sin(2 cos inverse (-3/5)) =

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We have : \sin (2 \cos^{- 1} (- \frac{3}{5})) We know \cos 53 = \frac{3}{5} \Rightarrow \cos (180 - 53) = - \cos 53 \Rightarrow \cos (180 - 53) = - \frac{3}{5} \Rightarrow (180 - 53) = \cos^{- 1} (- \frac{3}{5}) \Rightarrow \sin [2 \cos^{- 1} (- \frac{3}{5})] = \sin [2 (180 - 53)] = 2 \sin (180 - 53) . \cos (180 - 53) [\sin (2 A) = 2 sinA . cosA] = 2 \sin (53) [- \cos (53)] = 2 \times \frac{4}{5} \times [- \frac{3}{5}] = - \frac{24}{25} (Answer)

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