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\doctitle{The Sinusoidal Identity}

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\myrhead

\title{\doctitle}
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The sum of two sinusoidal functions of the same frequency is another
sinusoidal function with that frequency!
For any real constants $a$ and $b$,
\begin{align}
\label{three} a\cos\omega t+b\sin\omega t&=A\cos(\omega t-\phi)
\end{align}
where $A$ and $\phi$ can be described in at least three ways:
\begin{equation}\label{four}A=|a\mathbf{i}+b\mathbf{j}| \text{ and }\phi\text{ is the angle rotating }\mathbf{i}\text{ into dir}(a\mathbf{i}+b\mathbf{j});\end{equation}
\begin{equation}\label{five}A=\sqrt{a^2+b^2},\quad \phi=\tan^{-1} \frac ba;\end{equation}
\begin{equation}\label{six}a+bi=Ae^{i\phi}.\end{equation}
Conversely, we have 
\begin{equation}\label{seven}a=A\cos\phi \text{ and } b=A\sin\phi.\end{equation}
Geometrically this is summarized by the triangle in Figure~\ref{fig7.3.1}:

\TODO
\begin{figure}[h]
\centering
%  \includegraphics[width=.5\textwidth]{fig7.3.1}\\
  \caption{$a+bi = Ae^{i \phi}.$}\label{fig7.3.1}
\end{figure}

One proof of (\ref{three}) is a simple application of the cosine addition formula 
\[\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.\]
We will give an equivalent proof using Euler's formula and complex arithmetic.

The triangle in Figure~\ref{fig7.3.1} is the standard polar coordinates triangle. It shows $a+ib=Ae^{i\phi}$ or $a-ib=Ae^{-i\phi}$. Thus
\begin{align*}
A\cos(\omega t-\phi)&=\Repart(Ae^{i(\omega t-\phi)})\\
&=\Repart(e^{i\omega t}\cdot Ae^{-i\phi})\\
&=\Repart((\cos\omega t+i\sin\omega t)\cdot(a-ib))\\
&=\Repart(a\cos\omega t+b\sin\omega t+i(a\sin\omega t-b\cos\omega t))\\
&=a\cos\omega t+b\sin\omega t.
\end{align*}
We should stress the importance of the trigonometric identity
(\ref{three}). It shows that \emph{any} linear combination of
$\cos(\omega t)$ and $\sin(\omega t)$ is not only periodic of period
$\frac{2\pi}{\omega}$, but is also sinusoidal.  If you try to add
$\cos(\omega t)$ to $\sin(\omega t)$ ``by hand'', you will probably
agree that this is not at all obvious.

We will call $A\cos(\omega t-\phi)$ {\bf amplitude-phase form} and
$a\cos\omega t+b\sin\omega t$ {\bf rectangular} or {\bf Cartesian
  form}.  You should be familiar amplitude-phase form; we 
usually prefer it because both amplitude and phase have
geometric and physical meaning for us.  \hbcom{I changed ``geometrical''
  to ``geometric''.  Perhaps both are correct, but I find I have a
  strong preference.}