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Homerational sine and cosine

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# 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. $1^{{\underline{o}}}$. 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.

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

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

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

$\displaystyle\gcd(a,\,c)\;=\;\gcd(b,\,d)\;=\;1.$ | (1) |

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

$\displaystyle\frac{a^{2}}{c^{2}}\!+\!\frac{b^{2}}{d^{2}}\;=\;1.$ | (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.

## Mathematics Subject Classification

26A09*no label found*11D09

*no label found*11A67

*no label found*

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