Product-to-Sum and Sum-to-Product Identities

TL;DR

Product-to-Sum Identities

• \[\sin \alpha \cos \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) + \sin ( \alpha - \beta ) \}\]
• \[\cos \alpha \sin \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) - \sin ( \alpha - \beta ) \}\]
• \[\cos \alpha \cos \beta = \frac { 1 } { 2 } \{ \cos ( \alpha + \beta ) + \cos ( \alpha - \beta )\}\]
• \[\sin \alpha \sin \beta = - \frac { 1 } { 2 } \{ \cos ( \alpha + \beta ) - \cos ( \alpha - \beta ) \}\]

Sum-to-Product Identities

• \[\sin A + \sin B = 2\sin \frac{A+B}{2}\cos \frac{A-B}{2}\]
• \[\sin A - \sin B = 2\cos \frac{A+B}{2}\sin \frac{A-B}{2}\]
• \[\cos A + \cos B = 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}\]
• \[\cos A - \cos B = -2\sin \frac{A+B}{2}\sin \frac{A-B}{2}\]

It’s beneficial to learn not only the formulas but also their derivation processes.

Prerequisites

• Trigonometric Addition Formulas

Product-to-Sum Identities

• \[\sin \alpha \cos \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) + \sin ( \alpha - \beta ) \}\]
• \[\cos \alpha \sin \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) - \sin ( \alpha - \beta ) \}\]
• \[\cos \alpha \cos \beta = \frac { 1 } { 2 } \{ \cos ( \alpha + \beta ) + \cos ( \alpha - \beta )\}\]
• \[\sin \alpha \sin \beta = - \frac { 1 } { 2 } \{ \cos ( \alpha + \beta ) - \cos ( \alpha - \beta ) \}\]

Derivation

We use the Trigonometric Addition Formulas

\[\begin{align} \sin(\alpha+\beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \tag{1}\label{eqn:sin_add}\\ \sin(\alpha-\beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \tag{2}\label{eqn:sin_dif} \end{align}\]

Adding ($\ref{eqn:sin_add}$) and ($\ref{eqn:sin_dif}$), we get

\[\sin(\alpha+\beta) + \sin(\alpha-\beta) = 2 \sin \alpha \cos \beta \tag{3}\label{sin_product_to_sum}\] \[\therefore \sin \alpha \cos \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) + \sin ( \alpha - \beta ) \}.\]

Subtracting ($\ref{eqn:sin_dif}$) from ($\ref{eqn:sin_add}$), we get

\[\sin(\alpha+\beta) - \sin(\alpha-\beta) = 2 \cos \alpha \sin \beta \tag{4}\label{cos_product_to_dif}\] \[\therefore \cos \alpha \sin \beta = \frac { 1 } { 2 } \{ \sin ( \alpha + \beta ) - \sin ( \alpha - \beta ) \}.\]

Similarly, using

\[\begin{align} \cos(\alpha+\beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \tag{5}\label{eqn:cos_add} \\ \cos(\alpha-\beta ) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \tag{6}\label{eqn:cos_dif} \end{align}\]

Adding ($\ref{eqn:cos_add}$) and ($\ref{eqn:cos_dif}$), we get

\[\cos(\alpha+\beta) + \cos(\alpha-\beta) = 2 \cos \alpha \cos \beta \tag{7}\label{cos_product_to_sum}\] \[\therefore \cos \alpha \cos \beta = \frac { 1 } { 2 } \{ \cos(\alpha+\beta) + \cos(\alpha-\beta) \}.\]

Subtracting ($\ref{eqn:cos_dif}$) from ($\ref{eqn:cos_add}$), we get

\[\cos(\alpha+\beta) - \cos(\alpha-\beta) = -2 \sin \alpha \sin \beta \tag{8}\label{sin_product_to_dif}\] \[\therefore \sin \alpha \sin \beta = -\frac { 1 } { 2 } \{ \cos(\alpha+\beta) - \cos(\alpha-\beta) \}.\]

Sum-to-Product Identities

• \[\sin A + \sin B = 2\sin \frac{A+B}{2}\cos \frac{A-B}{2}\]
• \[\sin A - \sin B = 2\cos \frac{A+B}{2}\sin \frac{A-B}{2}\]
• \[\cos A + \cos B = 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}\]
• \[\cos A - \cos B = -2\sin \frac{A+B}{2}\sin \frac{A-B}{2}\]

Derivation

We can derive the Sum-to-Product Identities from the Product-to-Sum Identities.

Let \(\alpha + \beta = A, \quad \alpha - \beta = B\)

Solving these equations for $\alpha$ and $\beta$, we get

\[\alpha = \frac{A+B}{2}, \quad \beta = \frac{A-B}{2}.\]

Substituting these into ($\ref{sin_product_to_sum}$), ($\ref{cos_product_to_dif}$), ($\ref{cos_product_to_sum}$), and ($\ref{sin_product_to_dif}$) respectively, we obtain the following formulas:

\[\begin{align*} \sin A + \sin B &= 2\sin \frac{A+B}{2}\cos \frac{A-B}{2} \\ \sin A - \sin B &= 2\cos \frac{A+B}{2}\sin \frac{A-B}{2} \\ \cos A + \cos B &= 2\cos \frac{A+B}{2}\cos \frac{A-B}{2} \\ \cos A - \cos B &= -2\sin \frac{A+B}{2}\sin \frac{A-B}{2}. \end{align*}\]