My Math skills got real rustic over time so bear with a (hopefully) simple question:
I'm looking for a way to compute $n$ from this formula, where $\cot(x)$ is the cotangent.
$$\cot(n) = \frac{\pi}{4} = 0.785\ 398\ 163 \dots$$
Based on my source $n$ is $51^\circ 51^\prime 14.3251^{\prime\prime}$ (maybe the decimal points are a bit off, don't worry about it too much, it's all computed by hand), but I want to understand it myself and not just blindly trust it.
I tried $\tan(\frac{\pi}{4})$ and $\mathrm{acot}(\frac{\pi}{4})$ but both don't seem to do the trick (or I missed something).
Also does something like a natural or logarithmic $\cot(x)$ exist?
My book is about 100 years old, so maybe things are named differently today.
4 Answers
$\begingroup$$\mathrm{arccot}(x)$ is commonly defined over the domain $\mathbb R$by inverting the continuous $(0,\pi)$ portion of $\cot(x):$$$\mathrm{arccot}(x)=\begin{cases} \arctan\left(\frac1x\right)+\pi &\text{ when }x<0;\\ \frac {\pi}2 &\text{ when }x=0;\\ \arctan\left(\frac1x\right) &\text{ when }x>0, \end{cases}$$ As such, its range is $\left(0,\pi\right).$
Based on this definition, $\mathrm{arccot}$ and $\arctan$ are reflections of each other in the line$y=\frac{\pi}4.$ So, $$\mathrm{arccot}(x)=\frac {\pi}2-\arctan(x)\quad\text{ on }\mathbb R.$$
analytical proof:
$\arctan(x)$ and $\mathrm{arccot}(x)$ are both continuous on $\mathbb R,$ so $\arctan(x)+\mathrm{arccot}(x)$—which has derivative $0$ on $\mathbb R$ and value $(\frac{\pi}4+\frac{\pi}4=\frac{\pi}2)$ at $(x=1)$—is also continuous on $\mathbb R.$ As such, this sum identically equals $\frac{\pi}2.$
Do note that for negative inputs, $\mathrm{arccot}$ has two different definitions. For example, $\mathrm{arccot}(-\frac\pi4)$equals either$-0.91$or $2.24$. More information here.
Usually, cotangent (among the other trig functions) is not invertible. We have to restrict it in order to make it invertible. We typically show the interval $(0,\pi)$ for this inverse.
That is, the function $\cot : (0,\pi) \to \Bbb R$ has the inverse function $\mathrm{arccot}: \Bbb R \to (0,\pi)$.
Therefore, if $n \in (0,\pi)$ and $\cot(n) = \pi/4$, we have that
$$n = \mathrm{arccot}(\pi/4) \approx 0.905 \text{ rad} \approx 51.85^\circ = 51^\circ \; 51' \; 0'' $$
Of course, if you took out the approximation even further, you'd get something closer to your answer. Taking
$$\mathrm{arccot}(\pi/4) \approx 51.853974012777452454524489216985667095604636959922071782124197859$$
for instance gives (by this calculator anyhow)
$$51^\circ \; 51' \; 14.31''$$
That said, unless you tell how you obtained your value, it'd be hard to ascertain where you went wrong.
$\endgroup$ $\begingroup$The inverse function for $\cot(x)$ is denoted as $\operatorname{acot}(x)$ or $\operatorname{arccot}(x)$ or $\cot^{-1}(x)$ which is a particular restriction of $\cot(x)$ from $(0,\pi) \to \mathbb R$.
(credit Wikipedia)
An evaluation with wolfram returns
$$\operatorname{acot}\left(\frac \pi 4\right)\approx 0.90502\,(\text{radians})\quad(=51.85°) $$
$\endgroup$ $\begingroup$Yes, $\operatorname{arccot}$ is a function: it is the inverse of the function $f:(0,\pi)\to\Bbb{R}$ given by $f(x)=\cot x$. However, this function is seldom used, for the reason that will become apparent below. If $\cot n=\pi/4$, then$$ \frac{\cos n}{\sin n}=\frac{\pi}{4}\implies\frac{\sin n}{\cos n}=\tan n=\frac{4}{\pi}\implies n=\arctan(4/\pi) \approx 0.905\text{ rad}\approx 51.9°\, . $$Remark: usually, it is the inputs of trigonometric functions like $\tan$ which are given in terms of $\pi$. While there is nothing at all wrong with computing $\arctan(4/\pi)$, I wonder if this is the source of your confusion.
$\endgroup$
