Difference between revisions of "Arccot"
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There are two functions commonly called $\mathrm{arccot}$, which refers to inverse functions of the [[cotangent | $\mathrm{cot}$]] 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 $\mathrm{arccot_2} \colon \mathbb{R} \rightarrow \left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right) \setminus \{0\}$ which results from restricting cotangent to $\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)$. | There are two functions commonly called $\mathrm{arccot}$, which refers to inverse functions of the [[cotangent | $\mathrm{cot}$]] 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 $\mathrm{arccot_2} \colon \mathbb{R} \rightarrow \left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right) \setminus \{0\}$ which results from restricting cotangent to $\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)$. | ||
| − | + | <div align="center"> | |
| − | + | <gallery> | |
| − | + | File:Arccots.png|Graph of $\mathrm{arcsin}$ on $[-1,1]$. | |
| + | File:Complex ArcCot.jpg|[[Domain coloring]] of [[analytic continuation]]. | ||
| + | </gallery> | ||
| + | </div> | ||
=Properties= | =Properties= | ||
Revision as of 05:40, 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 $\mathrm{arccot_2} \colon \mathbb{R} \rightarrow \left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right) \setminus \{0\}$ which results from restricting cotangent to $\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)$.
- Arccots.png
Graph of $\mathrm{arcsin}$ on $[-1,1]$.
- Complex ArcCot.jpg
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 $$\dfrac{d}{dz} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1}.█$$
