The Sine of 1 Degree

Date: 01/07/2001 at 23:49:32
From: David Goldberg
Subject: Finding an exact value for sin(1 degree)

Dear Dr. Math,

I am an 11th grader in high school with great interest and passion for 
mathematics. My teacher has started talking about trigonometry, and 
the double- and half-angle formulas, and I realized that if the sine 
of 1 degree is known exactly (in terms of radicals), the sin of any 
angle can be found using other multiple angle formulas. I figure 
that since the sin(30) is known to be 1/2, all that is required is a 
general formula for sin(x/5) and sin (x/3), which can then be used 
to calculate sin(((30/5)/3)/2) = sin(1).

To find the formula for sin(x/a), I know that first the formula for 
sin(ax) must be found. I know the general angle-addition and 
double-angle formulas, and saw that they could be applied to find the 
sines of higher multiples of angles. The following is a series of 
expansions and then simplifications using the angle-addition formula 
to find a formula for sin(5x) in terms of sin(x) and cos(x).

You can read the work, which is somewhat cumbersome, or skip to the 
formula which is derived, and which I have tested.

     sin(5x) = sin(2x+3x)
             = sin(2x)cos(3x) + cos(2x)sin(3x)
             = 2*sin(x)cos(x)cos(x+2x) + (cos^2(x)-sin^2(x))sin(x+2x)
             = 2*sin(x)cos(x)(cos(x)cos(2x)-sin(x)sin(2x)) +
               (cos^2(x)-sin^2(x))*
               (sin(x)(cos^2(x)-sin^2(x))+cos(2x)sin(x)cos(x))
             = 2*sin(x)cos(x)*
               (cos^3(x)-sin^2(x)cos(x)-2*sin^2(x)cos(x)) +
               (cos^2(x)-sin^2(x))(3*cos^2(x)sin(x)-sin^3(x)) -
               10*sin^3(x)cos^2(x) + 5*cos^4(x)sin(x) + sin^5(x)
     sin(5x) = sin^5(x)-10*sin^3(x)cos^2(x)+5*cos^4(x)sin(x)

Thus, 

     sin (5*(x/5)) = 
               sin^5(x/5)-10*sin^3(x/5)cos^2(x/5)+5*cos^4(x/5)sin(x/5)

and

     sin(x) = sin^5(x/5)-10*sin^3(x/5)cos^2(x/5)+5*cos^4(x/5)sin(x/5)
     cos(x/5) = sqrt(1-sin^2(x/5))

By substitution and simplification:

     sin(x) = 16*sin^5(x/5) - 20*sin^3(x/5) + 5*sin(x/5)
     16*sin^5(x/5) - 20*sin^3(x/5) + 5*sin(x/5) - sin(x) = 0

Let r = sin(x/5), and a = sin(x). The above equation becomes:

     16r^5-20r^3+5r-a = 0

If r can be solved for in terms of a and the integer coefficients, 
then sin(x/5) can be expressed in terms of sin(x), and the sin(1) can 
be found exactly with radical values.

I have done the same for sin(x/3) and have reached an equation which I 
see is easily solvable using the general solution to third degree 
equations.

I know of Abel's impossibility theorem, which states that there is no 
general solution to 5th degree equations, and I know something about 
Galois theory, group theory, and algorithms. However, I was thinking 
that since finding the sine of 1 degree is such a simple and beautiful 
task, perhaps the equation would be solvable, for I know that certain 
fifth degree equations are indeed factorable and solvable. I thought 
this might be the case especially since the equation has no 4th degree 
or 2nd degree terms.

I would be very grateful if you could tell me if the equation is 
solvable exactly using radicals, and if so, perhaps guide me towards a 
solution.

Sincerely,
David Goldberg

Date: 01/08/2001 at 13:41:45
From: Doctor Schwa
Subject: Re: Finding an exact value for sin(1 degree)

Dear David,

That fifth degree equation certainly is solvable, but you've taken the 
hard road to getting the exact sin of 1 degree. I'd much rather solve 
a cubic - wouldn't you?

My steps for finding sin(1 degree) are like this:

You know sin 30. Find sin 18. You can do this with similar triangles, 
starting with a 36-72-72 triangle, drawing an angle bisector of one of 
the 72 degree angles, etc.

Then use the subtraction formula to get sin(12), then the half-angle 
formula to get sin(6) and sin(3), and finally use your sin(x/3) 
formula to get sin(1).

Give that a try and feel free to write us back if you'd like more 
hints along those lines.

Nice work!

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/   

Date: 01/11/2001 at 22:17:11
From: David Goldberg
Subject: Re: Finding an exact value for sin(1 degree)

Dear Dr. Math,

I know that

     sin(3x) = -4*sin^3(x) + 3*sin(x)

from the angle-addition formulas. Substituting x for x/3, I get:

     sin(x)  = -4*sin^3(x/3) + 3*sin(x/3)

I have verified this. Letting y = sin(x/3), I get the following cubic:

     -4*y^3 + 3*y = sin(x)

This can be written as:

     y^3 + (-3/4)y = -sin(x)/4

2 variables, s and t, are then introduced.

         3st = -3/4
     s^3-t^3 = -sin(x)/4

     t^6 + (-sin(x)/4)t^3 - ((-3/4)^3)/27 = 0

Let u = t^3.

     u^2 + (-sin(x)/4)u + .765625 = 0
     u = (sin(x)/4 + sqrt(sin^2(x)/16 - 3.0625))/2

From this point on, the term

     sqrt((sin^2(x)-49)/16) = sqrt(sin^2(x)-49)/4 

becomes a part of all further equations. As a final answer, 

     sin (x/3) = 
      (cbrt(sqrt(sin^2(x)-49)-sin(x))-cbrt(sqrt(sin^2(x)-49)+sin(x))/2

However, the problem is that the solution contains the term 
sqrt(sin^2(x)-49), which is always imaginary. Because of this, it 
would seem that the value is not a valid answer, for sin (x/5) is 
certainly a real number, if x is real. Is there any way around this?

Sincerely,
David Goldberg

Date: 01/12/2001 at 10:49:56
From: Doctor Schwa
Subject: Re: Finding an exact value for sin(1 degree)

Hi David,

This is exactly what frustrated mathematicians for about a century!

In the end, they had to admit the same thing I'm going to try to 
convince you of: sometimes the only way to get a solution to a cubic, 
a REAL solution, is to have COMPLEX numbers in the middle.

If you look at the cube roots, yes, each one is complex. But when you 
subtract them, since one of the numbers is of the form bi + a and the 
other of the form bi - a, you'll get just 2a ... a real number.

So, you'll just have to put up with having complex numbers in the 
middle of your work, and trust that they will all drop out in the end 
to give you a real solution.

Nice piece of work, David!

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/