Difference between revisions of "Arccot"
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| − | File:Arccots.png|Graph of $\mathrm{arccot}_1$ and $\mathrm{arccot}_2$ on $ | + | File:Arccots.png|Graph of $\mathrm{arccot}_1$ and $\mathrm{arccot}_2$ on $\mathbb{R}$. |
File:Complex ArcCot.jpg|[[Domain coloring]] of [[analytic continuation]] $\mathrm{arccot}$. | File:Complex ArcCot.jpg|[[Domain coloring]] of [[analytic continuation]] $\mathrm{arccot}$. | ||
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Revision as of 06:02, 31 October 2014
There are two functions commonly called $\mathrm{arccot}$, which refers to inverse functions of the $\mathrm{cot}$ function. First is the function $\mathrm{arccot_1}\colon \mathbb{R} \rightarrow (0,\pi)$ which results from restricting cotangent to $(0,\pi)$ and second is the function $\mathrm{arccot_2} \colon \mathbb{R} \rightarrow \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \setminus \{0\}$ which results from restricting cotangent to $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.
- Arccots.png
Graph of $\mathrm{arccot}_1$ and $\mathrm{arccot}_2$ on $\mathbb{R}$.
- Complex ArcCot.jpg
Domain coloring of analytic continuation $\mathrm{arccot}$.
Properties
Proposition: $$\dfrac{d}{dz} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1}$$
Proof: If $y=\mathrm{arccot}(z)$ then $\cot(y)=z$. Now use implicit differentiation with respect to $z$ to get $$-\csc^2(y)y'=1.$$ Substituting back in $y=\mathrm{arccos}(z)$ yields the formula $$\dfrac{d}{dz} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1}.█$$
