Answer
$\phi = 0.845~rad$
Work Step by Step
$\theta = \theta_m~cos~(\omega~t+\phi)$
$\frac{d\theta}{dt} = -\theta_m~\omega~sin~(\omega~t+\phi)$
When $t = 0$:
$\theta = \theta_m~cos~\phi = 0.040~rad$
$\frac{d\theta}{dt} = -\theta_m~\omega~sin~\phi = -0.200~rad$
We can divide the second equation by the first equation:
$-\omega~tan~\phi = -5.00$
$tan~\phi = \frac{5.00}{\omega}$
$\phi = tan^{-1}~\frac{5.00}{\omega}$
$\phi = tan^{-1}~\frac{5.00}{4.44}$
$\phi = 0.845~rad$
