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 $