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My question is really simple, why is the range of the $\sin^{-1}(x)$, $\cos^{-1}(x)$ and $\tan^{-1}(x)$ defined as $[-\pi/2,\pi/2]$, $[0,\pi]$ and $[-\pi/2,\pi/2]$ respectively? Is there some particular reason? We could choose another range for each inverse trigonometric function. For example, we can pick $[0,\pi]$ to be the range of $\sin^{-1}x$.

EDIT

I've understood why the range of the sine and cosine has to be $[-\pi/2,\pi/2]$ and $[0,\pi]$ respectively. I'm still wondering why can't we define the range of the tangent as $[0,\pi]$

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Why cannot $[0;\pi]$ be range of $\arcsin$?

Because at this range $\sin$ is not injective. It means that there exist $a,b\in[0;\pi],\;a\ne b$, that $\sin(a)=\sin(b)$. This is very inconvenient because $\arcsin$ would be multivalued. For one argument there would exist two values. That's why such range is selected that $\sin$ is injective and thus $\arcsin$ is a function.


Function $\sin(x)$ is periodic. It means that for every value $y$ there exist infinitely many arguments $x$ satisfying $y=\sin(x)$. Then, its inverse $\arcsin$ is multivalued. For every argument it takes infinitely many values. That's why we usually "split" it into branches, so that every branch is a function. It is usually done at points where tangent is vertical. Then we define such branches as $$...,\;\arcsin_{-1}(x),\;\arcsin_0(x),\;\arcsin_1(x),...$$

The branch $\arcsin_0(x)$ is the principal one, equivalent to $\arcsin(x)$.

The same thing as above goes with $\cos$.


Similar thing is with $\tan$. It is not injective, so we take such domain (this selected domain of $\tan$ is the range of $\arctan$) that $\tan$ is injective on it. It is $$...,\;[-3\pi/2;-\pi/2],\;[-\pi/2;\pi/2],\;[\pi/2,3\pi/2],\;...$$

Actually we can take domains such as $[0;\pi]$, but then, $\arctan$ would be not continuous.

Why the domain that is closest to $0$ is selected? Because it's convenient.

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  • $\begingroup$ ok, I understood why this range is more convenient to the sine inverse function. What about the inverse of the tangent? why can't it be $[0,\pi]$? $\endgroup$ – user42912 Dec 31 '15 at 13:19
  • $\begingroup$ the range of the tangent could be $[0,\pi]$ also, no? $\endgroup$ – user42912 Dec 31 '15 at 16:33
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    $\begingroup$ @user42912 yes, but it would be very problematic. It wouldn't be continuous at $x=0$ (see here). Also, which value ($0$ or $\pi$) choose for $x=0$? $\endgroup$ – Kamil Jarosz Dec 31 '15 at 16:36
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Inverse trigo functions are nothing but angles . Now trigo functions are periodic over $2π$ so there ranges are such also range of tan is $[-π/2,π/2]$ hope i have intrtpreted it correctly.

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You could make another choice. As I recall there are two different ways that calculus textbooks define $\sec^{-1}$.

Another answer. In complex analysis, each of these inverse trig functions may be written in terms of the complex (natural) logarithm. So take that definition, and use the principal value of the log to get the principal value for the inverse trig functions. Then restrict to the real line for baby use.

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