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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*}