\rbqquest{
What are the amplitude $A$, circular frequency $\omega$, and phase lag $\phi$ when 
\[\cos\left(\omega t\right)+\sqrt{3}\sin\left(\omega t\right)\]
is converted to amplitude-phase form $A\cos\left(\omega t-\phi\right)$?
}

\rbqchoice{
\begin{enumerate}[a)]
\item $2\cos\left(\omega t-\frac{\pi}4\right)$
\item $\sqrt{3}\left(\cos\left(\omega \left(t-\frac{\pi}{3}\right)\right)\right)$
\item $2\cos\left(\omega\left(t-\frac{\pi}3\right)\right)$
\item \label{answer7.4} $2\cos\left(\omega\left(t+\frac{\pi}3\right)\right)$
\item $\sqrt{3}\left(\cos\left(\omega t-\frac{\pi}3\right)\right)$
\item $\sqrt{3}\left(\cos\left(\omega t-\frac{\pi}4\right)\right)$
\item Don't know.
\end{enumerate}
}

\rbqans{
The answer is (\ref{answer7.4}) because $A=\sqrt{1^2 + \sqrt{3}^2} =
2$, and $\phi=\tan^{-1}\frac{\sqrt{3}}{1}=\frac{\pi}3$.

\hbcom{Is this correct?  I feel like $\phi$ shouldn't be inside
  parentheses.  Also, check the sign.  (Maybe there's an error in
  sess7.3.tex or maybe I'm missing something.)}
}