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Say I wish to find the sum of three arcus cotangents. I wanted to start with a smaller example, say $$\mathrm{arccot}{x} + \mathrm{arccot}{y}.$$

I know that $$ \mathrm{arccot}{x} = \begin{cases} \arctan{\frac{1}{x}} & \text{if $x>0$} \\ \pi + \arctan{\frac{1}{x}} & \text{if $x<0$} \end{cases} $$

And that $$ \arctan(x)+\arctan(y) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\ \pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\ -\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases} $$

Is this the right path to find the expression for $\mathrm{arccot}{x} + \mathrm{arccot}{y}$?

Simply consider every case of $x,y,xy$? How to get to the sum of three arcus cotangents from here?

Was also looking for a "ready" formulation online but I ultimately failed. So out of curiosity - are arcus cotangents that rarely used?

bajun65537
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  • Read here: https://math.stackexchange.com/questions/326334/a-question-about-the-arctangent-addition-formula – Bumblebee Nov 11 '21 at 09:28
  • Well as you can see in my post, I am looking for the expression for $\mathrm{arccot}{x} + \mathrm{arccot}{y}$. In the other post that you forwarded, the question, as well as the answer, was about the sum of arcus tangents and the intuition behind this expression. – bajun65537 Nov 11 '21 at 09:36
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    Sorry! I didn't read the post properly. – Bumblebee Nov 11 '21 at 09:39

1 Answers1

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I know that $$ \mathrm{arccot}(x) = \begin{cases} \arctan{\frac{1}{x}} & \text{if $x>0$} \\ \pi + \arctan{\frac{1}{x}} & \text{if $x<0$} \end{cases} $$

And a small addition: $$=\frac\pi2\quad\text{if }x=0.$$

Based on this definition (there are two common definitions of arccot), we have that on $\mathbb R,$ $$\arctan(x)\equiv\frac {\pi}2-\mathrm{arccot}(x).$$

and that $$\arctan(x)+\arctan(y) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\ \pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\ -\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$

Is this the right path to find the expression for $\mathrm{arccot }(x) + \mathrm{arccot }(y)$?

Substituting in the above identity and noting that $\arctan$ is an odd function: $$\mathrm{arccot }(x)+\mathrm{arccot }(y) = \begin{cases}\pi+\arctan\left(\dfrac{x+y}{xy-1}\right), &xy < 1 \\ \arctan\left(\dfrac{x+y}{xy-1}\right), &x>0,\; y>0,\; xy>1 \\ 2\pi + \arctan\left(\dfrac{x+y}{xy-1}\right), &x<0,\; y<0,\; xy > 1.\end{cases}$$

ryang
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  • Makes sense! Forgot about the $\arctan(x)\equiv\frac {\pi}2-\mathrm{arccot}(x)$ bit. – bajun65537 Nov 11 '21 at 12:25
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    @bajun65537 To be fair, this identity isn't that well-known; it is false (or rather, it is true only for $x=0$) when using Wolfram Alpha's definition of arccot, which is different from ours. – ryang Nov 11 '21 at 12:28
  • Hmm and could you clarify why we can write it this way? I mean why on the right-hand side we can have PLUS arctan instead of MINUS? At first, I just thought it does not make any difference but now I'm not so sure. – bajun65537 Nov 11 '21 at 12:39
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    @bajun65537 1. I've added an elaboration; in case it isnt sufficiently clear: an odd function $f$ is a function such that $f(-x)=-f(x).$ 2. Use this to check our derived identity. – ryang Nov 11 '21 at 12:42