Question

The value of $\sin {420}^{\circ }\cos {390}^{\circ }-\cos (-{660}^{\circ })\sin {330}^{\circ }$ is

Answer 1

As we know that $2\pi =360°$
So $\sin {420}^{\circ }\cos {390}^{\circ }-\cos (-{660}^{\circ })\sin {330}^{\circ }$
$=\sin {420}^{\circ }\cos {390}^{\circ }-\cos \left({660}^{\circ }\right)\sin {330}^{\circ }=\sin (2\pi +{60}^{\circ })\cos (2\pi +{30}^{\circ })-\cos (\frac{7\pi }{2}+{30}^{\circ })\sin (\frac{3\pi }{2}+{60}^{\circ })=\sin {60}^{\circ }\cos {30}^{\circ }-\sin {30}^{\circ }(-\cos {60}^{\circ })=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{1}{2}=1$