Hi Ferret Warlord,
If there is anything that is not easy to understand, please tell me right away, since I am not certain of the appropriate level of explanation in all cases.
In fact,
Niven's Theorem states that the only rational θ in degrees (0°<=θ<=90°) for which sin(θ) is rational is sin(0°)=0, sin(30°)=1/2, and sin(90°)=1. Likewise this implies that the only rational θ in degrees (0°<=θ<=90°) for which cos(θ) is rational is cos(0°)=1, cos(60°)=1/2, and cos(90°)=0. Note that rational degrees is the same as rational multiple of pi in radians. Note also that this theorem says nothing about tangent.
However, a
post on xkcd.com indicated a rather simple way to prove that, say, acos(1/5) is not a rational multiple of pi, better than what I posted, although it assumes something from field theory. (In my previous post, I used the
Binomial Theorem on (1/5+i√24/5)
m and then used some number theory to prove it).
First of all, θ is a rational multiple of pi if and only if cos(θ) + i*sin(θ) is a root of unity. So, let's say that we want to prove that acos(1/5) is not a rational multiple of pi. So we want to prove that 1/5+i*√24/5 is not a root of unity.
Now 1/5+i*√24/5 is an
algebraic number of degree 2, since if z=1/5+i*√24/5, then 5z-1=i*√24 and 25z
2-10z+1=-24 and so 1/5+i*√24/5 is a root of 25z
2-10z+25.
However, the algebraic degree of a
primitive nth root of unity is φ(n), where φ is the
Euler phi function (in other words, the number of positive integers between 1 and n inclusive that are relatively prime to n).
If n has prime factorization p1
a1p2
a2...pk
ak, then
φ(n)=(p1-1)p1
a1-1 (p2-1)p2
a2-1 ... (pk-1)pk
ak-1.
Using this, we can determine that the only values n for which φ(n)=2 are n=3,4,6. This gives the only possible nth roots of unity with n being one of these values are ±1, ±i, ±1/2±i*√3/2, none of which are equal to 1/5+i*√24/5. So 1/5+i*√24/5 is not a root of unity, and so acos(1/5) is not a rational multiple of pi.
In fact, we can prove a general case. To prove that asin(p/q) or acos(p/q) or atan(p/q) is not a rational multiple of pi except for a handful of exceptions for p,q, we just have to show that
√(q
2-p
2)/q+i*p/q or
p/q+i*√(q
2-p
2)/q or
q/√(p
2+q
2)+i*p/√(p
2+q
2), respectively,
is not a root of unity. The above quantities each have algebraic degree at most 4. The only values n for which φ(n) is at most 4 are n=1,2,3,4,5,6,8,10,12. This gives as the only possibilities for nth root of unity ±1, ±i, ±√2/2±i*√2/2, ±1/2±i*√3/2, ±√3/2±i*1/2, ±(√5+1)/4±i*√(2√5-10)/4, ±(√5-1)/4±i*√(2√5+10)/4. I leave the rest for you to work out. The asin case results in Niven's Theorem.
Edit: Corrected a mistake for the atan case.
