The sine of one degree

A sine wave is a circle seen from the side defined the sine of an angle as a height: walk round a circle of radius one through that angle, and the sine is how far above the starting level the walk has climbed. The definition is a picture, and the picture gives no number. For a few angles the number follows from geometry — $\sin 30° = \tfrac12$ from a half equilateral triangle, $\sin 45° = 1/\sqrt2$ from a half square — but a table of sines has an entry for every degree, and most of the entries do not fall out of any figure.

For two thousand years every such table was computed by hand, and every one of them ran into the same entry. This essay is about that entry, the sine of one degree: why it cannot be constructed, and the two ways it was computed anyway.

The chain that stops at three degrees #

A table of sines is built from a few angles whose sines are known exactly, together with rules that make new sines from old. The ancient tables were tables of chords rather than sines — the chord of an angle in a circle of radius one is twice the sine of half the angle, as every side measured by one diameter drew it — but the arithmetic is the same and is easier to follow in sines.

The regular pentagon and hexagon supply exact values: $\sin 72°$ involves $\sqrt5$, and $\sin 60° = \sqrt3/2$. The subtraction formula $\sin(a - b) = \sin a \cos b - \cos a \sin b$, which Ptolemy had in the form of his theorem about a quadrilateral in a circle, turns them into $\sin 12°$. The half-angle formula, $\sin^2(a/2) = (1 - \cos a)/2$, then halves as often as needed: $6°$, $3°$, $1\tfrac12°$, $\tfrac34°$. Every step takes a square root at most, so every sine on the chain can be written in square roots and drawn with ruler and compass. Adding and subtracting what has been reached gives every multiple of $3°$, and every halving of one.

Nothing on the chain is a whole number of degrees other than a multiple of three. To reach one degree from three requires dividing an angle by three, and the angle that will not divide by three proved that a general angle cannot be trisected with ruler and compass. The table-maker therefore has entries at $\tfrac34°$ and $1\tfrac12°$ and none at $1°$, and since every whole-degree entry not a multiple of three is built from $\sin 1°$ by the addition formula, the whole table depends on the one number the method cannot reach.

A cubic with three real roots #

The obstruction can be written as an equation. The triple-angle formula

$$\sin 3\theta = 3 \sin\theta - 4 \sin^3\theta$$

says that $x = \sin 1°$ satisfies

$$4x^3 - 3x + \sin 3° = 0,$$

and the number $\sin 3°$ is known. So the sine of one degree is a root of a cubic with known coefficients.

It is not the only root, and the hero figure shows the other two. Any angle whose triple has the same sine as $3°$ gives a root, and tripling $121°$ gives $363°$, tripling $241°$ gives $723°$ — both a whole number of turns past $3°$. The three angles are a third of a turn apart on the circle, the corners of an equilateral triangle, and their sines are $0.017452$, $0.857167$ and $-0.874620$, which add up to nought because the cubic has no $x^2$ term. Finding $\sin 1°$ means picking out the small root.

That a cubic with three real roots cannot be solved with square roots alone is the content of three real roots and no real radicals, and for this particular cubic there is a direct count. The cosine of one degree is a root of an irreducible polynomial whose degree over the rationals is half of Euler’s $\varphi(360)$, which is $48$. A number reachable by ruler and compass must have a degree that is a power of two, since each step can at most double it; forty-eight is sixteen times three.

The figure computes that degree for every whole-degree angle up to $45°$ and finds the pattern complete: an angle of $k$ degrees can be constructed exactly when $k$ is a multiple of three. The factor of three in the degree is the trisection, appearing as arithmetic. It is the same arithmetic as which polygons can be drawn: the angle $k°$ is a $360/k$-th of a turn, and $360$ has the prime $3$ in it twice, while Gauss’s criterion allows each odd prime at most once.

What one wrong entry would cost #

The dependence of a whole table on one entry is worth making concrete, because it explains why the table-makers spent so much effort on a single number. From $\sin 1°$ and $\cos 1° = \sqrt{1 - \sin^2 1°}$, the addition formula gives $\sin 2°$, then $\sin 4°$, $\sin 5°$ from $\sin 4°$ and $\sin 1°$, and so on round the quadrant, with the multiples of three supplied independently by the chain. An error $\varepsilon$ in $\sin 1°$ enters $\sin 2° = 2 \sin 1° \cos 1°$ as about $2\varepsilon$, and every entry built by repeated addition of one degree inherits a multiple of it that grows with the number of additions. A table of sines at every degree, computed this way, is no more accurate than its one-degree entry times the number of steps from the nearest multiple of three — at most two steps, which is why the multiples of three matter.

Entries between tabulated angles were interpolated, and linear interpolation over a step of $h$ radians errs by up to about $h^2/8$ times the size of the function’s second derivative. For Ptolemy’s chords, tabulated at steps of half a degree, that is a few millionths of the radius — about the size of the last sexagesimal place his table printed, one part in 3,600 of a unit in a radius of sixty. The table’s precision and its step were matched, and a finer table would need smaller steps, more precise interpolation and a more precise one-degree entry, all together.

Cardano’s formula hands the problem back #

Cubic equations had a formula from the sixteenth century, and it is natural to ask what it does here. Gerolamo Cardano’s formula writes a root of $x^3 + px + q = 0$ as $u + v$, where $u^3$ and $v^3$ are the two roots of a quadratic. For this cubic, scaled to $x^3 - \tfrac34 x + \tfrac14\sin 3° = 0$, that quadratic has no real roots.

The number whose cube root Cardano requires is $-\tfrac18\sin 3° + \tfrac{i}{8}\cos 3°$, a complex number of modulus one-eighth at an angle of $93°$. Its three cube roots have modulus one half and angles $31°$, $151°$ and $271°$, and twice their real parts are $\cos 31° = \sin 59°$, $\cos 151°$ and $\cos 271° = \sin 1°$ — the three roots again. The formula is correct, and it is useless for computing: taking the cube root of a complex number means dividing its angle by three, which is the original problem. Multiplying is turning explains why, since a cube root of a complex number turns through a third of its angle. Cardano called this the irreducible case, and it was the first place in history where complex numbers were forced on mathematicians who wanted only real answers.

Ptolemy’s bracket #

The table-makers did not need a formula. They needed a number to the precision of their table, and Ptolemy, in the second century, got one from an inequality.

The inequality is that when an angle grows, its sine grows proportionally less: for $0 < \beta < \alpha$ below a right angle,

$$\frac{\sin\alpha}{\sin\beta} < \frac{\alpha}{\beta}.$$

In modern terms this is the statement that $\sin x / x$ decreases, which the angle that is really an area made visible by comparing a triangle, a sector and a larger triangle; Ptolemy had it as a lemma about chords that goes back to Aristarchus. Applied with $\alpha = 1\tfrac12°$ and $\beta = 1°$ it gives $\sin 1° > \tfrac23 \sin 1\tfrac12°$, and applied with $\alpha = 1°$ and $\beta = \tfrac34°$ it gives $\sin 1° < \tfrac43 \sin\tfrac34°$. Both bounds are on the chain, so both are computable.

The bracket runs from $0.017451299$ to $0.017452794$, a width of one and a half millionths, and the true value $0.017452406$ sits inside. Ptolemy ran the argument on chords — on the chord of one degree, from the chords of $1\tfrac12°$ and $\tfrac34°$ — and in his units, a circle of radius sixty with sexagesimal fractions, both ends of the bracket came out as $1;2,50$. His table carried two sexagesimal places, so the bracket settled the entry completely and his value for the chord of one degree is correct to the last place he printed. An impossible construction had been sidestepped by not constructing anything: the number was trapped, and the trap was narrower than the table’s precision.

Doing the same thing again #

A bracket gives a fixed precision. A table-maker who wanted more had to bracket more finely, using constructible angles closer to one degree on either side, and the bracket narrows only in proportion to how close they come: its width is roughly the spacing of the two angles times the slope of $\sin x / x$, which near one degree is a few millionths per degree. Fifteen-place tables by this route would need constructible angles within a billionth of a degree of one degree on each side, reached by long chains of halvings and additions, each carried to fifteen places.

Jamshīd al-Kāshī, working at the observatory in Samarkand in the 1420s, found a better route through the cubic itself. Rewrite $4x^3 - 3x + \sin 3° = 0$ as

$$x = \frac{\sin 3° + 4x^3}{3}.$$

For the small root, the term $4x^3$ is tiny — about $0.00002$ — so the right side is very nearly $\tfrac13 \sin 3°$, a first guess. Putting the guess into the right side gives a better guess, and that one gives a better one still. Each step is one cube, one addition and one division by three, all of them operations a computer of the fifteenth century could carry out to as many places as he had patience for.

The gain per step is the slope of the right-hand side at the root, $4x^2 = 4\sin^2 1° \approx 0.00122$: each step multiplies the error by that factor and so adds almost three correct decimal places. Twenty steps reach sixty places, and the figure checks the result against an independent computation — the Taylor series $x - x^3/3! + x^5/5! - \cdots$ at $x = \pi/180$, with $\pi$ from Machin’s formula — to fifty-five places. Both routes give

$$\sin 1° = 0.017452406437283512819418978516\ldots$$

The iteration is a fixed-point method of exactly the kind the staircase that shows the whole orbit drew as a cobweb: a map whose graph crosses the diagonal at the answer, with slope so close to nought at the crossing that the cobweb collapses onto the point almost at once. Al-Kāshī did not have the language of contraction, but he had its substance: he noticed that the cubic, read as a rule for correcting a guess, corrects it enormously at every turn because the cubed term is so small. His value of $\sin 1°$ was accurate to the equivalent of some sixteen decimal places, far beyond any earlier table, and it was computed as part of the same programme that produced his value of $2\pi$ to sixteen decimal places, the most accurate for nearly two centuries.

Linear is enough, and Newton would be faster #

Each step of al-Kāshī’s iteration adds a fixed number of correct places — almost three — so the number of steps grows in proportion to the precision wanted. That is linear convergence, and it is the slow kind. Newton’s method applied to the same cubic would instead replace a guess $x$ by $x - f(x)/f'(x)$ with $f(x) = 4x^3 - 3x + \sin 3°$, and near a simple root the number of correct places roughly doubles at every step: three, six, twelve, twenty-four, forty-eight. Six steps of Newton reach the precision that twenty of al-Kāshī’s do.

Al-Kāshī’s choice was nevertheless the right one for his circumstances, and the reason is the cost of a step rather than the number of steps. A Newton step needs a division by $f'(x) = 12x^2 - 3$, a number of full precision, and long division of one many-place sexagesimal number by another was the most laborious operation in his arithmetic. His step divides only by three. When the contraction factor is as small as $0.00122$, the linear method is already fast, and each of its steps is cheap; the quadratic method would save steps and spend them back in divisions. A cubic method that is Newton’s in disguise followed the opposite trade, where each step does more work in order to need fewer.

There is a second, quieter advantage. Newton’s method converges quickly only once the guess is close; from a poor start it can jump to another root, as chaos on the line between two roots showed for exactly this kind of cubic. The fixed-point rule, by contrast, is a contraction on a whole interval containing the wanted root and no other, so it cannot be led astray by a bad start inside that interval. For a table-maker who could not afford to discover a wrong root after a week of arithmetic, that guarantee was worth more than speed.

Why the iteration converges to the right root #

A fixed-point iteration converges to a root only if the map is shrinking near it, and the cubic has three roots. Near the other two, the slope $4x^2$ is about $4 \times 0.73 \approx 2.9$ and $4 \times 0.76 \approx 3.1$ — larger than one — so the iteration started near them would be driven away. The rule $x \mapsto (\sin 3° + 4x^3)/3$ is a contraction only on an interval around nought of half-width about $\tfrac12$, which contains exactly one of the three roots. Started at $\tfrac13\sin 3°$, which is in that interval, it cannot find any root but the one wanted.

That is the selection the impossibility arguments said could not be made by radicals, made instead by a choice of starting point. The cubic’s three roots are algebraically indistinguishable — any statement about one of them that uses only the cubic’s coefficients is true of the others — which is why no formula in radicals can single out the small one without passing through complex numbers. The iteration distinguishes them by size, a property invisible to algebra and obvious to arithmetic.

Sixty places, and the cost left out #

The figures compute every number to sixty decimal places, with square roots, $\pi$ and the series all done in whole-number arithmetic, and the agreement of two independent routes is the check. What they cannot show is the cost of the computation in the fifteenth century, which was the real achievement: sexagesimal arithmetic by hand on numbers of ten or more places, where a single slip in a cube propagates into every later step. The iteration is forgiving of slips in one sense — an error in one step is corrected by the next, since the map contracts — but a slip in $\sin 3°$ itself is never corrected, and al-Kāshī’s $\sin 3°$ had to come from the square-root chain to the full precision of the final answer.

Nor do they show the history of the inequality. Ptolemy’s chord lemma is in the Almagest, but its earlier use by Aristarchus and Archimedes is partly a reconstruction from later sources, and the claim that Ptolemy’s bracket made his one-degree entry exact rests on the computation above rather than on any statement of his that the bracket was the reason.

Still open: how much a table of sines costs #

The sine of one degree is now a few lines of computation, by series or by the iteration. A question that is still being sharpened is how cheaply the sine can be computed to $n$ digits in general. Methods based on the arithmetic–geometric mean compute $\pi$ and the elementary functions to $n$ digits in about $\log n$ multiplications of $n$-digit numbers, and binary splitting of series does nearly as well, while the iteration here costs about $n/3$ steps — one per three digits — and the Taylor series fewer. Which of these is best depends on how multiplication of large numbers is done, and the cost of multiplication itself was settled only in 2019, when David Harvey and Joris van der Hoeven gave an algorithm taking $n \log n$ steps, the bound Arnold Schönhage and Volker Strassen had conjectured in 1971. Whether that bound is optimal — whether $n$-digit numbers can be multiplied faster than $n \log n$ — is not known, and with it the exact cost of a table of sines.