Author Topic: Inverse trig range (Read 1456 times) Tweet Share

Martoman · « **on:** February 14, 2010, 08:38:09 pm »

This is annoying. I have some arbitrary question: arcsin(cosx), with cosx domain restricted to [0,pi] and I want to find its implied domain and range. I know that the range of cosx is [-1,1] and also the domain of arcsin is [-1,1]. So its is defined. Now i know the domain of cosx = domain of the composition function = [0,pi].

This is the part I seem to fail at. I want to find the range of arcsin(cosx). So, I find the range of arcsin(x) for the domain of [0,pi] ? Then its [0,pi/2].... which fails according to textbook. THIS is the point which i need clarification on. Thanks.

Martoman · « **Reply #1 on:** February 15, 2010, 04:54:16 pm »

To any and all interested, I worked this out. To find the range, you sub in the range of the nested function to be the domain of the outer function. Then work out the range of the outer function with this restriction.

Martoman · « **Reply #2 on:** February 21, 2010, 12:27:02 am »

Yes... my last post made a lot of sense. here is what i meant.

We have $h(x) = cos(sin^{-1}(x))$

Call $f(x) = cos(x)$ and $g(x) = sin^{-1}(x)$ , do ignore the fact that x is being used twice here, this is just for a matter of simplicity.

We know that the range of the inner must be a subset or equal to the domain of the outer.
$R_g = [-\frac{\pi}{2}, \frac{\pi}{2}], and \; the D_f = [0, \pi]$

So lets restrict the range of $R_g$ to be something WITHIN $D_f$

So declare $R_g* = [0,\frac{\pi}{2}]$
THINK does this mean that it lies WITHIN $D_f$ Well of course, $R_g* = [0,\frac{\pi}{2}] \; is \; within \;D_f = [0, \pi]$ as $\frac{\pi}{2}$ is less than $\pi$

Now since we have a nicely defined composite function, the DOMAIN of the restricted inner function becomes our implied domain for the whole function.

So simply, we have the $R_g* = [0,\frac{\pi}{2}]$ which means that the domain is $D_g*=[0,1]$

NOW this is the part that is cool. To find the RANGE we sub in the range of $R_g*$ TO BE THE DOMAIN OF $f$ , that meaning that the domain of $f$ is no longer $D_f = [0, \pi]$ but is now $= R_g* = [0,\frac{\pi}{2}]$

Cool, now with this domain restriction, the $R_f = [0,1]$ This is the range of the whole function.

TrueTears · « **Reply #3 on:** February 21, 2010, 12:39:08 am »

Let $\sin^{-1}\left(\frac{x}{1}\right) = \theta$

$\cos(\theta) = \frac{\sqrt{1-x^2}}{1}$

Sketch the graph.

Done

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Author Topic: Inverse trig range (Read 1456 times) Tweet Share

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Inverse trig range

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