Searching for a surd-expression for sin 10º
David McNaughton

Note added subsequently:

After this article had been posted in the Internet, e-mails were received from Dave L. Renfro (and from other mathematicians). Expressing sin(10) in terms of surds is apparently known to be quite impossible - even after 'climbing up' to (say) 36th roots, or 108th roots, etc. etc.

However, my notes below will at least demonstrate why we cannot tackle the problem simply by satisfying <17> to <19> (which initially seemed such an obvious and logical approach):-
[Taken from the "Sixth Possibility" below; also see the "Tenth" one, and others]:-
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... i.e., we just need to find rational values of A, B, C and N which fit these three equations -

4A²BN + 4B²C + 4AC² - A = 0 .... <17>
8A³N² + 8B³N + 8C³+ 48ABCN - 6C + 1 = 0 .... <18>
4AB²N + 4BC² + 4A²CN - B = 0 .... <19>
[i.e., four unknowns with only three constraints] ...

... because if we succeed, the following formula would then apply:
sin(10) = A.³Ö(N²) + B.³ÖN + C ... <15>
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ABSTRACT: Sin 10º is one of the roots of 8t³=6t-1. A surd-expression would therefore have to fit that equation, but when matching up like terms it often proves impossible to balance the two sides. A few cases are investigated which did initially seem promising, but a detailed algebraic analysis shows that they are untenable.
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Note that the square root symbol is unfortunately rendered as Ö in Linux (so cube roots are portrayed as ³Ö).
Others represented differently, include the "approximately equals" sign, which becomes @.

Ancient and mediaeval mathematicians, architects and arabesque artists often found it convenient to create angles such as 60º, 45º and even 72º using just a ruler and compass. Whenever that is possible, trigonometric functions of such angles may be written in terms of square roots. In fact, by applying standard formulae we can then derive such trigonometric expressions for any multiple of 3º. However, there is no known method for constructing 10º, 20º or 40º with just a compass and an unmarked straight edge. In fact, I have heard from an authoritative source that it has been proved that those constructions are impossible.

Consider 30 as (10+10+10) and apply the standard trigonometric expansion for sin(3q) [in this instance, sin 30º]. We then see that sin 10º, abbreviated here as "t", is one of the roots of the equation
8t³ - 6t + 1 = 0 ... <1>
There are also two other quantities which satisfy it, namely sin 50º and sin (-70º).

Cubic equations like <1> may be tackled using Cardano's technique, which does indeed express the solutions in surd terms - but those also involve complex numbers. This happens even when all three roots are real - although with further manipulation the i-values do cancel out and disappear. At the same time, the surds simply give way to trigonometric functions - in this particular case leading directly to sin 10º, sin 50º and sin (-70º). In other words, Cardano's algorithm does not yield a surd-expression for sin(10) in real terms.

If there is one, we might suspect from <1> that cube roots should be involved in addition to square ones. In that case, it seems unlikely that an angle of 10º could be constructed using a compass and straight edge in just two dimensions.

In addition to regarding this as a search for a surd-equivalent for sin(10), it might be better to describe it as an attempt to discover more about the simple cubic equation <1>, i.e., by asking whether real surd-expressions exist for its roots. In this context, we could say that we are focussing on
y³ - 3y + 1 = 0 ... <1a>
which is of course obtained by substituting y = 2t in equation <1>.
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Rewriting <1> as
8t³ = 6t - 1 .... <1b>
helps appreciate the difficulty of finding a suitable surd-expression for sin(10) - if indeed one exists. When such an expression gets cubed, there will have to be algebraic 'compensation' and combining together of terms in order to produce the Right-Hand-Side of equation <1b>.

Besides, shown below is a circular interlocking relationship which demonstrates how a repeated application of (1-2sin²q) - [firstly on sin(10) ... then on the result, namely sin(70) ... and then on sin(50)] - 'leads back' to sin(10). Thus, any formula for sin(10) would have to yield to this process:
1 - 2sin²10 = cos(20) = sin(70) .... >
1 - 2sin²70 = cos(140) = -sin(50) .... >
1 - 2sin²50 = cos(100) = -sin(10) .... ^
It is also evident that sin(70) and sin (50) would possess their own formulae - with a structure comparable with the one belonging to sin(10). Indeed, that becomes even more obvious when we note the 'restrictions' illustrated by <20> and <21> below.
 

First possibility:

Try assuming that sin(10) = ... t = FÖG + H ... <2>
where F, G and H are to be non-zero rational numbers (including fractions). G should be positive (and not a perfect square), but F and H may be negative.

To satisfy equation <1b>, we therefore need:
8.[FÖG + H]³ = 6.[FÖG + H] - 1... <3>
Multiplying out and comparing the ÖG term on each side (and dividing out the unwanted solution "F = 0") yields
4F²G = 3 - 12H² ... <4>

Now go back to the expansion of <3> and compare the non-ÖG term on each side, giving:
24F²GH + 8H³ - 6H + 1 = 0 ... <5>

Substitute for 4F²G from <4> yields:
64H³ - 12H - 1 = 0 ... <6>
Cardano's technique for solving cubic equations shows that the three roots of <6> are:
H = (½).cos(20) ... or ... (½).cos(100) ... or ... (½).cos(140)
These candidate values of H are not rational quantities (see below) - so model <2> is not viable. [If they were rational, then trigonometric manipulation easily demonstrates that sin(10) would also have to be rational, thereby rendering formula <2> inappropriate]. Cos(100) is of course negative sin(10).

It is possible, incidentally, to substitute those three solutions for H into <4>.
With H = (½).cos(140), for example, we get FÖG = ±½.Ö3.cos(50), and the positive option does indeed produce sin(10) from equation <2>. This serves only as a check that our calculations are correct, even if we have not managed to discover a suitable expression for sin(10).

DLMcN.com/sin10-irr.html is a proof that sin(10) has to be irrational - and  the same argument may be applied to cos(20), cos(100) and cos(140).
 

Approximations

Despite the failure of model <2>, we can actually get quite close with
sin(10) @ 3·002 - 2Ö2
and can further improve it by fine-tuning the numbers, for example with
sin(10) @ 3¹/482 - 2Ö2 ... <7>
Of course, we can continue to make those values and coefficients as complicated as we wish, right up to the limit of accuracy of our computer or electronic calculator - whilst still retaining rational quantities for H and F. That would of course be a pointless exercise; this illustration is intended merely as a warning against doing it.

That value from <7> does indeed almost satisfy equation <1>, but when it is fed into <3>, the Ö2-terms and the others do not even come close to balancing.

In fact, we could even search for a purely rational expression, discovering for example:
sin(10) @ 4/23 .... or, with greater precision: sin(10) @ 2993/17236
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Second possibility:

Try writing sin(10) = ... t = Ö(FÖG + H) ... <8>

in which case sin(70) = cos(20) = 1 - 2FÖG - 2H
so sin(-70) = 2FÖG + 2H - 1 ... <9>
This equation <9> has essentially the same structure as equation <2>; i.e. the terms and coefficients possess the same characteristics, but carry different values.

Because sin(-70) also satisfies equations <1> and <1b>, our consideration of the "first possibility" above - shows that <9> is not possible, so pattern <8> will not work.
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Third possibility:

Let us introduce a second square-rooted quantity, i.e. hoping that
sin(10) = ... t = FÖN + LÖM + H ... <10>

We therefore require that
8.[FÖN + LÖM + H]³ = 6.[FÖN + LÖM + H] - 1

The ÖNÖM term on the left will not be matched by a similar one on the right, unless perhaps we allow surd-values for F, L and H. We must of course ignore examples where N or M is a perfect square.

We could therefore extend model <10> by adding a ÖNÖM term: this is considered below in the tenth possibility.
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Fourth possibility:

sin(10) = ... t = FÖG + L.³ÖG + H ... <11>
(where the ³Ö denotes of course "cube root of")

requiring
8.[FÖG + L.³ÖG + H]³ = 6.[FÖG + L.³ÖG + H] - 1 ... <12>

The ÖG.³ÖG term on the left will not be matched by a similar one on the right, unless perhaps we allow surd-values for F, L and H. We do not want G to be a perfect cube or square.

Even if F is zero, leaving a model of type
sin(10) = ... t = L.³ÖG + H ... <13>
there will still be a mismatch - with a ³Ö(G²) term on the left of <12> but without one on the right.
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Fifth possibility:

After rejecting that fourth possibility, and looking back at the second one above, it is also pointless to try
sin(10) = ... t = Ö(L.³ÖG + H) ... <14>
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Sixth possibility:

Because model <13> is inadequate, try assuming that
sin(10) = ... t = A.³Ö(N²) + B.³ÖN + C ... <15>
Thus, we have a "cube root of N-squared" term, a "cube root of N" term, and a term without any "N". A, B, C and N must all be rational; in addition, the model will not work if N is a perfect cube.

Equation <15> requires
8.[A.³Ö(N²) + B.³ÖN + C]³ = 6.[A.³Ö(N²) + B.³ÖN + C] - 1 ... <16>

This model will at least produce the necessary "algebraic compensation", enabling us to match up the two sides of <16>. Multiplying out and comparing like terms produces three equations with four unknowns:
4A²BN + 4B²C + 4AC² - A = 0 .... <17>
8A³N² + 8B³N + 8C³+ 48ABCN - 6C + 1 = 0 .... <18>
4AB²N + 4BC² + 4A²CN - B = 0 .... <19>

Optimistically, we should expect to find three sets of values of A, B, C and N consistent with equations <17>, <18> and <19>: i.e., those three sets of coefficients should correspond to each of the solutions to <1> ... the other roots being sin(50) and sin(-70). Supporting this expectation is the fact that those three different expressions, all of type <15>, will need to satisfy
8.sin(10) . sin(50) . sin(-70) = - 1 ... <20>
and
sin(10) + sin(50) + sin(-70) = 0 ... <21>
to be consistent with equation <1>.

In any event, simply by writing
sin(70) = cos(20) = 1 - 2sin²(10), it may be deduced that if sin(10) follows model <15>, and if A, B, C and N are all rational, then sin(70) and therefore sin(-70) also follow model <15> - although with different values of A, B and C, obviously. That same reasoning may then be extended to include sin(50).

For <21> to hold, it follows that
A1 + A2 + A3 = 0 ... <22>
where A1, A2 and A3 are the individual A-coefficients in <15> for each of the three roots of <1> ...
... and similarly B1 + B2 + B3 = 0 ... <23>
and C1 + C2 + C3 = 0 ... <24>

However, because we have four unknowns for just three equations <17>, <18> and <19>, there is a continuous spectrum of solutions - i.e., an infinite quantity. Thus, I personally find it difficult to believe that three and only three sets of rational numbers could exist to fit those four unknowns. As an analogy, we could perhaps visualise the relationship between just two equations with three unknowns. This can easily be pictured in three dimensions - as two intersecting surfaces. Their line of intersection contains an infinite number of values which satisfy the two equations.
 

At the same time, a surprisingly simple expression for N does emerge:
Multiplying equation <17> by B and <19> by A, and then subtracting, leads either to C=0 (examined below in the "eighth possibility"), or alternatively to
N = B³/A³... <25>

This may also be verified by substituting from <25> into equation <17>, and also into equation <19>: in both instances, that leads to equation <26> below.

Of course, perfect cube values of N are no use for constructing a model like <15>, because we would not then be entitled to collect up the ³Ö(N²) and ³ÖN terms of <16> in order to derive <17>, <18> and <19>.

In other words, model <15> is no use for expressing sin(10).
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This additional discussion (here, below) of model <15> should be regarded as an 'Appendix' of secondary importance. It showed me [before deriving <25>] that N was a perfect cube with specific permitted values of A, B and C.  Similar reasoning might be applicable, for example, to model <42> below.

Using equations <17> and <19>, we can immediately derive another one containing just A, B and C - by eliminating N. This yielded
(A + 2B²)(A - 2B²) = 4AC(AC + B²) ... <26>

Re-expressing <26> as a quadratic in B² gives
4B4 + 4B²AC + A²(4C² - 1) = 0 ... <26a>

B² and B are to be rational numbers, so the discriminant from <26a>, namely 4A²(1-3C²), has to be a perfect square. Remembering that we are also confining ourselves to rational values of A, we may deduce that:
Ö(1 - 3C²) is a rational quantity.

Inevitably, there is an infinite family of possible values of C to fulfill such a condition. Some examples are:
¹/2 . . . .4/7 . . . .3/14 . . . .5/14 . . . .4/13 . . . .8/19.

Such values may be fed in turn into equation <26a>, which is probably best regarded as a quadratic in B²/A. (after dividing throughout by A²). Dealing with B²/A as a single entity certainly reduces the algebraic complexity of the solution-process.

Thus, for example:
With C = 1/2 ... B²/A = - 1/2 ... (rejecting the option with B=0)
With C = 4/7 ... B²/A = - 3/14 ... or ... - 5/14
With C = 3/14 ... B²/A = 5/14 ... or ... - 4/7
With C = 4/13 ... B²/A = 7/26 ... or ... - 15/26
Incidentally, the possible values of B²/A are also "constrained", because (A²/B4 - 3) also has to be a perfect square. This may be demonstrated by rearranging <26> and analysing a suitable discriminant.
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Seventh possibility:

Following on from the sixth possibility, it is no use trying
sin(10) = ... t = Ö[A.³Ö(N²) + B.³ÖN + C] ... <27>
because (comparing with the second possibility), it is obvious that sin(70) and sin(-70) would then have to conform to equations <15> and <16>, which is not feasible.
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Eighth possibility:

sin(10) = ... t = A.³Ö(N²) + B.³ÖN ... <28>

This is a special case of the sixth possibilty, but with C=0.

Equations <17> and <19> then both become
4ABN = 1 ... <29>
 

Putting C=0 into <18> gives
8A³N² + 8B³N + 1 = 0 .... <30>

However, substituting for N from <29> into <30> produces complex values of B²/A.
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Ninth possibility:

sin(10) = ... t = P.³Ö(ÖN) + Q.³ÖN + R ... <31>

With this model there are problems with multiplying out and comparing the different terms of the modified version of <16>. There are too many equations for too few unknowns; and some terms simply cannot be matched up with others of similar type.
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Tenth possibility:

After the failure of our third possibility, let us look at
sin(10) = ... t = FÖN + LÖM + GÖNÖM + H ... <32>
(i.e., containing a ÖN in addition to a ÖM), requiring
8.[FÖN + LÖM + GÖNÖM + H]³ = 6.[FÖN + LÖM + GÖNÖM + H] - 1 ... <33>
yielding four equations with six unknowns. At first glance, the algebra looks rather formidable, but three of the equations are related by a 'symmetry' or 'pattern-repetition' similar to that which proved fruitful when combining <17> and <19> in the sixth possibility:

We multiply by L the equation (not shown here) obtained by collecting ÖN terms from <33>, then multiply by F the relationship derived from the ÖM terms, and subtract the two new equations. That yields:
Either:
FL + 3GH = 0 ... <34> > >

> > Or:
L²/F² = N/M ... <35>
- which would invalidate model <32>, because the distinction between N and M would disappear - sending us right back to model <2>. In other words, <35> may be ruled out here. However, we can still hope for something useful from <34>.
 

We may also multiply by G the equation obtained by collecting ÖN terms from <33>, then multiply by F the relationship (not shown here) derived from the ÖMÖN terms, and subtract. That produces:
Either:
FGN + 3LH = 0 ... <36> > >

> > Or:
F² = MG² ... <37>
- which is inconsistent with model <32>, because M must not be a perfect square ...
 

So we would have to accept <34> and <36>. Eliminating L between them - gives
N = 9H²/F²... <38>
- which is not permitted - because it is no use having N a perfect square.

Thus, model <32> will not work.
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Eleventh possibility:

sin(10) = ... t = Ö[FÖN + LÖM + GÖNÖM + H] ... <39>

This is the obvious 'follow-on' after the tenth one, and is not feasible - for the same reasons that the second and seventh possibilities were rejected.
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Twelfth and other possibilities:

If we now move up to expressions involving fourth roots, then we might consider:
sin(10) = ... t = A.ÖÖN+ B.ÖN + C ... <40>

It is probably advisable to include a N¾ term - which is of course ÖÖ(N³) or (ÖÖN)³
i.e., sin(10) = ... t = A.ÖÖN+ B.ÖN + D.N¾ + C ... <41>
 

However, it might be more sensible to proceed straight to sixth roots and (if necessary) ninth roots - obviously including all 'compound' fractional powers like
5/6 ...
or 2/9 ... 4/9 ... 5/9 and 7/9 ... etc.

Thus, I have indeed taken a preliminary look at a model which assumes that
sin(10) = ... t = A.N1/6 + B.N1/3 + C.N1/2+ D.N2/3+ F.N5/6 + H ... <42>
- but have not yet been able to draw any conclusions.
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David McNaughton
E-mail: DLMcN@yahoo.com

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