Question

$\sin 120^{\circ }\cos 150^{\circ }-\cos 240^{\circ }\sin 330^{\circ }$ is equal to :

• A. 1
• B. -1
• C. $\frac{2}{3}$
• D. $-\left(\frac{\sqrt{3}+1}{4}\right)$

Answer 1

Answer: (B)

$\sin 120^{\circ }\cos 150^{\circ }-\cos 240^{\circ }\sin 330^{\circ }$
$=-\cos 30^{\circ }\sin 60^{\circ }-\cos 60^{\circ }\sin 30^{\circ }$
$=-\left(\sin 60^{\circ }\cos 30^{\circ }+\cos 60^{\circ }\sin 30^{\circ }\right)$
$=-\sin \left(60^{\circ }+30^{\circ }\right)=-\sin 90^{\circ }$
$=-1$