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\begin{document}
\title{Derivation of sum and difference identities for sine and cosine}
\author{John Kerl}
\maketitle

The authors of your trigonometry textbook give a geometric derivation of the
sum and difference identities for sine and cosine.  I find this argument
unwieldy --- I don't expect you to remember it; in fact, I don't remember it.
There's a standard algebraic derivation which is far simpler.  The only catch
is that you need to use complex arithmetic, which we don't cover in Math 111.
Nonetheless, I will present the derivation so that you will have seen how
simple the truth can be, and so that you may come to understand it after you've
had a few more math courses.  And in fact, all you need are the following
facts:

\begin{itemize}
\item \emph{Complex numbers} are of the form $a+bi$, where $a$ and $b$ are real
	numbers and $i$ is defined to be a square root of $-1$.  That is, $i^2=-1$.
	(Of course, $(-i)^2=-1$ as well, so $-i$ is the other square root of $-1$.)
\item The number $a$ is called the \emph{real part} of $a+bi$;
	the number $b$ is called the \emph{imaginary part} of $a+bi$.  All the real
	numbers you're used to working with are already complex numbers --- they
	simply have zero imaginary part.
\item To add or subtract complex numbers, add the corresponding real and
	imaginary parts.  For example, $2+3i$ plus $4+5i$ is $6+8i$.
\item To multiply two complex numbers $a+bi$ and $c+di$, just FOIL out
	the product $(a+bi)(c+di)$ and use the fact that $i^2=-1$.  Then
	collect like terms.
\item The familiar exponential function $f(x)=e^x$ takes real-valued input.
	However, it can be extended to take complex-valued input.  All the usual
	rules for exponents apply, so
	$$
		e^{a+bi} = e^a e^{bi}.
	$$
	We compute $e^a$ as always --- this is the same exponential function
	as always.  The question is, what does it mean to raise $e$ to
	an imaginary power?  I assert to you that we write
	$$
		e^{bi} = \cos(b) + i \sin(b)
	$$
	where the cosine and sine functions are as usual.  This famous formula is
	called \emph{Euler's formula} (\emph{Euler} is pronounced \emph{Oiler}).
	You can read all about this formula on Wikipedia --- also see their
	nice article on the complex numbers.
\end{itemize}

Given these facts, we can simply write down what $e^{i(\alpha+\beta)}$ is: the
sum and difference formulas for sine and cosine fall out as a consequence.
Using the usual rules for exponents, we can write this as
$$
	e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta}.
$$
Now all we need to do is write out the two sides using Euler's formula.  The
left-hand side is
\begin{eqnarray*}
	e^{i(\alpha+\beta)} &=&
		\cos(\alpha+\beta) + i\sin(\alpha+\beta).
\end{eqnarray*}
Using the definition, FOILing, and collecting like terms, the right-hand side
is
\begin{eqnarray*}
	e^{i\alpha} e^{i\beta}
	&=& (\cos \alpha + i \sin \alpha) (\cos \beta + i \sin \beta) \\
	&=& (\cos \alpha \cos\beta - \sin\alpha\sin\beta)
		+ i(\sin\alpha\cos\beta + \cos\alpha\sin\beta).
\end{eqnarray*}
Equating real and imaginary parts of the left-hand side and the right-hand
side gives us, two for the price of one, the familiar sum identities for
sine and cosine:
\begin{eqnarray*}
\sin(\alpha+\beta) &=& \sin\alpha\cos\beta + \cos\alpha\sin\beta \\
\cos(\alpha+\beta) &=& \cos\alpha\cos\beta - \sin\alpha\sin\beta.
\end{eqnarray*}

Repeat this for $e^{i(\alpha-\beta)}$ to get the difference identities.  You
can do that --- just remember that cosine and sine are even and odd functions,
respectively, so $\cos(-\beta)=\cos(\beta)$ and $\sin(-\beta)=-\sin(\beta)$.

In summary, we have:
\begin{eqnarray*}
\sin(\alpha\pm\beta) &=& \sin\alpha\cos\beta \pm \cos\alpha\sin\beta \\
\cos(\alpha\pm\beta) &=& \cos\alpha\cos\beta \mp \sin\alpha\sin\beta.
\end{eqnarray*}
\end{document}
