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

AI Thread Summary
The discussion centers on the interpretation of the notation ##(mod ~ \pi)## in a mathematical formula related to inverse trigonometric functions. It is suggested that this notation indicates the addition or subtraction of integer multiples of ##\pi##, similar to how absolute values work. The inverse tangent function is highlighted as having unique values only within specific intervals, and the concept of ##\tan(\alpha + \pi\mathbb{Z}) = x## is discussed to illustrate the periodic nature of the tangent function. The conclusion reached is that the expression can differ by ##\pm \pi##, emphasizing the importance of defining the interval for angles. Overall, the notation reflects the need to account for periodicity in trigonometric functions.
vcsharp2003
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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
Inverse Trig Functions formula.png


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