Answer
The given statement makes sense.
Work Step by Step
One of the product-to-sum formulas is $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$. Therefore, according to the above-mentioned formula, the value of $\alpha $ is ${{13}^{\circ }}$ and the value of $\beta $ is ${{48}^{\circ }}$.
Thus, the expression can be evaluated as:
$\begin{align}
& \sin {{13}^{\circ }}\cos {{48}^{\circ }}=\frac{1}{2}\left[ \sin \left( {{13}^{\circ }}+{{48}^{\circ }} \right)+\sin \left( {{13}^{\circ }}-{{48}^{\circ }} \right) \right] \\
& =\frac{1}{2}\left[ \sin {{61}^{\circ }}+\sin \left( -{{35}^{\circ }} \right) \right]
\end{align}$
Now, applying the even-odd identity, which is $sin(-x)=-\sin x$, the expression can be further evaluated as shown below:
$\frac{1}{2}\left[ \sin {{61}^{\circ }}+\sin \left( -{{35}^{\circ }} \right) \right]=\frac{1}{2}\left( \sin {{61}^{\circ }}-\sin {{35}^{\circ }} \right)$
Therefore, $\sin {{13}^{\circ }}cos{{48}^{\circ }}$ can be expressed as $\frac{1}{2}\left( \sin {{61}^{\circ }}-\sin {{35}^{\circ }} \right)$.
