circumradii of $\triangle{IAB}$ Let $I$ be the incenter of $\triangle{ABC}$. Let $R$ be the radius of the circle that circumscribes $\triangle{IAB}$. Find a formula for $R$ in term of other elements $a, b, c, A, B, C, r, R$ of $\triangle{ABC}$. I need this formula in order to prove a geometric inequality.

Remember the extended law of sines:
$$R = \frac{a}{2\sin A} = \frac{b}{2\sin B} = \frac{c}{2\sin C}$$
The above gives the circumradius ($R$) for a triangle whose one angle is $A$ and the opposite side is $a$ and so on. Well, the labeling is a bit different here.

For our triangle $IAB$, the extended law of sines becomes: $$R = \frac{a}{2\sin\angle AIB}$$
So, we need to calculate $\angle AIB$, and we are done. This can be done by realizing that $IB$ is the angle bisector of $\angle A$ and $IA$ angle bisector of $\angle A$. So, $\angle AIB$ is:
$$180^{\circ} - (\angle IAB + \angle IBA) = 180^{\circ} - (\frac{\angle A}{2} + \frac{\angle B}{2}) = 180^{\circ} - (\frac{180^{\circ} - \angle C}{2}) = 90^{\circ} + \frac{C}{2} $$
Note that $\sin (90^{\circ} + \theta) = \cos\theta$. So we substitute back and our formula becomes:
$$R=\frac{AB}{2\cos\frac{C}{2}}$$
If you need proof of anything that I have used ask me. If you want to express $\cos\frac{C}{2}$ in terms of $\cos C$, you can do that by using the identity [and considering the positive root]:
$$\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}$$
