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

$\sin 120^{\circ} \cos 150^{\circ}-\cos 240^{\circ} \sin 330^{\circ}$ is equal to :
  1. 1
  2. -1
  3. $\frac{2}{3}$
  4. $-\left(\frac{\sqrt{3}+1}{4}\right)$

Solution

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

Asked in: AP EAMCET 2006

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