Answer
true
Work Step by Step
$\cos 90^{o}=0$,
Function Values of Special Angles:
$\displaystyle \cos 60^{o}=\frac{1}{2},\qquad \cos 30^{o}=\frac{\sqrt{3}}{2}$
$\displaystyle \sin 30^{o}=\frac{1}{2}, \quad \sin 60^{o}=\frac{\sqrt{3}}{2}$
$LHS= \cos 90^{o}=0$
$RHS= \displaystyle \frac{1}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{1}{2}=0=LHS$
The statement is true.
