Question

Find the value of $\sin {330}^{\circ }\cos {120}^{\circ }+\cos {210}^{\circ }\sin {300}^{\circ }$

Answer 1

Consider,$\sin {330}^{\circ }\cos {120}^{\circ }$

$=\sin ({360}^{\circ }-{30}^{\circ })\cos ({180}^{\circ }-{60}^{\circ })$

$=-\sin {30}^{\circ }\times -\cos {60}^{\circ }$

$=\sin {30}^{\circ }\cos {60}^{\circ }$

$=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Consider,$\cos {210}^{\circ }\sin {300}^{\circ }$

$=\cos ({180}^{\circ }+{30}^{\circ })\sin ({360}^{\circ }-{60}^{\circ })$

$=-\cos {30}^{\circ }\times -\sin {60}^{\circ }$

$=-\frac{\sqrt{3}}{2}\times \frac{-\sqrt{3}}{2}$

$=\frac{3}{4}$

$\therefore \sin {330}^{\circ }\cos {120}^{\circ }+\cos {210}^{\circ }\sin {300}^{\circ }=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1$