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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.

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    For the second part you'll also need to assume $x\ne0$.2012-02-24
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    Claim is false: $\sin 0^{\circ}$ and $\sin 0$ (in radians) are both algebraic.2012-02-24
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    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.2012-02-24
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    You may be interested in [this](http://math.stackexchange.com/questions/94478/sin-1-circ-is-irrational-but-how-do-i-prove-it-in-a-slick-way-and-tan1) and Hardy's comment there about Niven's Theorems and some links. Ofcourse, this comes for free with enlightening answers from various others.2012-02-24

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