2009 #7

    Hurkyl

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    Re: arcsin(sin(x))

    Did you mean all x in [-1,1]? (Or are you making an assertion about the complex Arcsin function?)
     
  9. Sep 20, 2009 #8
    Re: arcsin(sin(x))

    Thank you very much for all. :)
     
  10. Sep 20, 2009 #9
    Re: arcsin(sin(x))

    y=arcsin(sin(x))

    sin(y) = sin(x)

    y = x
     
  11. Sep 20, 2009 #10

    Hurkyl

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    Re: arcsin(sin(x))

    :confused:
     
  12. Sep 20, 2009 #11
    Re: arcsin(sin(x))

    You are right. For my equation you have to use complex arcsin. Wherever arcsin is defined, we have sin(arcsin(x)) = x , that is what we mean by arcsin.
     
  13. Sep 20, 2009 #12
    Re: arcsin(sin(x))

    y=arcsin[sin[x]] implies sin[y] = sin[x] and y=x. Forgetting the intervals, I don't the question had much to do with the interval which this valid for. It was just how does the graph look like y=x.
     
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