# Cannot understand what ##(mod ~ \pi)## means in the given formula

• vcsharp2003
In summary: Actually, using ##\mathbb Z## would have probably been better since some (many?) people do not include 0 in ##\mathbb N##.
vcsharp2003
Homework Statement
What is the meaning of ##(mod ~ \pi)## in the formula given below? This formula is taken from https://en.wikipedia.org/wiki/Inverse_trigonometric_functions
Relevant Equations
None

My understanding is that it probably means either ##+ \pi## or ## - \pi##, so that ##\pi## is either added to or subtracted from the first term, just like ##|x| = \pm x##.

vcsharp2003 said:
Homework Statement:: What is the meaning of ##(mod ~ \pi)## in the formula given below? This formula is taken from https://en.wikipedia.org/wiki/Inverse_trigonometric_functions
Relevant Equations:: None

View attachment 324323

My understanding is that it probably means either ##+ \pi## or ## - \pi##, so that ##\pi## is either added to or subtracted from the first term, just like ##|x| = \pm x##.
Look at the pictures here:
https://de.wikipedia.org/wiki/Arkusfunktion

##x \longmapsto \operatorname{arctan}(x)## is not only possible on its main branch between ##\pm\dfrac{\pi}{2}## but also below and above. We have ##\tan\left(\alpha + \dfrac{\pi}{2}\mathbb{Z}\right)=x,## i.e. infinitely many values of the tangent function map to the same function value. The inverse "function" can therefore be defined on any interval ##\operatorname{arctan}(x) \in \left[n\dfrac{\pi}{2},(n+1)\dfrac{\pi}{2}\right]## for some ##n\in \mathbb{Z}.##

fresh_42 said:
Look at the pictures here:
https://de.wikipedia.org/wiki/Arkusfunktion

##x \longmapsto \operatorname{arctan}(x)## is not only possible on its main branch between ##\pm\dfrac{\pi}{2}## but also below and above. We have ##\tan\left(\alpha + \dfrac{\pi}{2}\mathbb{Z}\right)=x,## i.e. infinitely many values of the tangent function map to the same function value. The inverse "function" can therefore be defined on any interval ##\operatorname{arctan}(x) \in \left[n\dfrac{\pi}{2},(n+1)\dfrac{\pi}{2}\right]## for some ##n\in \mathbb{Z}.##
Does this mean that we are adding an integer multiple of ##\pi## to the first term in the formula mentioned in question where an integer could be any positive or negative integer including 0?

vcsharp2003 said:
Does this mean that we are adding an integer multiple of ##\pi## to the first term in the formula mentioned in question where an integer could be any positive or negative integer including 0?
The inverse tangent function has only unique function values in an interval of length ##\pi##, usually, ##\left(-\dfrac{\pi}{2}\, , \,\dfrac{\pi}{2}\right).## Since ##\tan\left(\alpha+ \pi\mathbb{Z}\right)## is the same real number, we cannot uniquely determine one number as its inverse. We first have to determine in which interval our angles are. E.g. ##\operatorname{arctan} (\pi / 3)\approx 0.81.## Therefore
\begin{align*}
\operatorname{arctan}(\pi/3)+\operatorname{arctan}(\pi/3)&\approx 1.62 \\
&=\operatorname{arctan}\left(\dfrac{\dfrac{2\pi}{3}}{1-\dfrac{\pi^2}{9}}\right)\\
&=\operatorname{arctan}\left(\dfrac{6\pi}{9-\pi^2}\right)\\
&\approx \operatorname{arctan}(-21.676)\\
&\approx -1.5247
\end{align*}
which is obviously wrong. However, ##-1.5247 + \pi \approx 1.617## which is pretty close to ##1.62## if we consider all the approximations I made. So ##\mod \pi## means that the left- and right-hand side of our formulas can differ by ##\pm \pi.##

vcsharp2003
vcsharp2003 said:
Homework Statement:: What is the meaning of ##(mod ~ \pi)## in the formula given below? This formula is taken from https://en.wikipedia.org/wiki/Inverse_trigonometric_functions
Relevant Equations:: None

View attachment 324323

My understanding is that it probably means either ##+ \pi## or ## - \pi##, so that ##\pi## is either added to or subtracted from the first term, just like ##|x| = \pm x##.
In general, (mod ##\pi##) means ##\pm n \pi## for some ##n \in \mathbb N##.
UPDATE: I probably should have said ##+ i \pi## for some integer ##i \in \mathbb Z## since some (many?) people do not include 0 in ##\mathbb N## and I need to include 0.

Last edited:
vcsharp2003
FactChecker said:
In general, (mod ##\pi##) means ##\pm n \pi## for some ##n \in \mathbb N##.
Shouldn't ##n \in \mathbb Z##? Oh, I get it since you used a plus minus before ##n##.

fresh_42
vcsharp2003 said:
Shouldn't ##n \in \mathbb Z##?
Yes, but the ##\pm## does that.

vcsharp2003
vcsharp2003 said:
Shouldn't ##n \in \mathbb Z##? Oh, I get it since you used a plus minus before ##n##.
Actually, using ##\mathbb Z## would have probably been better since some (many?) people do not include 0 in ##\mathbb N## and I do need to include 0.

Last edited:
vcsharp2003

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