Question

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Answer 1

Answer:

$Put \: \sin^{ - 1} (x) = \theta. \: Then, \: x = \sin( \theta) \\ Now, \: \sin(3 \theta) = (3 \sin( \theta) - 4 \sin^{3} ( \theta) ) = (3x - 4 {x}^{3} ) \\ =>3 \theta = \sin^{ - 1} (3x - 4 {x}^{3} ) \\=>3 \sin ^{ - 1} (x) = \sin^{ - 1} (3x - 4 {x}^{3} ) \:[Since, \: \theta = sin^{-1}x]. \\ Hence, \: 3 \sin^{ - 1} (x) = \sin^{ - 1} (3x - 4 {x}^{3} ) .$