# Sine of factorial degrees

I was looking back at a post about the Soviet license plate game and was reminded of the amusing identity

sin (n!)° = 0

for n ≥ 6. Would it be possible to find sin (n!)° in closed form for all positive integers n?

For this post I’ll make an exception to my usual rule and assume all angles are in degrees. First of all, sin 5! = √3/2 so we’re off to a good start.

Next, let’s look at sin 4!. If we ask Mathematica, we get nowhere. Entering

    Sin[24 Degree]

returns Sin[24°]. I tried a few Simplify and Expand commands but couldn’t coax Mathematica into revealing a closed form.

It would seem we’re stuck, but we’re not. For reasons given here we can find sin 3: And if we can find sin 3, we can use a double angle identity to get sin 6, sin 12, and sin 24.

So we can compute sin n! for n ≥ 3. But we really are stuck at this point. The sine of 1 degree or 2 degrees cannot be computed in closed form . However, there are clever ways to numerically compute these values that go back to Medieval astronomers.

We can find sine of 6, 12, and 24 in closed form, so let’s actually do it.

    sin3 =  Sqrt[8 - Sqrt[10 - 2 Sqrt] - Sqrt (1 + Sqrt)]/4
sin6 = 2 sin3 Sqrt[1 - sin3^2]
sin12 = 2 sin6 Sqrt[1 - sin6^2]
sin24 = 2 sin12 Sqrt[1 - sin12^2]


We can apply Simplify to clean up these expressions. The sine of 12 degrees cleans up nicely, though the others are still complicated after asking Mathematica for a simplification. Depending on what you mean by “closed form.” I believe the sine of 1 degree can be written in terms of radicals, but the expression involves imaginary numbers that cannot be removed, even though the expression results in a real number.

Update: See the comments for a link to a paper confirming the footnote above and giving complex expressions for the sine of one degree.