Question

sin3a-cos2a при a=P:6

Answer 1

$sin(3\alpha)-cos(2\alpha)=\bigg| \alpha=\frac{\pi}{6} \bigg| = sin(3\cdot\frac{\pi}{6})-cos(2\cdot\frac{\pi}{6})=sin(\frac{\pi}{2} )-cos(\frac{\pi}{3})=\\\\=1-\frac{1}{2} =\frac{1}{2}$