This is merely a variation of your own proof, Matt, but I believe it makes things clearer.
The first step is to define $c:=1-a-b$. Then, your identity takes the form
$\dfrac{\Gamma\left(b\right)\Gamma\left(c\right)}{\Gamma\left(b+c\right)}+\dfrac{\Gamma\left(c\right)\Gamma\left(a\right)}{\Gamma\left(c+a\right)}+\dfrac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}=\pi^{1/2}\dfrac{\Gamma\left(\dfrac{a}{2}\right)}{\Gamma\left(\dfrac{1-a}{2}\right)}\cdot\dfrac{\Gamma\left(\dfrac{b}{2}\right)}{\Gamma\left(\dfrac{1-b}{2}\right)}\cdot\dfrac{\Gamma\left(\dfrac{c}{2}\right)}{\Gamma\left(\dfrac{1-c}{2}\right)}$
for $a+b+c=1$. This is symmetric in $a$, $b$, $c$, which means we are way less likely to go insane during the following computations.
Now, using the formula
$\dfrac{\Gamma\left(\dfrac{s}{2}\right)}{\Gamma\left(\dfrac{1-s}{2}\right)}=\pi^{-1/2}2^{1-s}\cdot\cos\dfrac{\pi s}{2}\cdot\Gamma\left(s\right)$,
the right hand side simplifies to
$\pi^{-1}\cdot 4\cdot\cos\dfrac{\pi a}{2}\cdot\Gamma\left(a\right)\cdot\cos\dfrac{\pi b}{2}\cdot\Gamma\left(b\right)\cdot\cos\dfrac{\pi c}{2}\cdot\Gamma\left(c\right)$
(here we used $2^{3-a-b-c}=2^{3-1}=4$), so the identity in question becomes
$\dfrac{\Gamma\left(b\right)\Gamma\left(c\right)}{\Gamma\left(b+c\right)}+\dfrac{\Gamma\left(c\right)\Gamma\left(a\right)}{\Gamma\left(c+a\right)}+\dfrac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}$
$ = \pi^{-1}\cdot 4\cdot\cos\dfrac{\pi a}{2}\cdot\Gamma\left(a\right)\cdot\cos\dfrac{\pi b}{2}\cdot\Gamma\left(b\right)\cdot\cos\dfrac{\pi c}{2}\cdot\Gamma\left(c\right)$.
Dividing by $\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(c\right)$ on both sides, we get
$\dfrac{1}{\Gamma\left(a\right)\Gamma\left(b+c\right)}+\dfrac{1}{\Gamma\left(b\right)\Gamma\left(c+a\right)}+\dfrac{1}{\Gamma\left(c\right)\Gamma\left(a+b\right)}=4\pi^{-1}\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.
Since $b+c=1-a$, $c+a=1-b$, $a+b=1-c$, this rewrites as
$\dfrac{1}{\Gamma\left(a\right)\Gamma\left(1-a\right)}+\dfrac{1}{\Gamma\left(b\right)\Gamma\left(1-b\right)}+\dfrac{1}{\Gamma\left(c\right)\Gamma\left(1-c\right)}=4\pi^{-1}\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.
Now, using the formula $\dfrac{1}{\Gamma\left(z\right)\Gamma\left(1-z\right)}=\pi^{-1}\sin{\pi z}$ on the left hand side, and dividing by $\pi^{-1}$, we can simplify this to
$\sin{\pi a}+\sin{\pi b}+\sin{\pi c}=4\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.
Since $a+b+c=1$, we can set $A=\pi a$, $B=\pi b$, $C=\pi c$ and then have $A+B+C=\pi$. Our goal is to show that
$\sin A+\sin B+\sin C=4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$
for any three angles $A$, $B$, $C$ satisfying $A+B+C=\pi$.
Now this can be proven in different ways:
1) One is by writing $C=\pi-A-B$ and simplifying using trigonometric formulae; this is rather boring and it breaks the symmetry.
2) Another one is using complex numbers: let $\alpha=e^{iA/2}$, $\beta=e^{iB/2}$ and $\gamma=e^{iC/2}$. Then,
$\sin A+\sin B+\sin C=4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$
becomes
(1) $\dfrac{\alpha^2-\alpha^{-2}}{2i}+\dfrac{\beta^2-\beta^{-2}}{2i}+\dfrac{\gamma^2-\gamma^{-2}}{2i} = 4\cdot\dfrac{\alpha+\alpha^{-1}}{2}\cdot\dfrac{\beta+\beta^{-1}}{2}\cdot\dfrac{\gamma+\gamma^{-1}}{2}$.
Oh, and $A+B+C=\pi$ becomes $\alpha\beta\gamma=2i$. Now proving (1) is just a matter of multiplying out the right hand side and looking at the $8$ terms (two of them, namely $\alpha\beta\gamma$ and $\alpha^{-1}\beta^{-1}\gamma^{-1}$, cancel out, being $i$ and $-i$, respectively).
3) Here is how I would have done it 8 years ago: We can WLOG assume that $A$, $B$, $C$ are the angles of a triangle (this means that $A$, $B$, $C$ lie in the interval $\left[0,\pi\right]$, additionally to satisfying $A+B+C=\pi$), because everything is analytic (or by casebash). We denote the sides of this triangle by $a$, $b$, $c$ (so we forget about the old $a$, $b$, $c$), its semiperimeter $\dfrac{a+b+c}{2}$ by $s$, its area by $\Delta$ and its circumradius by $R$. Then, $\sin A=\dfrac{a}{2R}$ (by the Extended Law of Sines) and similarly $\sin B=\dfrac{b}{2R}$ and $\sin C=\dfrac{c}{2R}$, so that $\sin A+\sin B+\sin C=\dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R}=\dfrac{a+b+c}{2R}=\dfrac{s}{R}$. On the other hand, one of the half-angle formulas shows that $\cos\dfrac{A}2=\sqrt{\dfrac{s\left(s-a\right)}{bc}}$, and similar formulas hold for $\cos\dfrac{B}2$ and $\cos\dfrac{C}2$, so that
$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$
$=4\cdot\sqrt{\dfrac{s\left(s-a\right)}{bc}}\cdot\sqrt{\dfrac{s\left(s-b\right)}{ca}}\cdot\sqrt{\dfrac{s\left(s-c\right)}{ab}}$
$=\dfrac{4s}{abc}\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.
Now, $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\Delta$ (by Heron's formula) and $\Delta=\dfrac{abc}{4R}$ (by another formula for the area of the triangle), so tis becomes
$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}=\dfrac{4s}{abc}\cdot\dfrac{abc}{4R}=\dfrac{s}{R}$.
This is exactly what we got for the left hand side, qed.
Best Answer
As $\cos x=\pm\sqrt{1-\sin^2 x}$ and $e^{ix}=\cos x +i\sin x$, and $\sin x=(e^{ix}-1/e^{ix})/2i$ then $\sin x$ is in a radical extension of $\mathbb{Q}$ iff $e^{ix}$ is. For rational $r$ with denominator $d$, $e^{2\pi i r}$ is a primitive $d$-th roots of unity. The extension of $\mathbb{Q}$ generated by a root of unity is a cyclotomic field. Every cyclotomic field is an abelian extension of $\mathbb{Q}$. (By the Kronecker-Weber theorem any abelian extension is contained in a cyclotomic field.) The Galois group of the $n$-th cyclotomic field is isomorphic to the multplicative group $(\mathbb{Z}/n\mathbb{Z})^*$
So we can obtain any $\sin r\pi$ for $r$ rational in terms of radicals, both in a trivial way if you allow an $n$-th root of unity as $1^{1/n}$, and also in a stricter sense, if you insist that you ascend through a chain of fields by adjoining at each stage a root of $x^m-a$ where this polynomial is irreducible over the previous field.
However if you insist on this more exacting definition, you will need radicals of non-real numbers unless $n$ is a product of distinct Fermat primes. All this is well-known.