Answer
The exact value of $\sin \left( 60{}^\circ -45{}^\circ \right)$ is $\frac{\sqrt{3}-1}{2\sqrt{2}}$.
Work Step by Step
Use the difference formula of sines and evaluate the expression as,
$\sin \left( 60{}^\circ -45{}^\circ \right)=\sin 60{}^\circ \cos 45{}^\circ -\cos 60{}^\circ \sin 45{}^\circ $
Substitute the values $\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\cos 60{}^\circ =\frac{1}{2},\text{ }\sin 45{}^\circ =\frac{1}{\sqrt{2}},\text{ and }\sin 30{}^\circ =\frac{\sqrt{3}}{2}$.
$\begin{align}
& \sin \left( 60{}^\circ -45{}^\circ \right)=\left( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2}\times \frac{1}{\sqrt{2}} \right) \\
& =\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}} \\
& =\frac{\sqrt{3}-1}{2\sqrt{2}}
\end{align}$
Hence, the exact value of $\sin \left( 60{}^\circ -45{}^\circ \right)$ is equivalent to $\frac{\sqrt{3}-1}{2\sqrt{2}}$.
