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Suppose I was working in radians and took

$$\sin \left(\frac{a\pi}{b}\right)$$

where both $a$ and $b$ are integers.

Is there a proof that the output of this function cannot be transcendental? Or, conversely, that

$$\arcsin \left(t \right)$$

where $t$ is some transcendental number, e.g., $e$, cannot be a rational angle $\left(\frac{a\pi }{b}\right)$?

Sadly, I couldn't get very far with this problem with my maths teacher so any information on the topic would be very useful.

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To start with, let's notice that we might just as well prove this fact for the cosines of rational multiples of $\pi$ as the sines. We just need to use the identity $\sin x=\cos\left(\frac{\pi}{2} - x\right)$, which gives us: $$\sin \frac{a\pi}{b}=\cos\left(\frac{\pi}{2} - \frac{a\pi}{b}\right)=\cos\left(\frac{(b-2a)\pi}{2b}\right)$$ As this is the cosine of a rational multiple of $\pi$, we get the result for sines more or less for free from the result for cosines.

Once this preliminary is out of the way, the key thing to notice is that $\cos(nx)$ can always be written as a polynomial function of $\cos x$. For example: \begin{align} \cos 2x &= 2 \cos^2 x - 1 \\ \cos 3x &= 4 \cos^3 x - 3 \cos x\\ \cos 4x &= 8 \cos^4 x - 8 \cos^2 x + 1 \\ \cos 5x &= 16 \cos^5 x - 20 \cos^3 x + 5 \cos x \end{align} and so on. (Formulas taken from here.)

Why does this matter? It means that we can find an explicit polynomial which has $\cos \frac{a\pi}{b}$ as a root. To see an example of this, let's look at $\alpha = \cos \frac{2\pi}{5}$. Using the last formula above, we know that $$ \cos (2\pi)=\cos\left(5 \cdot \frac{2\pi}{5}\right)=16 \cos^5 \frac{2\pi}{5}- 20 \cos^3 \frac{2\pi}{5} + 5 \cos \frac{2\pi}{5}=16\alpha^5-20\alpha^3+5\alpha $$ Since $\cos(2\pi)=1$, we know that $\alpha$ is a solution to the equation $16x^5-20x^3+5x=1$, or, to put it another way, a root of the polynomial $16x^5-20x^3+5x-1$. So, by definition, $\alpha$ is algebraic (i.e., not transcendental).

You can use multiple-angle formulas to do this in general. If you want to prove that $\beta=\cos\frac{a\pi}{b}$ is algebraic, you just need to use the formula for $\cos bx$. This will give you a polynomial expression for $\cos(a\pi)$ in terms of $\beta$; since we know that $\cos(a\pi)=\pm 1$, that's enough to prove that $\beta$ is algebraic.

So we just need to prove this key fact, and then we'll be done with cosines (and can move on to sines). This can be done by induction. For the base case, notice that $\cos x$ and $\cos 2x=2\cos^2 x -1$ are both polynomial functions of $\cos x$. For the inductive step, we'll suppose that $\cos nx$ and $\cos (n-1)x$ are polynomials, and use the identity given here for $\cos a + \cos b$, with $a=(n+1)x$, $b=(n-1)x$:

$$\cos(n+1)x + \cos(n-1)x=2\cos nx \cos x$$

This can be rearranged into the formula $\cos(n+1)x = 2 \cos x \cos nx - \cos(n-1)x$, which means that if $\cos nx$ and $\cos(n-1)x$ are polynomials in $\cos x$, then so is $\cos(n+1)x$. So the proof by induction is finished.

If you're interested in knowing more details about this kind of polynomial formula for $\cos nx$ in terms of $\cos x$, a useful keyword to search on would be "Chebyshev polynomials."

As a final note, we could try to do this proof directly for $\sin \frac{a\pi}{b}$, as there are multiple-angle formulas for $\sin$ which are similar to the ones for $\cos$. But it'd be a little bit trickier, because those formulas often involve both $\sin$ and $\cos$. For example, $\sin 2x=2\cos x \sin x$, and getting rid of the $\cos$ would involve taking a square root and doing a bunch of annoying case analysis involving signs.

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    $\begingroup$ This is a fabulous elementary answer to a seemingly intractable (by elementary means) problem. Answers like this are why I come to this Stack at all! Excellent! $\endgroup$ – Richard Rast Sep 20 '16 at 2:02
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    $\begingroup$ $\cos \frac{a\pi}{b}$ is always an algebraic number, as you say, because $ \frac{a\pi}{b}$ is trascendental. When $\alpha$ is algebraic, distinct of $1$ and $0$ what happen is exactly the contrary, $\cos \alpha$ is always trascendental (Gelfond-Schneider theorem). $\endgroup$ – Piquito Sep 21 '16 at 0:38
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    $\begingroup$ Given that the cos is algebraic, isn't the sin is algebraic by Pythagoras? $\endgroup$ – Neil Sep 21 '16 at 12:05
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    $\begingroup$ @Neil: That works. If you already know such facts as "the square of an algebraic number is algebraic", I believe it's about as simple as using $\sin x = \cos(\pi/2 - x)$. If you need to establish those facts, it's probably less simple. $\endgroup$ – Micah Sep 21 '16 at 15:06
  • $\begingroup$ +1. I just saw your answer and I like it. Hope you get 100 upvotes. $\endgroup$ – Xam Feb 9 '17 at 15:00
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Write:

$$z=e^{a\pi i/b}=\cos(a\pi/b)+i\cdot\sin(a\pi/b)$$

Then,

$$z^{2b}=e^{2a\pi i}=(e^{2\pi i})^{a}=1^a=1$$

$z$ is an algebraic number, as a root of the equation $x^{2b}-1=0$. Thus its real and imaginary parts ($\cos(a\pi/b)$ and $\sin(a\pi/b)$ respectively) are both algebraic.

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    $\begingroup$ This is in some sense the correct high-level approach (so, +1). However, "the real and imaginary parts of an algebraic number are also algebraic" is one of those obvious-sounding statements that take a surprising amount of machinery to prove. I don't know of a proof which is easier than proving that the sum of two algebraic numbers is algebraic, which means I don't know of a proof that doesn't rely on the theory of field extensions in some sense. I'm thinking this is probably a little beyond the OP... $\endgroup$ – Micah Sep 19 '16 at 21:03
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    $\begingroup$ @Micah: There's a "lowbrow" computational approach: if $\alpha$ is a root of $f$ and $\beta$ is a root of $g$ then $\alpha + \beta$ is a root of $\mathrm{Res_t}(f(t), g(x-t))$, so you can get there by developing the properties of resultants. (this formula is "obvious" once you know one of the various formulas for the resultant involving the roots of its arguments) $\endgroup$ – Hurkyl Sep 20 '16 at 0:46
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    $\begingroup$ Prepend a backslash to trig func names, so that they become LaTeX symbols, rendered with upright font: \sin,\cos→$\sin,\cos$, instead of just a sequence of letters in italics: sin,cos→$sin,cos$ looking like a multiplication: $s\cdot i\cdot n$... $\endgroup$ – CiaPan Sep 21 '16 at 6:54
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For each positive integer $b,$

\begin{align} \cos b\theta &= \frac{e^{ib\theta}+e^{-ib\theta}}{2} \\&= \frac{(\cos\theta + i \sin \theta)^b + (\cos \theta - i \sin \theta)^b}{2} \\&= \frac{\sum_{k=0}^b \binom{b}{k} (\cos^{b-k} \theta) (i^{k})(\sin^{k}\theta)+\sum_{k=0}^b\binom{b}{k} (\cos^{b-k} \theta) (-1)^{k}(i^{k})(\sin^{k}\theta)}{2} \\&= \frac{\sum_{\substack{0\le k\le b \\k \text{ even}}} 2 \binom{b}{k} (\cos^{b-k} \theta) ((-1)^{k/2})(\sin^{k}\theta)}{2} \\&= \sum_{\substack{0\le k\le b \\k \text{ even}}}(-1)^{k/2} \binom{b}{k} \cos^{b-k}\theta \sin^{k}\theta \\&= \sum_{\substack{0\le k\le b \\k \text{ even}}}(-1)^{k/2} \binom{b}{k} \cos^{b-k} \theta (1-\cos^{2}\theta)^{k/2}. \end{align}

Write $P_b(x) = \sum_{\substack{0\le k\le b \\k \text{ even}}}(-1)^{k/2} \binom{b}{k} x^{b-k} (1-x^2)^{k/2}.$

$P_b$ is a polynomial in one variable with integer coefficients, with the property that

$$\cos(b\theta) = P_b(\cos \theta)$$

for all $\theta.$

Now, if $\theta = a \pi/b$ where $a$ is an integer and $b$ is, without loss of generality, a positive integer, then

$$P_{b}(\cos \theta) = \cos (a \pi) = (-1)^a.$$

$P_{b}(x)-(-1)^a$ is not identically zero (since, if it were, the identity $P_{b}(\cos \theta)=\cos(b \theta)$ could not hold).

It follows that $\cos (\frac{a}{b}\pi) = \cos \theta$ is algebraic, since it's the root of a non-zero polynomial with integer coefficients.

Finally, $\sin (\frac{a}{b}\pi) = \cos (\frac{a}{b} \pi - \frac{\pi}{2}) = \cos(\frac{2a-b}{2b} \pi),$ which we already know is algebraic by the argument above, since $2a-b$ and $2b$ are integers.

If you want to exhibit a specific polynomial with integer coefficients that $\sin (\frac{a}{b}\pi)$ is a root of, $P_{2b}(x)-(-1)^b$ will work.

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    $\begingroup$ Keyphrase: "Chebyshev polynomial" $\endgroup$ – Hurkyl Sep 20 '16 at 0:42

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