rational sine and cosine

Theorem.  The only acute angles, whose sine and cosine are rational, are those determined by the Pythagorean triplets  $(a,b,c)$.

Proof. $1^{\underset{¯}{o}}$. When the catheti $a$, $b$ and the hypotenuse $c$ of a right triangle are integers, i.e. they form a Pythagorean triplet, then the sine $\frac{a}{c}$ and the cosine $\frac{b}{c}$ of one of the acute angles of the triangle are rational numbers.

$2^{\underset{¯}{o}}$. Let the sine and the cosine of an acute angle $\omega$ be rational numbers

$$\sin \omega =\frac{a}{c},\cos \omega =\frac{b}{d},$$

where the integers $a$, $b$, $c$, $d$ satisfy

$\gcd (a,c)=\gcd (b,d)=\mathrm{\hspace{0.33em}1}.$(1)

Since the square sum of sine and cosine is always 1, we have

${\displaystyle \frac{a^2}{c^2}}+{\displaystyle \frac{b^2}{d^2}}=\mathrm{\hspace{0.33em}1}.$(2)

By removing the denominators we get the Diophantine equation

$$a^2{d}^2+b^2{c}^2=c^2{d}^2.$$

Since two of its terms are divisible by $c^2$, also the third term $a^2{d}^2$ is divisible by $c^2$.  But because by (1), the integers $a^2$ and $c^2$ are coprime, we must have  $c^2\mid d^2$ (see the corollary of Bézout’s lemma).  Similarly, we also must have  $d^2\mid c^2$.  The last divisibility relations mean that  $c^2=d^2$,  whence (2) may be written

$$a^2+b^2=c^2,$$

and accordingly the sides $a,b,c$ of a corresponding right triangle are integers.