Note added subsequently:
After this article had been posted in the Internet, emails 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]:
_____________________________________________
... 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>
______________________________________________
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 ivalues 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 surdexpression 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 surdequivalent
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
surdexpressions
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>.
________________________________________________________________
Rewriting <1> as
8t³ =
6t  1 .... <1b>
helps appreciate the difficulty of finding a suitable surdexpression
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 RightHandSide of equation <1b>.
Besides, shown below is a circular interlocking relationship which demonstrates
how a repeated application of (12sin²q)
 [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 nonzero 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 nonÖG
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/sin10irr.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).
Despite the failure of model <2>, we
can actually get quite close with
sin(10) @ 3·002
 2Ö2
and can further improve it by finetuning 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 Ö2terms 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}
________________________________________________________
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.
________________________________________________________
Third possibility:
Let us introduce a second squarerooted 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 surdvalues 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.
__________________________________________________________
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 surdvalues 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.
__________________________________________________________
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>
___________________________________________________________________
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 Nsquared" 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
A_{1} + A_{2}
+ A_{3} = 0 ... <22>
where A_{1}, A_{2} and A_{3} are the individual
Acoefficients in <15> for each of the three roots of <1> ...
... and similarly B_{1}
+ B_{2} + B_{3}
= 0 ... <23>
and C_{1} + C_{2}
+ C_{3} = 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).

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>
Reexpressing <26> as a quadratic in B² gives
4B^{4} + 4B²AC
+ A²(4C²

1) = 0 ... <26a>
B² and B are to be rational numbers, so the
discriminant from <26a>, namely 4A²(13C²), 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 solutionprocess.
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/B^{4}
 3) also has to be a perfect square. This
may be demonstrated by rearranging <26> and analysing a suitable discriminant.
______________________________________________________________
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.
______________________________________________________________
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.
______________________________________________________________
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.
______________________________________________________________
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 'patternrepetition' 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.
______________________________________________________________
Eleventh possibility:
sin(10) = ... t = Ö[FÖN + LÖM + GÖNÖM + H] ... <39>
This is the obvious 'followon' after the tenth one, and is not feasible
 for the same reasons that the second and seventh possibilities were rejected.
______________________________________________________________
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 N^{¾}
term  which is of course ÖÖ(N³)
or
(ÖÖN)³
i.e., sin(10) = ... t = A.ÖÖN+
B.ÖN + D.N^{¾}
+ 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.N^{1/6
}+
B.N^{1/3}
+ C.N^{1/2}+
D.N^{2/3}+
F.N^{5/6}
+ H ... <42>
 but have not yet been able to draw any conclusions.
_______________________________________________________________________
David McNaughton
Email: DLMcN@yahoo.com
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