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Mathematics 13 Online
OpenStudy (kainui):

How do I simplify sin(2sin^-1(x))? I tried using this trig identity, sin2x=sinxcosx and got x(1-x^2)^1/2 but the answer wolfram alpha gave me was the same except that it was multiplied by 2. What have I done wrong?

OpenStudy (lgbasallote):

sin^-1 (x) means arcsin of x \[\sin [\sin^{-1} (x)] = x\]

OpenStudy (dumbcow):

sin(2x) = 2sin(x)cos(x)

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