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Math Help - Trig Theory Question

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    Trig Theory Question

    In order to have an inverse a normal function must be one to one, What must happen to the trigonometric functions in order for them to have inverses.

    Thanks alot
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  2. #2
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    Quote Originally Posted by aussiekid90
    In order to have an inverse a normal function must be one to one, What must happen to the trigonometric functions in order for them to have inverses.

    Thanks alot
    Actually, the proper terminology is a bijective function but I will not go into that. (If the function is countinous and one-to-one then it has an inverse).

    Yes, dreadfully true it is that the trigonometric functions have no inverses because they are not one to one. However, what we can do is restrict the domain. We define the y=\sin^{-1}(x) (not reciprocal but rather inverse) as that \sin(y)=x,-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}, notice that this is the inverse on this interval.

    Checking the requirements for the interval [-\pi/2,\pi/2] we see that,
    \sin(\sin^{-1}(x))=\sin^{-1}(\sin(x))=x.

    Similar things are done to the other 5 trigonomteric functions, however, some domains are restricted differently than others. For example, the domain of \cos^{-1}(x) is 0\leq x\leq\pi
    Last edited by ThePerfectHacker; March 9th 2006 at 06:23 PM.
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