Trig Theory Question

**aussiekid90** · March 9th 2006, 02:59 PM

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

**ThePerfectHacker** · March 9th 2006, 06:20 PM

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 $0\leq x\leq\pi$

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