A) 0
B) 1
C) \[\frac{1}{2}\]
D) \[-1\]
E) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
\[cos\text{ }480{}^\circ .sin\text{ }150{}^\circ +sin\text{ }600{}^\circ .\,cos\text{ }390{}^\circ \] \[=[\cos (3\pi -60{}^\circ )\sin (\pi -30{}^\circ )+\sin (3\pi +60{}^\circ )\] \[\times \cos (2\pi +30{}^\circ )]\] \[=-\cos 60{}^\circ \sin 30{}^\circ +(-\sin 60{}^\circ )\cos 30{}^\circ \] \[=-\frac{1}{2}.\frac{1}{2}+\left( -\frac{\sqrt{3}}{2} \right)\left( \frac{\sqrt{3}}{2} \right)=-\frac{1}{4}-\frac{3}{4}=-1\]
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