Question

Find the value of the trigonometric functions of sin $(-\frac{11\pi }{3})$.

Answer 1

sin $\left(\frac{-11\pi }{3}\right)$ = sin $\left(\frac{-11\times {180}^0}{3}\right)$

= sin $(-{660}^0)$ = sin $(-2\times {360}^0+{60}^0)$

= sin ${60}^0=\frac{\sqrt{3}}{2}$