Show that if $x$ is rational, then $\sin x$ is algebraic number when $x$ is in degrees and $\sin x$ is non algebraic when $x$ is in radians.

Details: so we have $\sin(p/q)$ is algebraic when $p/q$ is in degrees, that is what my book says. of course $\sin (30^{\circ})$, $\sin 45^{\circ}$, $\sin 90^{\circ}$, and halves of them is algebraic. but I'm not so sure about $\sin(1^{\circ})$.

Also is this is an existence proof or is there actually a way to show the full radical solution.

One way to get this started is change degrees to radians. x deg = pi/180 * x radian. So if x = p/q, then sin (p/q deg) = sin ( pi/180 * p/q rad). Therefore without loss of generality the question is show sin (pi*m/n rad) is algebraic. and then show sin (m/n rad) is non-algebraic.

  • $\begingroup$ For the second part you'll also need to assume $x\ne0$. $\endgroup$ – Henning Makholm Feb 24 '12 at 17:28
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    $\begingroup$ Claim is false: $\sin 0^{\circ}$ and $\sin 0$ (in radians) are both algebraic. $\endgroup$ – Arturo Magidin Feb 24 '12 at 17:29
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    $\begingroup$ Hint for the first part: Instead of considering $\sin(\frac{\pi}{180}x)$, view it as the real part of $z=-ie^{\frac{\pi}{180}xi}$ and consider $z^{180q}$ to see that $z$ is algebraic. Then the sine, being $\frac{z}{2}+\frac{\bar z}{2}$, is also algebraic, because the algebraic numbers are closed under addition. $\endgroup$ – Henning Makholm Feb 24 '12 at 17:31
  • $\begingroup$ You may be interested in this and Hardy's comment there about Niven's Theorems and some links. Ofcourse, this comes for free with enlightening answers from various others. $\endgroup$ – user21436 Feb 24 '12 at 17:45

Lindemann-Weierstrass theorem implies that for $\alpha$ non-zero algebraic, $\sin \alpha$ is transcendental.

  • $\begingroup$ well sin (pix) , pix is not algebraic since pi is not algebraic. That is sufficient to prove sin(pi*x) is algebraic? Are you saying that for any 'a' transcendental sin(a) is algebraic? I should qualify 'a' to be a real transcendental. so is sin(e) algebraic? $\endgroup$ – bob thornton Feb 24 '12 at 18:48
  • $\begingroup$ @bobthornton: No, I am saying for any $a$ algebraic (and non-zero), $\sin a$ is transcendental. The other portion of your question was answered by bgins. No clue when $a$ is transcendental (except rational multiples of $\pi$). $\endgroup$ – Aryabhata Feb 24 '12 at 18:58

$\sin\left(\frac{p}{q}\pi\right)=\sin\left(\frac{p}{q}180^\circ\right)$ is always algebraic for $\frac{p}{q}\in\mathbb{Q}$: Let $$ \alpha=e^{\frac{i\pi}{q}}=\cos\frac{\pi}{q}+i\sin\frac{\pi}{q}. $$ Then $\alpha^q+1=0$, i.e. $\alpha$ is an (algebraic) $2q^\text{th}$ root of unity, i.e. it is a root of $x^{2q}-1$. Hence, so is its power $\alpha^p$ and reciprocal/conjugate power, which for $p$ an $q$ in lowest terms are roots of $x^q-(-1)^p=0$. Therefore, so too are $$ \cos\frac{p\pi}{q}=\frac{\alpha^p+\alpha^{-p}}{2} \qquad\text{and}\qquad \sin\frac{p\pi}{q}=\frac{\alpha^p-\alpha^{-p}}{2i}, $$ by the closure of the algebraic numbers as a field.

Ivan Niven gives a nice proof at least that $\sin x$ is irrational for (nonzero) rational $x$. As @Aryabhata points out, the Lindemann-Weierstrass theorem gives us that these values of $\sin$ and $\cos$ are transcendental (non-algebraic), by using the fact that the field extension $L/K$ of $L=\mathbb{Q}(\alpha)$ over $K=\mathbb{Q}$ has transcendence degree 1.

  • $\begingroup$ Why is a^q +1 = 0 algebraic? I thought algebraic number means it is the root of a polynomial with integer coefficients. this might be beyond the scope of my knowledge. But your alpha is a root of unity , and is a complex number. $\endgroup$ – bob thornton Feb 24 '12 at 18:19
  • $\begingroup$ I edited my post. Is it clear now? $\alpha$ is a root of $x^q+1=0$ and hence also $x^{2q}-1=0$. $\endgroup$ – bgins Feb 24 '12 at 18:23
  • $\begingroup$ @bob: The relevant polynomial is $x^q + 1$, not $\alpha^q + 1$. As $\alpha$ is a root, and $sin$ can be expressed linearly in $\alpha$, we have that $sin$ is algebraic at that value too. $\endgroup$ – davidlowryduda Feb 24 '12 at 18:23

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