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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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