Subsection B.2.7 Identities — Extras
¶Subsubsection B.2.7.1 Sums to Products
Consider the identities
\begin{align*}
\sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha) \sin(\beta) &
\sin(\alpha-\beta) &= \sin(\alpha)\cos(\beta) - \cos(\alpha) \sin(\beta)
\end{align*}
If we add them together some terms on the right-hand side cancel:
\begin{align*}
\sin(\alpha+\beta) + \sin(\alpha-\beta) &= 2\sin(\alpha)\cos(\beta).
\end{align*}
If we now set \(u=\alpha+\beta\) and \(v = \alpha-\beta\) (i.e. \(\alpha=\frac{u+v}{2}, \beta=\frac{u-v}{2}\)) then
\begin{align*}
\sin(u) + \sin(v) &= 2\sin\left(\frac{u+v}{2}\right)\cos\left(\frac{u-v}{2}\right)
\end{align*}
This transforms a sum into a product. Similarly:
\begin{align*}
\sin(u) - \sin(v) &= 2\sin\left(\frac{u - v}{2}\right)\cos\left(\frac{u + v}{2}\right)\\
\cos(u) + \cos(v) &= 2\cos\left(\frac{u + v}{2}\right)\cos\left(\frac{u - v}{2}\right)\\
\cos(u) - \cos(v) &= -2\sin\left(\frac{u + v}{2}\right)\sin\left(\frac{u - v}{2}\right)
\end{align*}
Subsubsection B.2.7.2 Products to sums
Again consider the identities
\begin{align*}
\sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha) \sin(\beta) &
\sin(\alpha-\beta) &= \sin(\alpha)\cos(\beta) - \cos(\alpha) \sin(\beta)
\end{align*}
and add them together:
\begin{align*}
\sin(\alpha+\beta) + \sin(\alpha-\beta) &= 2\sin(\alpha)\cos(\beta).
\end{align*}
Then rearrange:
\begin{align*}
\sin(\alpha)\cos(\beta)&= \frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{2}
\end{align*}
In a similar way, start with the identities
\begin{align*}
\cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) &
\cos(\alpha-\beta) &= \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)
\end{align*}
If we add these together we get
\begin{align*}
2\cos(\alpha)\cos(\beta) &= \cos(\alpha+\beta) + \cos(\alpha-\beta)\\
\end{align*}
while taking their difference gives
\begin{align*} 2\sin(\alpha)\sin(\beta) &= \cos(\alpha-\beta) - \cos(\alpha+\beta) \end{align*}Hence
\begin{align*}
\sin(\alpha)\sin(\beta)&= \frac{\cos(\alpha-\beta) - \cos(\alpha+\beta)}{2}\\
\cos(\alpha)\cos(\beta)&= \frac{\cos(\alpha-\beta) + \cos(\alpha+\beta)}{2}
\end{align*}