I recently saw Lindemann's proof that $\pi$ is transcendental by using the fact that $e^{i\pi} = -1$, and this made me realize that the Lindemann-Weierstrass theorem implies that the $\cos$ , $\sin$ of all rational angles is transcendental.
Do we know anything about the other direction, that is, which transcendental numbers give algebraic answers for the trig functions.
Thanks 🙂
Best Answer
I don't have much depth to offer here, but I do find the question quite engaging; so I fooled around with it for awhile and was able to affirm my hunch that, at the very least, the cosine of any rational multiple of $\pi$ is algebraic over $\Bbb Q$; the calcuations are pretty simple though a little "grungy"; here's what I've got:
Set
$\theta = \dfrac{p \pi}{q}, \; 0 \ne p, 0 < q \in \Bbb Z; \tag 1$
then
$e^{i\theta} = \cos \theta + i \sin \theta = \cos \left (\dfrac{p \pi}{q} \right ) + i \sin \left ( \dfrac{p \pi}{q} \right ); \tag 2$
$(e^{i\theta})^q = e^{iq \theta} = \cos q \theta + i \sin q \theta$ $= \cos q \left ( \dfrac{p \pi}{q} \right ) + i \sin q \left ( \dfrac {p \pi}{q} \right ) = \cos p \pi + i \sin p \pi = (-1)^p; \tag 3$
$(e^{i\theta})^q = (\cos \theta + i\sin \theta)^q = \displaystyle \sum_0^q i^{k}\dfrac{q!}{k!(q - k)!} \cos^{q - k} \theta \sin^k \theta; \tag 4$
we may break this sum into two sums, over even and odd $k$, which are then its real (even) and imaginary (odd) parts; we will effect this operation by introducing a new index $j$ such that
$\displaystyle \sum_0^q i^{k}\dfrac{q!}{k!(q - k)!} \cos^{q - k} \theta \sin^k \theta = \sum_{j = 0}^{ 2j \le q} i^{2j} \dfrac{q!}{(2j)!(q - 2j)!} \cos^{q - 2j} \theta \sin^{2j} \theta$ $+ \displaystyle \sum_{j = 0}^{ 2j + 1 \le q} i^{2j + 1} \dfrac{q!}{(2j + 1)!(q - 2j - 1)!} \cos^{q - 2j - 1} \theta \sin^{2j + 1} \theta; \tag 5$
since $i^{2j} = (-1)^j$ and $i^{2j + 1} = (-1)^j i$, the sums on the right may be re-written and we have
$(e^{i\theta})^q = \displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} \cos^{q - 2j} \theta \sin^{2j} \theta$ $+ \displaystyle i \sum_{j = 0}^{ 2j + 1 \le q} (-1)^j \dfrac{q!}{(2j + 1)!(q - 2j - 1)!} \cos^{q - 2j - 1} \theta \sin^{2j + 1} \theta; \tag 6$
we see from (3),
$(e^{i\theta})^q = (-1)^p, \tag 7$
that $(e^{i\theta})^q$ is real; therefore, the second sum on the right of (6) vanishes; being a real multiple of $i$, this entire sum represents the imaginary part of $(e^{i\theta})^q$; we therefore need not deal further with this second sum, and will from here on only focus on the first, which is the real part of $(e^{i\theta})^q$; thus,
$(e^{i\theta})^q = \displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} \cos^{q - 2j} \theta \sin^{2j} \theta = (-1)^p; \tag 8$
our next step is to use the identity
$\sin^2 \theta = 1 - \cos^2 \theta \tag 9$
to eliminate the $\sin^{2j} \theta$ factors in (8), replacing them by
$\sin^{2j} \theta = (\sin^2 \theta)^j = (1 - \cos^2 \theta)^j, \tag{10}$
whence
$(e^{i\theta})^q = \displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} (\cos^{q - 2j} \theta) (1 - \cos^2 \theta)^j = (-1)^p, \tag{11}$
or
$(e^{i\theta})^q - (-1)^p = \left (\displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} (\cos^{q - 2j} \theta) (1 - \cos^2 \theta)^j \right ) + (-1)^{p + 1} = 0, \tag{12}$
which shows that $\cos \theta = \cos (p\pi / q)$ satisfies the polynomial
$P_{p, \; q}[x] = \left (\displaystyle \sum_{j = 0}^{ 2j \le q} (-1)^j \dfrac{q!}{(2j)!(q - 2j)!} x^{q - 2j} (1 - x^2)^j \right ) + (-1)^{p +1} \in \Bbb Q[x], \tag{13}$
and is hence algebraic over $\Bbb Q$. In fact, it is easy to see that the coefficients of $P_{p, \; q}[x]$ are integers, so that indeed
$P_{p,\; q}[x] \in \Bbb Z[x]. \tag{14}$
Now it should come as no surprise that the "results" derived here are well-known, and deeply analyzed; see this wikipedia article on Chebyshev_polynomials.