Answer
The statement makes sense.
Work Step by Step
We know tatt the statement says that the sum and difference formulas for cosines and sines are used to derive the product-to-sum formulas. This can be well understood with the help of an example as shown below.
The sum and difference formulas for sines are $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ and $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $.
Add the left and right sides of these two identities as follows:
$\begin{align}
& \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\
& =2\sin \alpha \cos \beta
\end{align}$
Now, multiply both sides by $\frac{1}{2}$ as given below:
$\begin{align}
& \frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]=\frac{1}{2}\cdot 2\sin \alpha \cos \beta \\
& =\sin \alpha \cos \beta
\end{align}$
Thus, one of the product-to-sum formulas is derived, which is $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$.
