__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 30**º** = 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 = 18**º**,
then **sin 3q = cos 2q.**

Letting y = sin q, and using standard formulae
to expand that last equation gives
**4y**³ **- 2y**² **-3y + 1 =
0.**

Dividing out the unwanted solution for y=1 (i.e. for q=90**º**),
leaves
**4y**² **+ 2y - 1 = 0**

leading to y = (-1 ± Ö5) / 4.

Other identities involving multiples of 18**º
**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 3**º**
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 10**º**
and sin 20**º** 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 10**º** (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 50**º**
and sin (-70**º**).

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 10**º**,
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 http://dlmcn.com/sin10-surd.html

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.

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

And if a represents p/7
radians, i.e. 25.7142857**º**...,
**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

E-mail: DLMcN@yahoo.com

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