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Math Help - inverse function

  1. #1
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    Smile inverse function

    \mathfrak{hello}
    what is the domain of this function :
    f(x)=Arcsin(sin(x))
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  2. #2
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    Quote Originally Posted by Raoh View Post
    what is the domain of this function :
    f(x)=Arcsin(sin(x))
    domain is all real numbers
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  3. #3
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    Smile

    thanks
    but we have arcsin(x) :  \left [\frac{-\pi}{2},\frac{\pi}{2}   \right ]\mapsto \left [ -1,1 \right ]
    i don't get it,would you explain it to me,thanks a lot.
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  4. #4
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    the domain of the arcsin function is [-1,1]

    the range of the sine function is [-1,1]

    the domain of the sine function is all reals.


    look at the graph ...
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  5. #5
    Senior Member Twig's Avatar
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    Hi!

    Domain is all real numbers.

    This is because arcsin(x) : [-1,1] \to [-\frac{\pi}{2},\frac{\pi}{2}] and we have sin(x) : \mathbb{R} \to [-1,1]

    Therfore, the inner function, sin(x) , will have codomain equal to the domain of the outer function.
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  6. #6
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    Smile

    thanks to both of you.
    just to make sure,i'm going to write to you what is in my papers :
    \forall x\in \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ], Arcsin(sin(x)) = x
    that why i got confused
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  7. #7
    Senior Member Twig's Avatar
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    Quote Originally Posted by Raoh View Post
    thanks to both of you.
    just to make sure,i'm going to write to you what is in my papers :
    \forall x\in \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ], Arcsin(sin(x)) = x
    that why i got confused

    Ok, this is because if you let x=\frac{3\pi}{4} , then sin(x)=\frac{1}{\sqrt{2}}

    But arcsin(\frac{1}{\sqrt{2}})=\frac{\pi}{4} , not \frac{3\pi}{4}

    This is why it is only true when x\in [-\frac{\pi}{2},\frac{\pi}{2}]
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  8. #8
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    Smile

    Awesome,thank you
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