Trigonometric expressions in terms of square roots

David McNaughton

Note that the square root symbol is unfortunately rendered as Ö in Linux

The well-known trigonometric values which can be written using square roots are of course:
sin 45º = 1/Ö2
sin 60º = Ö3/2 and
tan 60º = Ö3
along with others which are closely related
(remembering too that sin 30º = ½)

Maths textbooks will sometimes also mention that
sin 75º = (Ö3+1) / 2Ö2 and
sin 15º = (Ö3-1) / 2Ö2
because these can easily be derived by expanding sin (45±30)
- as (sin 45 . cos 30 ± cos 45 . sin 30).

More concise are
tan 15º = 2 - Ö3 and
sec 15º = Ö6 - Ö2

Multiples of 18 degrees

Less widely known are
sin 18º = cos 72º = (Ö5-1) / 4  and
sin 54º = cos 36º = (Ö5+1) / 4

How can we demonstrate these?
If for instance q = 18º, then sin 3q = cos 2q.
Letting y = sin q, and using standard formulae to expand that last equation gives
4y³ - 2y² -3y + 1 = 0.

Dividing out the unwanted solution for y=1 (i.e. for q=90º), leaves
4y² + 2y - 1 = 0
leading to y = (-1 ± Ö5) / 4.

Other identities involving multiples of 18º may of course be evaluated using Pythagoras's theorem, but they turn out to be clumsier, e.g.
cos 18º = Ö(5 +Ö5) / 2Ö2 and
tan 72º = Ö(5 + 2Ö5)

Other angles

Multiples of angles as small as 3º can of course be tackled
by expanding combinations such as sin (18-15).

And halved quantities are easily dealt with using formulae
like cos 2q = 1 - 2sin²q. In this way we find for example that
sin 22½º = ½Ö(2 - Ö2) and
tan 22½º = Ö2 - 1

Indeed, there is no limit to the extent to which we can indefinitely keep halving angles like that, although the expressions will become progressively clumsier.

Multiples of 10 degrees

If we can find a way of computing quantities like sin 10º and sin 20º in terms of surds, then it would obviously be possible to extend them to trigonometric functions of any integral number of degrees (and indeed half-degrees, quarter-degrees, etc.) ...
... for example by working on sin (20-18).

By considering 30 as (10 + 10 + 10) and applying standard trig. formulae,
we can show 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).

Cardano's algorithm enables the evaluation of cubic equations like this one. If indeed it is possible to write the answers as surds, then these would probably comprise a horrendous mixture of cube and higher roots. (Nevertheless, see the "curiosities" mentioned below).

Incidentally, it was the quest for a solution to cubic equations which first persuaded 16th century mathematicians to take seriously the square root of minus one (written as "i", or occasionally as "j"). Even when all three answers are real, it is usually necessary during the manipulation process to work with i-values (which get eliminated later).

Accordingly, when attempting to solve the above equation for sin 10º, I was faced with the task of finding
the cube root of (Ö3i/2 - ½), which is of course
cos 40º + i.sin 40º (along with other possible values) - but unfortunately the surds have now become lost. In other words, the calculations would certainly have produced correct answers, but of no help in the context of this essay.

Some unsuccessful attempts to express sin(10) in terms of cube roots and square roots are discussed in

It has been suggested that the ancient Babylonian mathematicians divided the circle into 360 degrees because that figure was close to the number of days in a year. However, if instead they had decided on (say) 240, or 480 angular parts, then few people would have bothered to search for surd-expressions of sines and cosines of the equivalent of our ten degrees, and one degree; perhaps then they would never have stumbled upon the "curiosities" in the next section! Indeed, this is why (in the next section) I deliberately searched for a couple of expressions involving multiples of p/7 radians.

A few curiosities

tan²20º + tan²40º + tan²80º = 33
tan 20º . tan 40º . tan 80º = Ö3

And if a represents p/7 radians, i.e. 25.7142857º...,
tan a . tan 2a . tan 3a = Ö7
tan²a + tan²2a + tan²3a = 21

Expressions like these are deduced from equations whose roots are the terms (like those above) being multiplied or added. The key therefore lies in inspecting the coefficients in the original equation - because some of them are of course equal to the sum or product of the roots.

In the previous section, for example, we could have noticed that:
8.sin 10º . sin 50º . sin 70º = 1
and even more thought-provoking:
sin 10º + sin 50º = sin 70º... <2> ... because this will be the clue for solving the geometry problem in Figure 1 below, which would probably be difficult using conventional techniques.

Also intriguing are
sin²36º + sin²72º = 1¼
sin 20º + sin 40º = sin 80º .... <3>
sin 15º + sin 45º = sin 75º .... <4>

This last identity <4> can of course be confirmed using the known surd expressions for the angles concerned. Furthermore, it could be used to construct another diagram like Figure 1 (still having to prove AQ equal to CP) , but with angles of 30º, 30º and 15º replacing the 20º, 40º and 10º. Because of its symmetry, this alternative problem could probably be solved and verified geometrically.

Expressions <2>, <3> and <4> could have been deduced using the standard trigonometric formula
sin(C) + sin(D) = 2.sin(C/2 + D/2).cos(C/2 - D/2)

And they all follow the pattern
sin(y) + sin(60º - y) = sin(60º + y) - which is of course obvious just by expanding the terms.

Another family of identities may be defined by
sin(45º + y) - sin(45º - y) = Ö2sin(y)
- producing for example
sin(50º) - sin(40º) = Ö2sin(5º)
together with an infinite number of others.

David L. McNaughton

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