## Trigonometry (11th Edition) Clone

$\cos 90^{o}=0$, 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.