Tricky trigonometry question

This problem can be solved with a simple geometric construction:

Consider a right triangle ABC, where the angle A is called alpha

As BC is 4 and AB is 3, then tan(alpha) is by definition 4/3. Consider now the angle bisector AD, according to the angle bisector theorem AC/AB = CD/BD. CD can therefore be written as 5x and BD as 3x, x being a number belonging to R. But CD+BD=4, so x=1/2, it follows that CD=5/2 and BD=3/2.

By Pythagoras' theorem AD is therefore equal to $\sqrt(3^2+(3/2)^2$ = $3\sqrt(5)/2$ and cos(alpha/2)=AB/AD= $2\sqrt(5)/5$ .

The answer is therefore 2+5+5=12

Let $\theta = \arctan \frac 43$ . Then we have:

$\begin{aligned} \tan \theta & = \frac 43 & \small \color{#3D99F6}{\text{Let }t = \tan \frac \theta 2} \\ \frac {2t}{1-t^2} & = \frac 43 \\ 6t & = 4-4t^2 \\ 4t^2 + 6t - 4 & = 0 \\ 2t^2 + 3t - 2 & = 0 \\ (2t-1)(t+2) & = 0 \\ \implies t & = \frac 12, \ - 2 \\ \tan \frac \theta 2 & = \frac 12, \ - 2 \end{aligned}$

Now, we have:

$\begin{aligned} \cos \left(\frac {\arctan \frac 43}2 \right) & = \cos \frac \theta 2 = \begin{cases} \dfrac 2{\sqrt{1^2+2^2}} = \dfrac {2\sqrt 5}5 \\ - \dfrac 2{\sqrt{1^2+2^2}} = - \dfrac {2\sqrt 5}5 \end{cases} \end{aligned}$

$\implies a + b + c = 2 + 5 + 5 = \boxed{12}$ .