When is $\sin x$ an algebraic number and when is it non-algebraic? 
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. 
 A: Lindemann-Weierstrass theorem implies that for $\alpha$ non-zero algebraic, $\sin \alpha$ is transcendental.
A: $\sin\left(\frac{p}{q}\pi\right)=\sin\left(\frac{p}{q}180^\circ\right)$
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.
