Sine defined for a triangle inscribed in a circle with a diameter of one Let a circle be drawn with a diameter of one (and thus a radius of one half). Then let a triangle with vertices A, B, and C be inscribed in the circle (i.e. points A, B, and C are arbitrary points on the circle). 
Then a, the side of the triangle opposite angle A is equal to sin(A)
Likewise, b=sin(B) and c=sin(c). I have attempted to find or devise a proof of this, but I don't know where to start!

Although lab bhattacharjee has already said, we have to use the Law of Sines. If you aren't familiar with it or its proof, see the link. I will tell you how to proceed in a detailed manner. 

Here we have our $\triangle ABC$ and its circumscribed circle with center $O$. We now construct a diameter $BOD$. So, $\angle BAC=\angle BDC$ and $\angle BCD=90^{\circ}$. Now,
$$\sin\angle A=\sin\angle BDC=\frac{a}{2r}$$
Where, $a=BC$ and $r$ is the radius. You can similarly draw conclusions for $\angle B$ and $\angle C$. This gives rise to what we call the, extended law of sines:
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=D$$
Where $D$ is the diameter of the circumradius. It is a very useful theorem, and applying to your triangle gives:
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=1$$
Done! There is one caveat though, we did not prove the extended law of sines for right triangles [it should be obvious] and obtuse triangles. However, we can it do it similarly, and I leave the proof as an exercise for you.    
