Find the value of $\sin {240}^{\circ }$.

Find the value of given trigonometric expression

Given, $\sin {240}^{\circ }$can be expressed as, $\sin {240}^{\circ }=\sin ({180}^{\circ }+{60}^{\circ })$

Since $$\sin$$ is a periodic function of time period $$2\pi$$, also negative in third quadrant or $\sin (\pi +\theta )=-\sin \theta$,where$\pi ={180}^{\circ }$

Therefore,

$$\begin{gathered} \sin {240}^{\circ }=\sin ({180}^{\circ }+{60}^{\circ }) \\ =-\sin {60}^{\circ } \\ =-\frac{\sqrt{3}}{2} \\ \therefore \sin {240}^{\circ }=-\frac{\sqrt{3}}{2} \end{gathered}$$
Hence, the value of $\sin {240}^{\circ }$is$-\frac{\sqrt{3}}{2}$.