Question

The value of, $\sin {600}^o.\cos {330}^o+\cos {120}^o\sin {150}^o,$

Answer 1

$=\sin {600}^{\circ }\cos {330}^{\circ }+\cos {120}^{\circ }\sin {150}^{\circ }=\sin ({540}^{\circ }+{60}^{\circ })\cos ({360}^{\circ }-{30}^{\circ })+\cos ({180}^{\circ }-{60}^{\circ })\sin ({180}^{\circ }-{30}^{\circ })=(-\sin {60}^{\circ }\left)\right(\cos {30}^{\circ })+(-\cos {60}^{\circ }\left)\right(\sin {30}^{\circ })=-(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2})=-2$