Finding Exact Trig Values for All Angles

[Mentallic]

Most would have memorized the exact trigonometric values for those special angles (30^o,45^o,60^o), but these angles are only special due to the convenience of the geometry of the isosceles triangle. If I were asked to find the value of, say, cos30^o without use of these triangles, I would be out of luck doing so.
With some study in complex numbers, our class was able to find the trig values for 15^o and 75^o by a method other than sin(A+B) ~ cos(A+B).
So now we know all the trig values for angles that are multiples of 15^o, which I see as being a large restriction.

Eventually, by some unconventional way using complex polynomials, we found the trig values to 50^o. This was a great breakthrough for us, but it is seemingly impossible to try and apply the same method to find other obscure angles.

Is it possible to endeavour in finding the trig values (mostly sine and cosine) for any angles?

 

[lurflurf]

let t stand for the computation of any trig function
I use radians
What is t(5pi/18)?
You know t(pi/12)
the other main easy one is t(pi/10)
with the difference formula one then know how to handle t(pi/60) and also all multiples of pi/60
t(pi/2^n) n=1,2,3,... is trivial by repeated use of half angle formula
it is a trivial fact that t(r*pi) is algebraic when r is rational
thus we can compute t when we can solve the polynomial
if we can solve cubics we know t(x/3) when we know t(x)
now comes the question of which polynomails we can solve, which depends on which tools we can use
we can use Galois theory to know how powerful particular tools are
classically we restrict our tools to solution by radical ie /*-+ and roots
thus we cannot solve the general quintic though we can reduce to
x^5+ax+b=0 so if we can solve that (bring radical) we can solve the quintic
on the other hand if we do not allow complex numbers we cannot do t(pi/9) or t(pi/180)
it is a matter of which functions we allow in our expansion and a question of what we want to do with our exact expressions

Also we do not need triangles we can compute with algebra, though one could argue that they are the same in some sense or that triangles give more insight.
right angles help
pi/2=pi/4+pi/4
pi/2=pi/12+5pi/6
pi/2=pi/6+pi/3
pi/2=pi/2+pi/2
pi/2=pi/10+2pi/5
pi/2=pi/5+3pi/10
pi/2=pi/2+0

 

[MathematicalPhysicist]

The simple answer is that with geometry we can compute every angle's cosine and sine, the work is messy, but it's possible.

 

[HallsofIvy]

If you mean "exact" values in terms of radicals, no that is not true. We can, with a lot of work, find exact values for rational multiples of $\pi$ but not for irrational multiples.

 

[MathematicalPhysicist]

The proof of this is in algebra correct, Halls?

Can you prove this within the framework of euclidean geometry?

 

[Mentallic]

it is a trivial fact that t(r*pi) is algebraic when r is rational
thus we can compute t when we can solve the polynomial
if we can solve cubics we know t(x/3) when we know t(x)
now comes the question of which polynomails we can solve, which depends on which tools we can use

So we are now restricted to which polynomials we can solve with Galois' theorem. Rather than taking any real polynomial and expressing its roots in terms of trigonometric values...

If you mean "exact" values in terms of radicals, no that is not true. We can, with a lot of work, find exact values for rational multiples of $\pi$ but not for irrational multiples.

Yes in terms of radicals. Does this include all rational multiples of $\pi$ or only a select few which is what I have been able to find so far on the net? I would like to know what this hard work involves (of course I don't expect these values to handed out on a platter).

The simple answer is that with geometry we can compute every angle's cosine and sine, the work is messy, but it's possible.

Hopefully it can also be done with algebra?

 

[Ben Niehoff]

It is possible to find an exact expression, in radicals, for the sine of 1 degree.

Not every rational multiple of $\pi$ can be worked out, because the degree of the polynomial equation becomes too high.

 

[qntty]

You might be interested in this wikipedia article on constructing polygons which is reduced to computing the exact value of $\cos{\frac{2 \pi}{n}}$ to construct an n-gon.

 

[lurflurf]

http://www.mathpages.com/HOME/kmath186.htm

more

https://nrich.maths.org/discus/messages/117730/144228.html?1214753730

The nice link of qntty's show we need the denominator of the form
n=2^i*3^j*5^k*17^l*257^m*65537^n
where i=0,1,2,...
j,k,l,m,n=0,1
to use only square roots
if higher roots are allowed we can handle all r*pi when r is rational
ie
2pi/9=2pi/3^2
or
2pi/360=2pi/(2^3*3^2*5)

in particular and unhelpfully
cos(r*pi/n)+i*sin(m*pi/n)=(-1)^r
cos(r*pi/n)=Re{(-1)^r}
sin(r*pi)=Im{(-1)^r}

since (-1)^r=exp(r*log(-1))=exp(r*pi*i)

of course most people would not consider that satisfatory and would proceed to do horrible algebra until the answer looked impressive

So we are now restricted to which polynomials we can solve with Galois' theorem. Rather than taking any real polynomial and expressing its roots in terms of trigonometric values...


Yes in terms of radicals. Does this include all rational multiples of $\pi$ or only a select few which is what I have been able to find so far on the net? I would like to know what this hard work involves (of course I don't expect these values to handed out on a platter).


Hopefully it can also be done with algebra?

You hint at the inverse problem of finding the polinomials that can be solved with trig.
If I recall correctly a real cubic with positive discriminant has 3 real roots and can be solved with cosine ie
x^3+px+q=0 has solutions
2sqrt(-p/3)cos(t-2pi/3)
2sqrt(-p/3)cos(t)
2sqrt(-p/3)cos(t+2pi/3)
where t=(1/3)arccos(.5q(-3/p)^-1.5)


The hard work involves tedious algebra

 

[Mentallic]

Thank you lurflurf, that cleared up a lot. I especially like the fact that the cosines correspond exactly with the cubics we solved

<snip>
can be solved with cosine ie
x^3+px+q=0 has solutions
2sqrt(-p/3)cos(t-2pi/3)
2sqrt(-p/3)cos(t)
2sqrt(-p/3)cos(t+2pi/3)
where t=(1/3)arccos(.5q(-3/p)^-1.5)

Ben Niehoff, I was curious what you meant by:

Not every rational multiple of can be worked out, because the degree of the polynomial equation becomes too high.

and luckily enough, one of the links presented in this topic explain what you meant by it. While very small, rational angles, less than a degree even have exact values, some of the larger angles don't.