Asked by mtd
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
answer is (a cosx) and I don't know how to get there
Answers
Answered by
Reiny
√(a^2 -u^2) where u = asinx
= √(a$ - a^2(sinx)^2)
= √(a^2(1 - (sinx)^2)
= a√(cosx)^2
= a cos x
= √(a$ - a^2(sinx)^2)
= √(a^2(1 - (sinx)^2)
= a√(cosx)^2
= a cos x
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