1. ## Inverse circular functions

State (i) the implied domain and (ii) the range of each of the following.

a) $\displaystyle y=sin^{-1}(x^2)$
b) $\displaystyle y=tan^{-1}(2x)$
c) $\displaystyle y=tan^{-1}(4-x)$
d) $\displaystyle y=cos^{-1}(x^2-1)$

I don't know how to do all of the above. What are the steps for undertaking the processes?

Perhaps someone can explain and then attempt one or two and then I could try to do the rest and they someone else could check whether it's right or wrong?

Anyway, whatever the case, how do you do the above?

Any help/guidance to do the above will be appreciated!

2. Originally Posted by Joker37
State (i) the implied domain and (ii) the range of each of the following.

a) $\displaystyle y=sin^{-1}(x^2)$
b) $\displaystyle y=tan^{-1}(2x)$
c) $\displaystyle y=tan^{-1}(4-x)$
d) $\displaystyle y=cos^{-1}(x^2-1)$

I don't know how to do all of the above. What are the steps for undertaking the processes?

Perhaps someone can explain and then attempt one or two and then I could try to do the rest and they someone else could check whether it's right or wrong?

Anyway, whatever the case, how do you do the above?

Any help/guidance to do the above will be appreciated!
The implied domain: the real x-values for which the function is valid.
The range: The possible y values that the function can take.

a) $\displaystyle y=sin^{-1}(x^2)$
Firstly, we know that in any situation, $\displaystyle -1\leq~Sin(\theta)\leq~1$
The arcsin function takes the value obtained for $\displaystyle Sin(\theta)$ and generates the $\displaystyle \theta$ value.
If $\displaystyle Sin(\theta)=x$ then $\displaystyle \theta=Sin^{-1}(x)$

So, as $\displaystyle -1\leq~Sin(\theta)\leq~1$

$\displaystyle Sin^{-1}(\theta)$ Is only defined for $\displaystyle -1<\theta<1$
$\displaystyle -1\leq~x^2\leq 1$
$\displaystyle -1\leq~x^2\leq 1$
$\displaystyle x^2<1$
$\displaystyle -1<x<1$

To find the range then. You know that the minimum value obtained for $\displaystyle x^2=0$ and the maximum for $\displaystyle x^2=1$

$\displaystyle Sin^{-1}(1)=90$
$\displaystyle Sin^{-1}(0)=0$

So $\displaystyle 0\leq~y\leq90$

Not entirely certain on this but that was my attempt!