Question #3d879

1 Answer
yumean1119
Sep 28, 2017

#x#=#sqrt(1-k^2)# when #0≦k≦1#.

Explanation:

In this equation, #x#=#sin{arcsin(x)}#
=#sin#{#pi/2#-#arcsin(k)#}
=#cos{arcsin(k)}#.

By difinition it follows #-pi/2#≦#arcsin(k)#≦#pi/2# and
#cos{arcsin(k)}# is non-negative.
So, #x#=#cos{arcsin(k)}#=#sqrt(1-sin^2{arcsin(k)}#
=#sqrt(1-k^2)#.
You may think domain of #k# is #-1≦k≦1# but this is wrong.

If you solve this equation for #k# instead of #x#, #k=sqrt(1-x^2)#
and #k# must be non-negative too.
The true domain is #0<=k<=1#.

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