Math (Trigonometry)

square root(a^2 -u^2) where a>0, let u=(a sinx) where -pi/2<x<pi/2

answer is (a cosx) and I don't know how to get there

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asked by mtd
  1. √(a^2 -u^2) where u = asinx
    = √(a$ - a^2(sinx)^2)
    = √(a^2(1 - (sinx)^2)
    = a√(cosx)^2
    = a cos x

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    posted by Reiny

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