For a low pass filter, derivephase of the transfer function, Vout/Vin
A quick Google search does provide the answer, but maybe you are after a somewhat easy and intuitive way to find the answer.

Well, the phase response is the argument (or phase) of the transfer function. Manipulate the transfer function just like a complex number.

Remember that \$1 = e^0\$, so the inverse of a complex number will negate the phase (that is, \$\arg{}(1/c)\ = - \arg{}(c)\$). That's because if you were to write \$c\$ in exponential form, the exponents subtract when using division.

The argument of a complex number \$a + bj\$ is \$\arctan\left(b/a\right)\$. So mapping \$a = 1\$ and \$b = RC\omega\$, one obtains your solution:

$$\arg(H(j\omega))=-\arctan\left({RC\omega}\right)$$

Just remember that transfer functions are basically functions that "generates" complex numbers that tells the amplitude and phase of the function for all \$j\omega\$. Thus, to obtain the magnitude and phase response of a transfer function you basically use the appropriate tools to obtain the absolute/argument of the complex function.