Every side measured by one diameter

An angle that does not care where it stands found that a chord is seen at the same angle from everywhere on the opposite arc. One number for every chord through a point turned that constancy into a constant product. This essay turns it into a constant ratio — the one that makes the constancy numerical, and that ties the inscribed angle to the sine function directly.

The law of sines says that in any triangle, with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$,

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$

It is usually proved by dropping altitudes and left as a statement that three ratios agree. The better statement names what they agree on: each ratio is the diameter of the circle through the triangle’s three corners. That is the extended law of sines, and it is the inscribed angle theorem made into a measurement.

The proof is one diameter #

Every triangle has a circle through its three corners, its circumcircle, with some radius $R$. Take the side $a = BC$ and the angle $A$ opposite it. Draw the diameter from $B$, and call its other end $D$.

Two things are now true, both from the earlier essay. The angle at $D$ in the triangle $BCD$ is the angle at which $D$ sees the chord $BC$ — and $D$ is on the same arc as $A$, so that angle equals $A$. And the angle at $C$ in the triangle $BCD$ stands on the diameter $BD$, so by Thales it is a right angle. The triangle $BCD$ is therefore right-angled, with hypotenuse $BD = 2R$ and an angle equal to $A$ at $D$. In a right-angled triangle the side opposite an angle is the hypotenuse times the sine of that angle:

$$a = 2R \sin A.$$

The same argument at the other two sides gives $b = 2R\sin B$ and $c = 2R\sin C$, and the law of sines follows, with its common value named. The figure measures it: the side is 8.95, the angle 63.50°, and 8.95 divided by the sine of 63.50° is 10, the diameter. When $A$ is obtuse, $D$ lands on the other arc and the angle at $D$ is $180° - A$, whose sine is the same, so nothing changes.

The sine was a half-chord #

The formula $a = 2R\sin A$ says something about the sine itself. In a circle of diameter one, every chord is the sine of the angle it subtends from the circumference.

That is historically the right way round. Ptolemy’s Almagest, around 150, contains a table not of sines but of chords: for each angle at the centre of a circle of radius 60, the length of the chord it cuts off, in steps of half a degree. A chord subtending an angle $\theta$ at the centre subtends $\theta/2$ at the circumference, so Ptolemy’s chord of $\theta$ is $120\sin(\theta/2)$, and his table is a sine table in disguise. Indian astronomers a few centuries later tabulated half the chord of the doubled angle — the jyā — which is exactly the sine, and the word came into Latin as sinus through a misreading of its Arabic transliteration.

So the sine was never a ratio in a triangle first. It was a length in a circle, and a sine wave is a circle seen from the side for the same reason: the height of a point going round a circle is half a chord. The law of sines is the statement that the triangle ratio and the circle length are the same number, because a triangle’s corners always lie on a circle.

Nine ratios, one number #

The law is easy to check in a single triangle and more convincing across several.

Three quite different triangles — one acute, one with a nearly flat angle, one with a long side — give nine ratios, and every one is ten. The shape of the triangle changes which angles are large and which sides are long, and the ratio of each side to the sine of its opposite angle is unaffected, because the only thing it depends on is the circle, and all three triangles share one size of circle.

That makes the law a tool for finding circles. Given a triangle’s side and opposite angle, the circumradius is $a/(2\sin A)$; given two angles and any side, every other length follows. It also explains a fact that is otherwise surprising: in a triangle, the largest angle is opposite the longest side. The sine increases from $0°$ to $90°$ and is symmetric about $90°$, and $a = 2R\sin A$, so a longer side needs a sine closer to one — a larger acute angle, or an obtuse angle whose supplement is smaller than the others, and in either case the largest angle of the three.

Where the centre of the circle sits #

The proof drew a diameter from $B$ and used the far end $D$. Where $D$ lands depends on the triangle, and following it answers a question the formula hides: where the circumcentre is — one of the several centres a triangle has, and, with the orthocentre, one of the two that can leave the triangle altogether.

If the angle at $A$ is acute, $D$ is on the same arc as $A$, the angle at $D$ equals $A$, and the centre of the circle lies on the same side of $BC$ as $A$. If $A$ is a right angle, $BC$ itself is a diameter, $\sin A = 1$, and the formula says $a = 2R$: the circumcentre is the midpoint of the hypotenuse, which is Thales’s theorem stated as a length. If $A$ is obtuse, $D$ lands on the arc on the far side of $BC$, the quadrilateral $ABDC$ has opposite angles adding to $180°$, and the angle at $D$ is $180° - A$. The sine does not distinguish an angle from its supplement, so $a = 2R\sin A$ still holds — and the circumcentre has crossed $BC$ and lies outside the triangle.

So the three kinds of triangle correspond to three positions of the circumcentre — inside, on the longest side, outside — and to three values of the largest angle’s sine relative to the diameter: less than one on either side of a right angle, and exactly one at it. The formula is uniform across the three cases only because the sine of an angle and the sine of its supplement coincide, which is the circle’s symmetry about the chord’s perpendicular bisector showing up in a function.

The tangent is the limit of a chord #

The inscribed angle theorem has a limiting case that the law of sines uses at its edge, and it is worth watching the limit happen.

As the apex slides towards $B$, the side of the angle that joins it to $B$ gets shorter and turns, and when the apex arrives it has become the tangent line at $B$. The angle never changed along the way — it was 70° from every apex — so the angle between the chord and the tangent is 70° too. This is Euclid’s alternate segment theorem, and it is the version of the inscribed angle that has no apex left: the angle between a chord and the tangent at one end equals the angle the chord subtends from the far side of the circle.

In the law of sines it covers the degenerate triangle. A triangle whose corner $A$ has slid onto $B$ has a side $BC$ still of length $2R\sin A$, with $A$ now read as the tangent–chord angle, and the formula survives the limit without change. That kind of continuity — a formula that holds at the edge because the quantity it measures never jumped — is what makes the tangent a chord rather than a separate object.

Ptolemy’s theorem is the addition formula #

Ptolemy did not merely tabulate chords; he needed a way to compute them, and the tool he used is his theorem on cyclic quadrilaterals: in a quadrilateral whose corners lie on a circle, the product of the diagonals equals the sum of the products of opposite sides. The map that trades circles for lines proved it in a line by inversion. With chords written as sines it becomes something else.

Make one diagonal a diameter of length 1. The triangles on either side of it are right-angled at $B$ and $D$, by Thales, so with angles $\alpha$ and $\beta$ at $A$ their sides are $\cos\alpha$ and $\sin\alpha$, and $\cos\beta$ and $\sin\beta$. The other diagonal $BD$ subtends the angle $\alpha + \beta$ at $A$, so by the law of sines, in a circle of diameter 1, its length is $\sin(\alpha + \beta)$. Ptolemy’s theorem — diagonals’ product equals the sum of opposite sides’ products — is then

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.$$

The addition formula for the sine is Ptolemy’s theorem. Ptolemy used it in exactly this way: from the chords of $72°$ and $60°$, which the pentagon and hexagon give, he computed the chord of their difference, $12°$, then halved repeatedly with a half-angle version of the same argument, and built his table half a degree at a time. The formula the modern reader learns as trigonometry was, for fifteen hundred years, a theorem about four points on a circle.

The circumradius from three lengths #

The law of sines gives the circumradius from a side and an angle. Combining it with the area gives it from the sides alone.

The area of a triangle is half of two sides times the sine of the angle between them, $K = \tfrac12 bc\sin A$. Substitute $\sin A = a/(2R)$ from the law of sines and the sine disappears: $K = abc/(4R)$, or

$$R = \frac{abc}{4K}.$$

The table checks it against six random triangles, each time fitting the circle through the three corners directly and comparing. With Heron’s formula for the area in terms of the sides, the circumradius is then a function of the three side lengths and nothing else — the circle through a triangle is decided by three numbers, which is the same kind of statement as the nine-point circle being decided by the triangle, made quantitative.

How far apart the two centres sit #

A triangle has two natural circles: the circumcircle through its corners, radius $R$, and the incircle touching its sides, radius $r$, centred at the incentre $I$ where the angle bisectors meet. Leonhard Euler found in 1765 that the distance between their centres is fixed by the two radii alone:

$$OI^2 = R^2 - 2Rr.$$

The proof puts the power of a point and the law of sines together. The power of $I$ with respect to the circumcircle is $OI^2 - R^2$, and it can be measured along any chord through $I$ — take the bisector from $A$, which meets the circumcircle again at $L$, the midpoint of the arc $BC$. The power is then $-IA \cdot IL$, by the constant product. Now measure both pieces. The incircle touches the side $AB$ at a point whose distance from $I$ is $r$, and the bisector makes an angle $A/2$ with that side, so $IA = r/\sin(A/2)$. The chord $LB$ is seen from $A$ at the angle $A/2$, so by the law of sines $LB = 2R\sin(A/2)$; and a short angle chase shows the triangle $LBI$ is isosceles, so $IL = LB$. The product is

$$IA \cdot IL = \frac{r}{\sin(A/2)} \cdot 2R\sin(A/2) = 2Rr,$$

the sine cancels, and $OI^2 - R^2 = -2Rr$. The formula checks against random triangles to fourteen decimal places.

Its consequence is an inequality. A squared distance cannot be negative, so $R^2 \ge 2Rr$, that is, the circumradius is always at least twice the inradius — with equality exactly when the two centres coincide, which happens only for the equilateral triangle. Every other triangle has its incircle smaller than half its circumcircle, by an amount the distance between the centres records.

A circle of rational distances #

The law of sines has a consequence in arithmetic. A chord of a circle of diameter one is the sine of half the angle it spans at the centre. If the angles are chosen so that their sines and cosines are fractions, the chords are fractions too.

The angles whose cosine and sine are both fractions are exactly the angles of right triangles with whole-number sides, which every triple on one circle found are as plentiful as the rational points on the circle. Place points at twice such angles. The distance between two of them is the sine of the difference of their angles, and the sine of a difference of angles with rational sines and cosines is rational by the addition formula. So any number of points can be placed on a circle with every pairwise distance a fraction — and scaling up by a common denominator, a whole number.

Nothing like that is possible on a line in any interesting way, and off lines and circles it is severely limited. Paul Erdős and Norman Anning proved in 1945 that an infinite set of points with all pairwise distances whole numbers must lie on a line. For rational distances, József Solymosi and Frank de Zeeuw showed in 2010 that any algebraic curve other than a line or a circle holds only finitely many points of such a set. The circle is special, and the reason is the law of sines.

What the circles cannot show #

The obtuse case, and the sign. Every triangle in the figures is drawn with its corners where the argument is cleanest. When the angle at $A$ is obtuse the diameter’s far end $D$ lands on the arc with $A$’s supplement, and the proof needs the fact that an angle and its supplement have the same sine. The figures include an obtuse triangle in the ratio table; they do not draw the argument for it.

Why Ptolemy’s theorem holds. The quadrilateral figure measures the four sides and two diagonals and checks the equation. It does not prove Ptolemy’s theorem, which needs inversion or a construction with similar triangles; it shows that, granted the theorem, the addition formula is the theorem written in sines.

Where the rational points come from. The rational-distance figure uses four Pythagorean triangles. That there are infinitely many, and that they give infinitely many points, is the parametrisation of the circle by lines of rational slope, which the figure assumes.

Still open: rational distances everywhere #

A circle can carry infinitely many points at rational distances from each other; so can a line. Whether there is a set of points dense in the whole plane — meeting every disc, however small — with every pairwise distance rational is not known. Erdős and Stanisław Ulam asked it in the 1940s, and it remains open.

It is not even known whether there is a point at rational distance from all four corners of a unit square. Every approach so far shows that such sets are rare on curves — the Solymosi–de Zeeuw theorem says a rational-distance set can put only finitely many points on any curve but a line or a circle — and a dense set would have to put infinitely many points near every curve there is. Some conjectures in number theory would rule it out; none is proved.

One diameter for every side #

The inscribed angle makes a chord’s viewing angle constant, and the law of sines makes the constant numerical: a chord is its circle’s diameter times the sine of the angle it subtends. For a triangle that means every side over the sine of the opposite angle is the circumcircle’s diameter — nine ratios from three triangles, every one of them ten. Combined with the constant product of the previous theorem it fixes even the distance between a triangle’s two centres.

That single formula is why the sine began as a half-chord in Ptolemy’s table, why his theorem on four concyclic points is the addition formula, why the circumradius is $abc/4K$, and why a circle can hold any number of points at rational distances from one another — while whether the plane can hold a dense set of them is still unknown.