Posted by mtd on .
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

Math (Trigonometry) 
Reiny,
√(a^2 u^2) where u = asinx
= √(a$  a^2(sinx)^2)
= √(a^2(1  (sinx)^2)
= a√(cosx)^2
= a cos x