# Implied domain

• Mar 13th 2010, 04:19 AM
Stroodle
Implied domain
Hi there, I'm having a little trouble with working out the implied domain of this equation:

$\displaystyle y=tan[2Sin^{-1}(x)]$

I thought that since the range of $\displaystyle 2Sin^{-1}(x)$ must be a subset or equal to the domain of $\displaystyle tan(x)$, then $\displaystyle -\frac{\pi}{2}<2Sin^{-1}(x)<\frac{\pi}{2}$. Making the implied domain $\displaystyle x\in \left (-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right )$

But apparently this is wrong...

• Mar 13th 2010, 04:46 AM
Hello Stroodle
Quote:

Originally Posted by Stroodle
Hi there, I'm having a little trouble with working out the implied domain of this equation:

$\displaystyle y=tan[2Sin^{-1}(x)]$

I thought that since the range of $\displaystyle 2Sin^{-1}(x)$ must be a subset or equal to the domain of $\displaystyle tan(x)$, then $\displaystyle -\frac{\pi}{2}<2Sin^{-1}(x)<\frac{\pi}{2}$. Making the implied domain $\displaystyle x\in \left (-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right )$

But apparently this is wrong...

The range of $\displaystyle \arcsin (x)$ is $\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. So $\displaystyle -\pi \le 2\arcsin(x) \le \pi$. These are therefore potential values for the domain, but you must remember that $\displaystyle \tan \theta$ has problems when $\displaystyle \theta = \pm\frac{\pi}{2}$.

• Mar 13th 2010, 04:55 AM
Stroodle
Ahh, now I get it. Thanks for your help.

I actually thought that it was $\displaystyle Tan(x)$ not $\displaystyle tan(x)$.

If this was the case, would my answer/method have been correct?

Thanks again.
• Mar 13th 2010, 10:53 AM
Hello Stroodle
Quote:

Originally Posted by Stroodle
Ahh, now I get it. Thanks for your help.

I actually thought that it was $\displaystyle Tan(x)$ not $\displaystyle tan(x)$.

If this was the case, would my answer/method have been correct?

Thanks again.

Sorry, is this some meaning of Tan that I don't know about? What's the difference between Tan x and tan x ?

• Mar 13th 2010, 01:13 PM
Stroodle
Oh. Our teacher taught us that $\displaystyle Tan(x)$ is the same as $\displaystyle tan(x)$ except that its domain is restricted to $\displaystyle x\in \left (-\frac{\pi}{2},\frac{\pi}{2}\right )$
• Mar 13th 2010, 01:46 PM
Krizalid
what a ridiculous thing, it's just a matter of upper and lowercase, it's just absurd.
• Mar 13th 2010, 11:01 PM
Domain of tan x
Has anyone else come across this 'convention'?

• Mar 13th 2010, 11:06 PM
Stroodle
Just did a quick google and found this Complex Analysis/Elementary Functions/Inverse Trig Functions - Wikibooks, collection of open-content textbooks and http://docs.google.com/viewer?a=v&q=...jqEaNUEA&pli=1

I guess this is what my teacher was talking about...

Back to my question though; if the domain of $\displaystyle tan(x)$ was restricted to $\displaystyle x\in \left ( -\frac{\pi}{2},\frac{\pi}{2}\right )$ would my original answer be correct?

• Mar 14th 2010, 12:53 AM
Hello Stroodle
Quote:

Originally Posted by Stroodle
Just did a quick google and found this Complex Analysis/Elementary Functions/Inverse Trig Functions - Wikibooks, collection of open-content textbooks and Powered by Google Docs

I guess this is what my teacher was talking about...

Back to my question though; if the domain of $\displaystyle tan(x)$ was restricted to $\displaystyle x\in \left ( -\frac{\pi}{2},\frac{\pi}{2}\right )$ would my original answer be correct?

$\displaystyle \text{Tan }(2\arcsin(x))$
would indeed be $\displaystyle \left ( -\frac{\pi}{2},\frac{\pi}{2}\right )$