Two AC potential differences (v_{1}=V_{1} sin left(omega t+phi_{1} ight)) and (v_{2}=V_{2} sin left(omega t+phi_{2} ight)) are added

Two AC potential differences \(v_{1}=V_{1} \sin \left(\omega t+\phi_{1}\right)\) and \(v_{2}=V_{2} \sin \left(\omega t+\phi_{2}\right)\) are added according to

\[ \begin{aligned} v_{\text {sum }} & =V \sin (\omega t+\phi)=v_{1}+v_{2} \\ & =V_{1} \sin \left(\omega t+\phi_{1}\right)+V_{2} \sin \left(\omega t+\phi_{2}\right), \end{aligned} \]

with \(\omega=10.0 \mathrm{~s}^{-1}, V_{1}=5.00 \mathrm{~V}\), and \(V_{2}=8.00 \mathrm{~V}\). At \(t=0, v_{1}(t=0)=4.50 \mathrm{~V}\) and \(v_{2}(t=0)=3.70 \mathrm{~V}\).

(a) Use phasors to determine values for \(V\) and \(\phi\) in the expression \(v_{\text {sum }}=V \sin (\omega t+\phi)\) and a value for \(v(t=0.0500 \mathrm{~s})\).

(b) Repeat using trigonometry, and then compare your results from the two methods.