Appendix B Trigonometric Identities
Fundamental relationships of trigonometric functions can be deduced from the remarkable formula of Euler :
\begin{equation}
e^{i\theta} = \cos(\theta) + i\sin(\theta) \quad (\theta
\in \mathbb{R})\tag{B.1}
\end{equation}
where \(i \in \mathbb{C}\) is a complex number with \(i^2 = -1\text{.}\) For instance, since complex conjugation corresponds to reflection along the real axis, \(\overline{e^{i\theta}} = e^{i(-\theta)}\) for any \(\theta\text{,}\) so
\begin{equation*}
\cos(\theta) - i\sin(\theta) \equiv \cos(-\theta)+i\sin(-\theta).
\end{equation*}
From identity we conclude that sine is an odd function and cosine is an even function, that is
\begin{equation}
\sin(-\theta) \equiv -\sin(\theta)\ \text{and}\ \cos(-\theta)
\equiv \cos(\theta).\tag{B.2}
\end{equation}
Also,
\begin{align*}
1 & = e^{i\theta}e^{-i\theta}\\
& =
(\cos(\theta) + i\sin(\theta))(\cos(\theta)-i\sin(\theta)\\
& = \cos^2(\theta) + \sin^2(\theta).
\end{align*}
That is
\begin{equation}
\sin^2(\theta) + \cos^2(\theta) \equiv 1.\tag{B.3}
\end{equation}
Now, whenever \(\sec(\theta)\) is defined, i.e. \(\cos(\theta)
\neq 0\text{,}\) by dividing \(\cos^2(\theta)\) on both sides of (B.3), we obtain another form of the identity
\begin{equation}
\tan^2(\theta) + 1 \equiv
\sec^2(\theta).\tag{B.4}
\end{equation}
Compare the real and imaginary parts on both sides of the equation
\begin{align*}
\cos(A+B) & + i\sin(A+B) = e^{i(A+B)}
= e^{iA}e^{iB}\\
&= (\cos(A) +
i\sin(A))(\cos(B) + i\sin(B))\\
&=
\cos(A)\cos(B)-\sin(A)\sin(B)\\
&\phantom{=}+i(\sin(A)\cos(B)+\sin(B)\cos(A)).\\
&=\cos(A)\cos(B)-\sin(A)\sin(B)\\
&\phantom{=}+i(\sin(A)\cos(B)+\sin(B)\cos(A))
\end{align*}
we get the following angle-sum identities.
\begin{gather}
\sin(A+B) \equiv \sin(A)\cos(B)+\sin(B)\cos(A)\tag{B.5}\\
\cos(A+B) \equiv \cos(A)\cos(B)-\sin(A)\sin(B)\tag{B.6}
\end{gather}
\begin{equation}
\sin(A-B) \equiv \sin(A)\cos(B)-\sin(B)\cos(A)\tag{B.7}
\end{equation}
\begin{equation}
\cos(A-B) \equiv \cos(A)\cos(B)+\sin(A)\sin(B)\tag{B.8}
\end{equation}
Before moving on to the sum-product identities, we offer another way of deducing the identities above of one of them (actually any one of them). Suppose you know the identity (B.5)
\begin{equation*}
\sin(A+B) \equiv \sin(A)\cos(B)+\sin(B)\cos(A)
\end{equation*}
Now think of \(A\) as the variable and \(B\) as a constant. Then differentiating with respect to \(A\) immediately yields,
\begin{align*}
\cos(A + B) & \equiv \cos(A)\cos(B)
+\sin(B)(-\sin(A)) \\
& \equiv \cos(A)\cos(B) -\sin(B)\sin(A).
\end{align*}
This is just (B.6). The other two identities can be obtained by replacing \(B\) by \(-B\) just as before.
We will now deduce a couple identities relating sum and product of trigonometric functions. Halving the sum of (B.5) and (B.7) yields
\begin{equation}
\frac{1}{2}(\sin(A+B) + \sin(A-B)) \equiv
\sin(A)\cos(B).\tag{B.9}
\end{equation}
\begin{equation}
\frac{1}{2}(\cos(A+B) + \cos(A-B))
\equiv \cos(A)\cos(B)\tag{B.10}
\end{equation}
\begin{equation}
\frac{1}{2}(\cos(A-B) - \cos(A+B))
\equiv \sin(A)\sin(B)\tag{B.11}
\end{equation}
By setting \(B=A\) in (B.5), we get
\begin{equation}
\sin(2A) \equiv 2\sin(A)\cos(A).\tag{B.12}
\end{equation}
\begin{equation}
\cos(2A) \equiv \cos^2(A)-\sin^2(A) \equiv 2\cos^2(A) -1
\equiv 1-2\sin^2(A).\tag{B.13}
\end{equation}
And hence,
\begin{equation}
\cos^2(A) \equiv \frac{1+\cos(2A)}{2}, \qquad \sin^2(A) \equiv
\frac{1-\cos(2A)}{2}\tag{B.14}
\end{equation}
Here is a way taken from [1] of deducing the angle-sum formulas without using complex numbers. 
In one coordinate system, the coordinates of \(A\) and \(C\) are \((1,0)\) and \((\cos(\alpha + \beta), \sin(\alpha+\beta))\text{,}\) respectively. In the coordinate system in which \(B=(1,0)\text{,}\) \(A = (\cos(\alpha), -\sin(\alpha))\) and \(C=(\cos(\beta), \sin(\beta))\text{.}\)
Now by computing the square of the distance between \(A\) and \(C\) in these two coordinate systems, we conclude that
\begin{align*}
(\cos(\alpha + \beta)-1)^2 + (\sin(\alpha + \beta))^2
\equiv (\cos(\beta)-\cos(\alpha))^2\\
&\phantom{\equiv}+ (\sin(\beta)-\sin(\alpha))^2
\end{align*}
Expanding both sides and then simplifying by the Pythagorean identity, one arrives to the angle-sum formula for cosine.
\begin{gather*}
\cos(\alpha + \beta) \equiv \cos(\alpha)\cos(\beta) -
\sin(\alpha)\sin(\beta).
\end{gather*}
