# How to I make arctrig functions grow arbitrrily large while they maintain cyclical properties?

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I'm running into a problem where I want to represent a series of radians as A*arccos(Bx+C)+D, like for instance pi/2, 3*pi/4, pi...18*pi/12.. or pi/2, pi, 3pi/2, 2pi, pi/2, pi, 3pi/2....ect. But, because the domain of arccos is limited and then becomes some imaginary hyperbolic trig function outside of the domain 0<x<pi, I'm having trouble figuring out how to do that. Like normally, you could represent a series of cyclical numbers like 1,-1,1,-1 as cos(pi*x). I want to do the same thing with radians using arccos(x) where it just keeps repeating periodically as x grows arbitrarily large

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|Bx+C|≤1 will give you angles. > 1 leads to hyperbolic function.

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Yeah...that's more or less what I said within the first post except the exact domain of arcos(x) is [0, pi].

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Can you do:

DY = number of digits in Y

Arccos(Y/10^DY)

?

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I don't know what you mean exactly, but it could lead to the right idea. If you increase by a smaller and smaller number that never increases past pi, then arccos should grow indefinitely large, but it does so logarithmically according to the hyperbolic trig function so it's still not cyclical. the only thing I could think that works is arccos(cos(pi*x))*pi or something like that, but I never see anything like that anywhere.

Edited by MWresearch
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Yeah...that's more or less what I said within the first post except the exact domain of arcos(x) is [0, pi].

The domain is [-1,1]. The range is [0,pi].

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I don't know what you mean exactly, but it could lead to the right idea. If you increase by a smaller and smaller number that never increases past pi, then arccos should grow indefinitely large, but it does so logarithmically according to the hyperbolic trig function so it's still not cyclical. the only thing I could think that works is arccos(cos(pi*x))*pi or something like that, but I never see anything like that anywhere.

It will convert arbitrary values into the range arccos is looking for.

123456 / 106 = 0.123456

-123 / 103 = -0.123

# of Digits to left of decimal = ⌊log10(|x|)⌋+1

arccos(x / 10⌊log10(|x|)⌋+1 )

Edited by Endy0816
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Yeah still not working for what I'm trying to do, maybe there's no way.

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Are you wanting to keep your result normal via eliminating the imaginary bit?

That's the only other option really.

Edited by Endy0816
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I don't know, I just want to represent an indefinite series of radians as some function of arccos x. If there's some way to do that with archosh(x) then that's fine.

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