I don't have a particularly good reason to want to do this, and I'm just asking out of curiosity.

I am looking for the coordinates of point $\pmb B$, a point on the circumference of a circle. If I know the following:

  1. The equation of the circle (the coordinates of the center $\pmb O$, and its radius $\pmb R$)
  2. The coordinates of a point on the circumference, point $\pmb A$
  3. The central angle $\widehat{AOB}$ (I know the angle might be useless without trigonometry)
  4. The length of the arc $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}$
  5. Points $\pmb A$ and $\pmb B$ are both in the upper half of the circle, and the angle $\widehat{AOB}$ is less than $\frac{\pi}{4}$ radians (for this specific example)

Is it possible to find the ccoordinates of $\pmb B$ without using any trigonometric functions (sin,cos,tan,etc)?

I'm not sure how to create an image for this, otherwise I would've included one for clarity.

Edits: I realize that the arc length can be calculated from the central angle and the radius, so point 4 is redundant. Someone else pointed out that there would be two possible points given the angle and arc length. Would it be possible to find either one or both without using trig?

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    $\begingroup$ If you know 1 and 3, 4 is redundant $\endgroup$ Commented Jun 7, 2022 at 22:19
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    $\begingroup$ Interesting, albeit random question. $\endgroup$ Commented Jun 7, 2022 at 22:26
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    $\begingroup$ Would it be ok to use series expansions for the trig functions, and so write the coordinates of $B$ as a converging, infinite sum? $\endgroup$
    – apg
    Commented Jun 7, 2022 at 22:32
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    $\begingroup$ Suppose $A$ is the point $(R,0)$ then $B$ is $(R \cos \widehat{AOB}, R \sin \widehat{AOB})$, so the question is basically asking whether it's possible to calculate the trig functions without using them. $\endgroup$
    – dxiv
    Commented Jun 7, 2022 at 22:38
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    $\begingroup$ I believe that by "origin" you meant the circle's centre. By definition, the origin is the reference point for all coordinates $\endgroup$ Commented Jun 7, 2022 at 23:04

5 Answers 5


The trig functions provide a dictionary between the arc measurement of an arbitrary$^\dagger$ point on a circle (or similarity class of a right triangle described by an angle) and the rectangular coordinates of the point on the circle (or ratio of side lengths of the triangle). Nothing more, nothing less.

You have one type of information—namely, the radius and angle—and you're asking about the other. Any possible tool to accomplish this is equivalent to using the trig functions (possibly obfuscated a bit).

$^\dagger\!$ Caveat: for "special" angles, the values of the trig functions (i.e., ratios of sides of a triangle with those angles or coordinates of a rotated point) can be found using the Pythagorean theorem and algebra, but they are like dust among the continuum of possible angles.


The answer to your title question is yes (stereographic projection). However, given the information you have, no, not really, because one of the pieces of information you're given is the length of the arc joining the two points on the circumference (equivalently, the angle), and these notions are essentially baked into the definition of the trig (or inverse trig) functions.

Nevertheless, I'll describe in more detail the stereographic projection, because I think it's a good way of looking at the circle (parametrization using rational functions), and by considering arclengths, as calculated with respect to this parametrization, one essentially 'discovers' the $\arctan$ function (typically, this is described, in what I feel is a rather ad-hoc way, in first-year calculus courses as Weierstrass' half-tangent angle substitution $t=\tan\frac{\theta}{2}$, but I prefer looking at the geometry first and then arriving at the calculus).

Note that by translating the center of the circle, and by rescaling, we may as well assume that we're dealing with a the unit circle centered at the origin.

We'll consider stereographic projection 'from the west'. This means we fix the point $(-1,0)$, the 'west pole' of $S^1$. For any point $(x,y)$ with $x\neq -1$ (i.e other than the west pole), imagine drawing a line through this point, and the west pole. This ray will intersect the $y$-axis at exactly one point. The $y$-coordinate of this point is what we shall call the stereographic projection of the point $(x,y)$ from the west pole. This is an extremely simple geometric approach for identifying the circle minus a point with the real line (and this easily generalizes to higher dimensions).

If we work this word-description out in terms of formulae, we get the following: let $U=\{(x,y)\in S^1\,:\, x\neq -1\}$, and consider $\sigma:U\to \Bbb{R}$, \begin{align} \sigma(x,y)=\frac{y}{1+x}, \end{align} and the inverse is $\sigma^{-1}:\Bbb{R}\to U$, \begin{align} \sigma^{-1}(t)=\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right). \end{align} Notice the significance of stereographic projection: it has a very simple geometric meaning, it is invertible, and both $\sigma$ and its inverse are rational functions. As examples, for $t=0$, we get $\sigma^{-1}(0)=(1,0)$, for $t=-1$ we get $\sigma^{-1}(-1)=(0,-1)$, and for $t=+1$, we get $\sigma^{-1}(1)=(0,1)$, and so on.

Now, fix two parameter values $-\infty<t_1<t_2<\infty$. Let us calculate the length of the arc joining the points $\sigma^{-1}(t_1)$ and $\sigma^{-1}(t_2)$: \begin{align} L_{t_1,t_2}&=\int_{t_1}^{t_2}\|(\sigma^{-1})'(t)\|\,dt =\int_{t_1}^{t_2}\frac{2}{1+t^2}\,dt, \end{align} where the second equality is a straight-forward differentiation. Of course, we would now recognize this immediately as $2[\arctan(t_2)-\arctan(t_1)]$. But, imagine you were never introduced to trigonometry. This would then be a pretty natural way of developing the subject (simple geometry of lines intersecting circles and other lines, and then using calculus to figure out arclengths).

In your question, you're given the coordinates of point $A$ (meaning you know $t_1$ and $\sigma^{-1}(t_1)$), and you know the arclength between $A$ and $B$ (so you know $L$). In order to figure out the coordinates of the point $B$, using this approach, one would have to invert the integral (i.e we've now arrived at a definition for $\tan$).

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    $\begingroup$ Of course, this approach may seem roundabout and indirect, but that's only because we're extremely familiar with the usual presentations of trigonometry. If we were to start from scratch, then (besides the complex exponential) defining $\arctan$ via an integral and defining the other trig functions from here is an extremely efficient way of defining and proving all the basic facts of trigonometry $\endgroup$
    – peek-a-boo
    Commented Jun 7, 2022 at 23:44

Consider the circle centered at the origin of unit radius. Consider the rotation matrix $$M = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \\ \end{bmatrix}$$ The point $A$ is rotated to a new point anticlockwise rotated by angle $\theta$, called $B$, via:

$$ \mathbf{B} = \mathbf{M} \mathbf{A} $$

Via this method, with $A=(0,1)$, then $$\mathbf{B}=(-\sin(\theta),\cos(\theta)).$$

Using the series expansions for the trig functions, one can instead write $$\mathbf{B}=\left(-\sum_{i=0}^{\infty}\frac{(-1)^i\theta^{2i+1}}{(2i+1)!},\sum_{i=0}^{\infty}\frac{(-1)^i\theta^{2i}}{(2i)!})\right)$$ So one can at least approximate the point via the partial sums, for example for small angles. Though, these will obviously converge to their respective trig functions if you want the exact point. A bit of a trick, but at least you avoid the trig, as requested.

Writing series expansions for transcendental functions, like $\sin(\theta)$ and $\cos(\theta)$, though these were done later via it is thought Euler or perhaps earlier, is what allowed Newton to work on his first results in calculus, see Section VI.14 Isaac Newton, Princeton Companion of Mathematics, Princeton University Press, 2008.


We know that the equation of a circle is $$(x-h)^2 + (y-k)^2 = R^2$$ which we can write as $$y = \sqrt{R^2 - (x-h)^2} + k$$ which means that $$\frac{dy}{dx} = - \frac{x-h}{\sqrt{R^2 - (h-x)^2}}$$ also we know that the arc of a function between two points A(Ax,Ay) and B(Bx,By) is $$\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB} = \int_{Ax}^{Bx}\sqrt{1+(\frac{dy}{dx})^2}dx=R*\widehat{AOB}$$ so just solve the integral and you will find Bx and a function of h,R,AOB,Ax $$Bx=f(h,R,AOB,Ax)$$ and then put Bx in the circle equation and get By

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    $\begingroup$ "Solving the integral" is essentially equivalent to evaluating a trigonometric function. $\endgroup$ Commented Jun 7, 2022 at 22:55

I'm sure this is not the answer you intended, but you didn't specify which coordinate system you wish to work with, although Cartesian is implied.

If the circle is centred at the origin and point $A$ has polar coordinates $(R,\theta)$, point $B$ will have coordinates $(R, \theta \pm \widehat{AOB})$, depending if $\widehat{AOB}$ represents a clockwise or counter-clockwise rotation.

However, if the circle is centred around some arbitrary point $C$ with Cartesian coordinates $(x_C, y_C)$ and you absolutely want to avoid trig functions for some reason, you'll be forced to work with two coordinate systems: one to specify the circle's centre and one to specify points on the circle. Such a system would have very limited applications.


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