Given that the general solution of the equation of motion $m \bar{x}=-k x$ for the harmonic oscillator is $x(t)=a \cos \omega t+b \sin \omega t$, where $\omega=\sqrt{k / m}$, (i) show that the solution can be written in the form $x(t)=A \cos (\omega t-\delta)$, where $A$ is the amplitude of the vibration and $\delta$ is the phase angle, and express $A$ and $\delta$ in terms of $a$ and $b$; (ii) find the amplitude and phase angle for the initial conditions $x(0)=1, \dot{x}(0)=\omega$,