Difference between revisions of "Arccot"
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File:Arccots.png|Graph of $\mathrm{arccot}_1$ and $\mathrm{arccot}_2$ on $\mathbb{R}$. | File:Arccots.png|Graph of $\mathrm{arccot}_1$ and $\mathrm{arccot}_2$ on $\mathbb{R}$. | ||
| − | File: | + | File:Complexarccotplot.png|[[Domain coloring]] of $\mathrm{arccot}$. |
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Revision as of 23:05, 15 May 2016
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 $\mathbb{R}$.
Domain coloring of $\mathrm{arccot}$.
Properties
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1},$$ where $\mathrm{arccot}$ denotes the inverse cotangent function.
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 $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1},$$ as was to be shown. █
References
References
Which is the correct graph of arccot x?
