Answer
$\dfrac{3-3\sqrt 2}{2}$
Work Step by Step
We know that the given trigonometric values are:
$\sin 60^{\circ}=\dfrac{ \sqrt 3}{2}$ and $\cos 45^{\circ}=\dfrac{\sqrt 2}{2}$
We evaluate the given expression to obtain:
$3 \sin^2 60^{\circ}- 3 \cos 45^{\circ} =2(\dfrac{\sqrt 3}{2})^2- 3 (\dfrac{\sqrt 2}{2})$
or, $= 2 \times \dfrac{3}{4} - \dfrac{3 \sqrt 3}{2}$
or, $=\dfrac{3-3\sqrt 2}{2}$
