Sine of five degrees

Today’s the first day of a new month, which means the exponential sum of the day will be simpler than usual. The exponential sum of the day plots the partial sums of

\sum_{n=0}^N \exp\left( 2\pi i \left( \frac{n}{m} + \frac{n^2}{d} + \frac{n^3}{y} \right ) \right )

where md, and y are the month, day, and (two-digit) year. The n/d term is simply n, and integer, when d = 1 and so it has no effect because exp(2πn) = 1. Here’s today’s sum, the plot formed by the partial sums above.

exponential sum for August 1, 2018

It’s possible in principle to work out exactly each of the points are on the curve. In order to do so, you’d need to compute

\exp\left(\frac{2\pi i}{72}\right) = \cos\left(\frac{2\pi}{72}\right) + i \sin\left(\frac{2\pi}{72}\right) = \cos(5^\circ) + i \sin(5^\circ)

Since we can use the Pythagorean theorem to compute sine from cosine and vice versa, we only need to know how to compute the sine of five degrees.

OK, so how do we do that? The sine and cosine of 45 degrees and of 30 degrees are well known, and we can use the trig identity for sin(A – B) to find that the sine of 15 degrees is (√3 – 1)/ 2√2. So how do we get from the sine of 15 degrees to the sine of 5 degrees? We can borrow a trick from medieval astronomers. They did something similar to our work above, then solved the trig identity

\sin 3\theta = 3 \sin\theta - 4\sin^3\theta

as a cubic equation in sin θ to find the sine of 1/3 of an angle with known sine.

If we ask Mathematica to solve this equation for us

    Solve[3 x -4 x^3 == Sin[15 Degree], x]

we get three roots, and it’s the middle one that we’re after. (We can evaluate the roots numerically and tell that the middle one has the right magnitude.) So Mathematica says that the sine of 5 degrees is

\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{-\sqrt{2}+\sqrt{6}+i \sqrt{8+4 \sqrt{3}}}}{4\ 2^{2/3}}+\frac{1-i \sqrt{3}}{2 \sqrt[3]{2 \left(-\sqrt{2}+\sqrt{6}+i \sqrt{8+4 \sqrt{3}}\right)}}

Unfortunately, this is the end of the line for Mathematica. We know that the expression above must be real, but Mathematica is unable to find its real part. I tried running FullSimplify on this expression, and aborted evaluation after it ran for 20 minutes or so. If anybody wants to pick up where Mathematica left off, let me know your solution. It may be possible to continue with Mathematica by giving the software some help, or it may be easier to continue with pencil and paper.

Update: Based on the comments below, it seems this is about as far as we can go. This was a surprise to me. We’re solving a cubic equation, less than 5th degree, so the roots are expressible in terms of radicals. But that doesn’t mean we can eliminate the appearance of the i‘s, even though we know the imaginary part is zero.

6 thoughts on “Sine of five degrees

  1. Using first ComplexExpand and then FullSimplify gets you an expression that depends on cosine and sine of 25 degrees. Don’t know if this is useful or not. (30 – 5 = 25 so maybe)

  2. Using

    Reduce[3 x – 4 x x x == Sin[Pi / 12] && 0 < x < 2 Pi / 72]

    Mathematica says that sin(2π/72) is a root of the irreducible polynomial 4096 x^{12}-12288 x^{10}+13824 x^8-7168 x^6+1680 x^4-144 x^2+1.

  3. Using maxima: solve([3*x-4*x^3 = (sqrt(3) – 1) / (2 * sqrt(2))], [x])

    The first answer of the triplet is:

    x=((-1)/2-(sqrt(3)*%i)/2)*((sqrt(2*sqrt(3)+4)*%i)/2^(9/2)-(3^(1/2)-1)/(2*(2^(1/2))^7))^(1/3)

    Calling realpart on that gives (oh for MathJax in the comments):
    x=((sqrt(3)*sin((%pi-atan(sqrt(2*sqrt(3)+4)/(sqrt(3)-1)))/3))/(2*(sqrt(3)/256+(sqrt(3)-1)^2/512+1/128)^(1/6))+((-1)*cos((%pi-atan(sqrt(2*sqrt(3)+4)/(sqrt(3)-1)))/3))/(2*(sqrt(3)/256+(sqrt(3)-1)^2/512+1/128)^(1/6)))/4+((sqrt(3)/256+(sqrt(3)-1)^2/512+1/128)^(1/6)*sqrt(3)*sin((%pi-atan(sqrt(2*sqrt(3)+4)/(sqrt(3)-1)))/3))/2+((-1)*(sqrt(3)/256+(sqrt(3)-1)^2/512+1/128)^(1/6)*cos((%pi-atan(sqrt(2*sqrt(3)+4)/(sqrt(3)-1)))/3))/2

    which when evaluated is 0.08715574274765811 which is sin 5 degrees. Is this what you meant, or did I misunderstand you?

  4. This is indeed exactly a case of (generalized) casus irreducibilis.
    In the field ℚ(√2, √3), you are trying to solve 3 x – 4 x³ = (√3 – 1)/(2√2), whose disrciminant is 108(2+√3) > 0.
    The equation does not have any rational solutions (in ℚ(√2, √3) ; the proof is left to the reader), hence three distincts real solutions.
    And unfortunately, by this criterion, none of which are expressibles with nested radicals (of any degree) in ℚ(√2, √3) (hence radicals in ℚ). All three require some radicals of complex numbers to be expressed in closed forms.

Leave a Reply

Your email address will not be published. Required fields are marked *