Question

# Find the value of $$\sin\left ( -\frac{-11\pi}{3} \right )$$

Solution

## We know that values of $$\sin x$$ repeats after an interval of $$2\pi$$.so, $$\sin\left ( -\dfrac{11\pi}{3} \right )=-\sin\left ( \dfrac{11\pi}{3} \right )$$                                  $$[\because \sin(-\theta)=-\sin\theta]$$                      $$=-\sin\left ( 4\pi-\dfrac{\pi}{3} \right )$$                      $$-\sin\left ( 2\times2\pi-\dfrac{\pi}{3} \right )$$(So, this angle will be in IV quadrant in which $$\sin$$ is negative.)                      $$=\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$$Hence, $$\sin\left ( -\dfrac{11\pi}{3} \right )=\dfrac{\sqrt{3}}{2}$$Mathematics

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