# Answer to Question #20186 in Trigonometry for liz

Question #20186

cos[arcsin(2√x/1+x)]

Expert's answer

we have that x>0 and 2sqrt(x)<=1+x for every x>0

cos(t)=sqrt(1-sin^2(t))

cos [arcsin(2vx/1+x)]=sqrt( 1-sin^2[arcsin(2vx/1+x)])=sqrt( 1- ( sin[arcsin(2vx/1+x)])^2 )=sqrt( 1- (2vx/(1+x))^2)=

=sqrt(1-4x/(1+x)^2)

cos(t)=sqrt(1-sin^2(t))

cos [arcsin(2vx/1+x)]=sqrt( 1-sin^2[arcsin(2vx/1+x)])=sqrt( 1- ( sin[arcsin(2vx/1+x)])^2 )=sqrt( 1- (2vx/(1+x))^2)=

=sqrt(1-4x/(1+x)^2)

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