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.

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- Mar 9th 2006, 02:59 PMaussiekid90Trig 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 - Mar 9th 2006, 06:20 PMThePerfectHackerQuote:

Originally Posted by**aussiekid90**

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 $\displaystyle y=\sin^{-1}(x)$ (not reciprocal but rather inverse) as that $\displaystyle \sin(y)=x,-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$, notice that this is the inverse on this interval.

Checking the requirements for the interval $\displaystyle [-\pi/2,\pi/2]$ we see that,

$\displaystyle \sin(\sin^{-1}(x))=\sin^{-1}(\sin(x))=x$.

Similar things are done to the other 5 trigonomteric functions, however, some domains are restricted differently than others. For example, the domain of $\displaystyle \cos^{-1}(x)$ is $\displaystyle 0\leq x\leq\pi$