Answer
The proof is below.
Work Step by Step
We first use the distance formula to find:
$ A = \sqrt{a^2 + (-b)^2} \\ A = \sqrt{a^2+b^2}$
We now are asked to find the phase constant. To do so, we set the derivative of the equation equal to zero:
$0=-asin\omega t - b cos\omega t $
This becomes:
$asin \phi = b cos \phi $
$atan \phi = b $
$\phi = tan^{-1}(\frac{b}{a})$
