# Trig Theory Question

• Mar 9th 2006, 02:59 PM
aussiekid90
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
• Mar 9th 2006, 06:20 PM
ThePerfectHacker
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 :eek: 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 $\displaystyle y=\sin^{-1}(x)$ (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 $\displaystyle 0\leq x\leq\pi$