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Is $\sin(1)$ algebraic over $\mathbb{Q}$?

At the moment I have no idea how to proceed. Could you tell me how to solve it?

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Interesting problem! May I know for which course this is homework? (or textbook where this problem has been taken from, just curious) –  Prism Jun 17 '13 at 7:44
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@Prism: Are you implying the NSA doesn't have access to his documents? –  NikolajK Jun 17 '13 at 7:50
    
@NickKidman: I am not allowed to disclose this information. ;) $\*$pretends to be NSA official in his revolving chair$\*$ –  Prism Jun 17 '13 at 7:56
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@Prism, you can find proof of this fact in "An introduction to the theory of numbers" by Hardy & Wright. What is interesting, that for every $n\in\mathbb{N}$ neither $\sin n$ nor $\cos n$ are algebraic (simple consequence from transcendentality of $\sin 1$). –  Bartek Pawlik Jun 17 '13 at 8:00
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2 Answers 2

As explained in usersujo’s answer, the answer is YES if $1$ means $1$ degree.

The answer is NO, however, if $\theta=1$ means $1$ radian.

Because then $\sin(\theta)=\sqrt{1-\cos^2(\theta)}$ would be algebraic also, and hence $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ would be algebraic. But we know $e^{i\theta}$ is not algebraic by the Lindemann–Weierstrass theorem

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I assume by 1 you mean $1^{\circ}$. We know that by De Moivre's theorem, $\cos(90) + i\sin(90) = (\cos(1) + i\sin(1))^{90},$ but

$$\cos(90) + i\sin(90)= i\Rightarrow (\cos(1) + i\sin(1))^{90} = i.$$

Now expand $(\cos1 + i\sin 1)^{90}$ using binomial theorem, and consider the real part. You will get a polynomial in $\cos(1)$, in which every power of $\cos(1)$ is even. So, substitute $\cos^2(1)=1-\sin^2(1)$, and you get a polynomial in $\sin(1)$ which equals to $0$. Hence $\sin(1)$ is algebraic over $\mathbb Q$.

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That’s if you assume 1 means 1 degree, not 1 radian. By default mathematicians assume 1 means 1 radian. –  Ewan Delanoy Jun 17 '13 at 7:46
    
@EwanDelanoy : Thank you. I have edited my previous answer. –  mathmansujo Jun 17 '13 at 8:18
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