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.
Best Answer
$\sin\left(\frac{p}{q}\pi\right)=\sin\left(\frac{p}{q}180^\circ\right)$ 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.