Answer
The required solution is $2\sin \left( {{45}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)$, $\frac{\sqrt{6}}{2}$.
Work Step by Step
One of the sum-to-product formula is $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$. So, in this question, according to the above-mentioned formula, the value of $\alpha $ is ${{75}^{\circ }}$ and the value of $\beta $ is ${{15}^{\circ }}$.
Now, the expression can be evaluated as provided below:
$\begin{align}
& \sin {{75}^{\circ }}+\sin {{15}^{\circ }}=2\sin \left( \frac{{{75}^{\circ }}+{{15}^{\circ }}}{2} \right)\cos \left( \frac{{{75}^{\circ }}-{{15}^{\circ }}}{2} \right) \\
& =2\sin \left( \frac{{{90}^{\circ }}}{2} \right)\cos \left( \frac{{{60}^{\circ }}}{2} \right) \\
& =2\sin \left( {{45}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)
\end{align}$
Thus, the values as per the trigonometry table are put to solve the expression further and to find the exact value:
$\begin{align}
& 2\sin \left( {{45}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)=2\left( \frac{\sqrt{2}}{2} \right)\left( \frac{\sqrt{3}}{2} \right) \\
& =\frac{\sqrt{6}}{2}
\end{align}$
Hence, the provided expression can be written as $2\sin \left( {{45}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)$. And the product’s exact value is $\frac{\sqrt{6}}{2}$.
