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What is a simple proof that $\sin 1^\circ$ is irrational?

I have no idea how to solve this problem.

Blue
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Blaj Ionuţ
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    that question is about $\sin(1)$ radians, not degrees – Kenta S Dec 04 '19 at 23:28
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    Welcome to Math.SE! The community prefers/expects questions to include something of what the asker knows. "I have no idea" is almost-certainly inaccurate; if this is a textbook exercise, you could say something about the lesson it follows. In any case, this information helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already know or talking over your head. (It also helps convince people that you aren't simply trying to get them to do your homework for you. "I have no idea" tends to give the wrong impression.) – Blue Dec 04 '19 at 23:36

2 Answers2

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Assume $\sin (1^\circ)$ is rational. Then both $\cos^2(1^\circ)=1-\sin^2(1^\circ)$ and $\cos(2^\circ)=1-2\sin^2(1^\circ)$ are also rational. Since $\cos((2^n)^\circ)=2\cos^2((2^{n-1})^\circ)-1,$ inductively, all the $\cos((2^n)^\circ)$ are rational, for $n\ge 1.$

Now, we have \begin{equation} \begin{split} \frac{\sqrt3}2&=\cos(30^\circ)\\ &=\cos(32^\circ-2^\circ)\\ &=\cos(32^\circ)\cos(2^\circ)+\sin(32^\circ)\sin(2^\circ)\\ &=\cos(32^\circ)\cos(2^\circ)+2^6\cos(16^\circ)\cos(8^\circ)\cos(4^\circ)\cos(2^\circ)\cos^2(1^\circ)\sin(1^\circ)\\ \end{split} \end{equation}

The LHS is irrational, while the RHS is rational, a contradiction.

Kenta S
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1

This follows from the fact that $2 \sin \alpha \pi = i(e^{-\alpha \pi} - e^{\alpha \pi})$ is an algebraic integer, where $\alpha = \frac{1}{180}$.

Therefore, if $\sin \alpha \pi$ were rational, then $2 \sin \alpha \pi $ would be a rational number that is an algebraic integer.

Since $\mathbb Z$ is integrally closed, it would follow that $2 \sin \alpha \pi$ is indeed an integer. This is not possible, as it lies between $0$ and $1$.

WhatsUp
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