Answer
The required solution is $-2\sin \frac{\pi }{6}\cos \frac{\pi }{4}$, $-\frac{\sqrt{2}}{2}$.
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 $\frac{\pi }{12}$ and the value of $\beta $ is $\frac{5\pi }{12}$.
Now, the expression can be evaluated as provided below:
$\begin{align}
& \sin \frac{\pi }{12}-\sin \frac{5\pi }{12}=2\sin \left( \frac{\frac{\pi }{12}-\frac{5\pi }{12}}{2} \right)\cos \left( \frac{\frac{\pi }{12}+\frac{5\pi }{12}}{2} \right) \\
& =2\sin \left( \frac{\frac{-4\pi }{12}}{2} \right)\cos \left( \frac{\frac{6\pi }{12}}{2} \right) \\
& =2\sin \left( -\frac{4\pi }{24} \right)\cos \left( \frac{6\pi }{24} \right) \\
& =2\sin \left( -\frac{\pi }{6} \right)\cos \left( \frac{\pi }{4} \right)
\end{align}$
Thus, applying the even-odd identity, which is $sin(-x)=-\sin x$, the expression can be further evaluated as given below:
$2\sin \left( -\frac{\pi }{6} \right)\cos \left( \frac{\pi }{4} \right)=-2\sin \frac{\pi }{6}\cos \frac{\pi }{4}$
Then, the respective values are put to find the product’s exact value:
$\begin{align}
& -2\sin \frac{\pi }{6}\cos \frac{\pi }{4}=-2\left( \frac{1}{2} \right)\left( \frac{\sqrt{2}}{2} \right) \\
& =-\frac{\sqrt{2}}{2}
\end{align}$
Hence, the provided expression can be written as $-2\sin \frac{\pi }{6}\cos \frac{\pi }{4}$. And the product’s exact value is $-\frac{\sqrt{2}}{2}$.
