In calculation of Maximum power transferred for reactive circuits why is this done?
First I'll set the two equations equal to each other. 
$$ \frac{1}{2}| I|^2R = \frac{1}{2} V_mI_mcos(\theta_v-\theta_i) $$

Since \$V = IR\$, we can substitute the \$V_m\$ in the right hand side for \$I_mR\$. which becomes:

$$ \frac{1}{2}| I|^2R = \frac{1}{2} I^2R cos(\theta_v-\theta_i) $$

Since there are no reactive components to the circuit, the current and the voltage are perfectly in sync, which means:

$$ \theta_v = \theta_i $$
when you subtract the two, you will get zero, and so:
$$ cos(\theta_v-\theta_i) = cos(0) = 1$$
when plugging that into the equation it becomes 
$$ \frac{1}{2}| I|^2R = \frac{1}{2} I^2R $$

Since (for real numbers) \$x^2 = |x|^2 \$, and current is only a real number. 
$$ \frac{1}{2} I^2R = \frac{1}{2} I^2R $$
We can show that the two equations you are calling into question are in fact the same equation, one is simply simplified to account for the entirely real nature of a non-reactive circuit.