Recall the following trigonometric identity
$\begin{align*}\cos x + \cos y = 2 \cos \left( \frac{x - y}{2}\right) \sin \left( (x+ y)/2 \right)\end{align*}$
Let $x = 2 \pi f_A t$ and let $y = 2 \pi n f_D t$ . Here, $n$ corresponds to the harmonic, $f_A = 440.000 \mbox{ Hz}$ and $f_D = 73.416 \mbox{ Hz}$ .
Then, the superposition of the two notes is
$\begin{align*}\cos 2 \pi f_A t + \cos 2 \pi n f_D t = 2 \cos \left( 2 \pi \frac{f_A + f_D}{2}\right) \cos \left( 2 \pi t (f_A...$
So, the beat frequency is $(f_A- f_D)/2$ . The following table shows the value of $(f_A- f_D)/2$ for various values of $n$ .
\begin{center}\begin{tabular}{|c|c|}
\hline
n & $(f_A- f_D)/2$ (in Hz)\\
\hline
1 & 183.292 \\
\hline
2 & 146.584 \\
\hline
3 & 109.876 \\
\hline
4 & 73.168 \\
\hline
5 & 36.46 \\
\hline
6 & 0.248 \\
\hline
7 & 36.956 \\
\hline
\end{tabular}\end{center}
So, clearly the smallest number of beats occurs when $n = 6$ . Recall that a beat occurs whenever the second cosine factor equals $+1$ or $-1$ , therefore, the number of beats per second is $2\cdot0.248 = 0.496$ . Therefore, answer (B) is correct.

Solution to 1996 Problem 80