Answer
The required solution is $2\sin \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)$, $-\frac{\sqrt{2}}{2}$.
Work Step by Step
One of the sum-to-product formula is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$. 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:
$\begin{align}
& \cos {{75}^{\circ }}-\cos {{15}^{\circ }}=-2\sin \left( \frac{{{75}^{\circ }}+{{15}^{\circ }}}{2} \right)\sin \left( \frac{{{75}^{\circ }}-{{15}^{\circ }}}{2} \right) \\
& =-2\sin \left( \frac{{{90}^{\circ }}}{2} \right)\sin \left( \frac{{{60}^{\circ }}}{2} \right) \\
& =-2\sin \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)
\end{align}$
Thus, the values as per the trigonometry table are put to solve the expression further and to find the exact value:
$\begin{align}
& -2\sin \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)=-2\left( \frac{\sqrt{2}}{2} \right)\left( \frac{1}{2} \right) \\
& =-\frac{\sqrt{2}}{2}
\end{align}$
Hence, the provided expression can be written as $2\sin \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)$. And the product’s exact value is $-\frac{\sqrt{2}}{2}$.
