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I am using extensively trigonometric functions when an angle is given in degrees.
Some of these functions like sine or cosine have rational values, for example, the well known example is that $\cos(\theta) =0.6 $ and $\sin(\theta) =0.8 $.

However besides the case of multiplicity of $90^\circ$ it seems there are no rational numbers $\theta$ with simultaneously rational values of sine and cosine.

  • Is it possible somehow to prove that for rational values of an angle given in degrees there are no values simultaneously rational of sine and cosine functions, beside obvious case of multiplicities of $90^\circ$?
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Niven's Theorem: If $x/\pi$ (in radians) and $\sin x$ are both rational, then the sine takes values $0$, $\pm 1/2$, and $\pm 1$.

Obviously, angle in radians is a rational multiple of $\pi$ iff angle in degrees is rational.

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  • $\begingroup$ Hmm, how to translate this theorem into degrees ? $\endgroup$ – Widawensen Jun 28 '17 at 9:48
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    $\begingroup$ @Widawensen, angle in degrees$/360$ = angle in radians$/2\pi$. $\endgroup$ – Martín-Blas Pérez Pinilla Jun 28 '17 at 9:51
  • $\begingroup$ I'm starting to understand, so it's enough to prove this Niven theorem (thank you for presenting it) - I wonder whether the proof of it is very complicated.. $\endgroup$ – Widawensen Jun 28 '17 at 10:04
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    $\begingroup$ @Widawensen, the proof is nontrivial but elementary: proofwiki.org/wiki/Niven's_Theorem. $\endgroup$ – Martín-Blas Pérez Pinilla Jun 28 '17 at 10:08
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    $\begingroup$ @Widawensen There is a different proof (maybe comparable comlpexity) that I prefer in "When is the (co)sine of a rational angle equal to a rational number?": arxiv.org/abs/1006.2938 $\endgroup$ – Mark S. Jun 28 '17 at 11:53
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Let us take the Pythagorean theorem $a^2 + b^2 = c^2$ and divide both sides by $c^2$ to get $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$. Now, we can use the well-known relationship between sine and cosine: $\sin^2(x) + \cos^2 (x) = 1$, and let $\sin(x) = \frac{a^2}{c^2}$, and $\cos(x) = \frac{b^2}{c^2}$.

Since the hypotenuse of a triangle is always longer than the two legs, $\frac{a^2}{c^2}, \frac{b^2}{c^2} < 1$. Therefore, there exists a bijection between one triplet of $a,b,c$ and $\frac{a^2}{c^2}, \frac{b^2}{c^2}$, and at least one value of $\frac{a^2}{c^2}, \frac{b^2}{c^2}$ for $\sin(x)$.

One example using a $3,4,5$ triangle shows that $\cos(x) = \frac{9}{25}$, and using the second equation $\sin(x) = \frac{16}{25}$. As a bonus, since there are infinitely many Pythagorean triples, this shows that there are infinitely many values of $\sin(x)$ and $\cos(x)$ that are rational.

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    $\begingroup$ Thank you Toby for this answer but also $x$ expressed in degrees should be rational. I don't know whether this is possible even though there are infinitely many Pythagorean triples .. $\endgroup$ – Widawensen Jun 28 '17 at 10:08
  • $\begingroup$ For my example using the 3-4-5 triangle, I know there is at least one solution according to Wolfram Alpha: wolframalpha.com/input/?i=arcsin(16/25), wolframalpha.com/input/?i=arccos(9/25) $\endgroup$ – Toby Mak Jun 28 '17 at 10:13
  • $\begingroup$ Sorry to everyone who expected my older answer, I found a way to shorten it massively by jumping straight into the Pythagorean theorem, rather than working through the identity $\sin^2 (x) + \cos^2 (x) = 1$ and using rational numbers. $\endgroup$ – Toby Mak Jun 28 '17 at 10:14
  • $\begingroup$ @Widawensen I have to show that $\cos^{-1} (x)$ and $\sin^{-1} (x)$ are rational if $x$ is, which I cannot do with the knowledge that I have. Anyways, the previous answer probably expressed it better than I could have $\endgroup$ – Toby Mak Jun 28 '17 at 10:25
  • $\begingroup$ Yes, this Niven's theorem with the provided link to its proof seems to answer the question very well, but your answer is also very informative so let's treat it as a partial one.. $\endgroup$ – Widawensen Jun 28 '17 at 10:28

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