\rbqquest{

\begin{figure}[h]
\centering
%  \includegraphics[width=.5\textwidth]{figname}\\
  \caption{Mystery sinusoid.}\label{sinusoid}
\end{figure}

The graph of a sinusoidal function is displayed. The problem is to
express it in the {\em standard form}
$$f(t) = A \cos(\omega t - \phi).$$
}

\rbqchoice{
\begin{enumerate}[a)]
\item $f(t) = 2\cos\left(4\pi t+\frac{\pi}4\right)$
\item $f(t) = 2\cos\left(\frac{\pi}{4}t+\frac{\pi}4\right)$
\item $f(t) = 2\cos\left(4\pi t -\frac{\pi}4\right)$
\item \label{answer7.2} $f(t) = 2\cos\left(\frac{\pi}{4}t-\frac{\pi}{4}\right)$
\item $f(t) = 2\cos\left(4t+1\right)$
\item $f(t) = 2\cos\left(4t-1\right)$
\end{enumerate}
}

\rbqans{
The answer is (\ref{answer7.2}).

\hbcom{I've put a bunch of words in to glue the equations together.
  Please double check my descriptions, especially of $t_0$.}

On the graph, the vertical distance between a ``peak'' and a
``trough'' of the graph is $4$, so the amplitude is $A = 2$.  The
function's period is $P=8$, and we know $P = 2\pi/\omega$, so the
angular frequency is $\omega =
\frac{\pi}{4}$.  The graph is ``shifted to the left'' one unit, so the
delay is 
$t_0 = \phi/\omega = -1$ and the phase lag is $\phi =
-\frac{\pi}{4}$.  Hence the equation of the sinusoid is:
$$f\left(t\right)=2\cos\left(\frac{2\pi}{8}\left(t+1\right)\right)=2\cos\left(\frac{\pi}{4}t+\frac{\pi}{4}\right).$$
}
