Question #3d879

1 Answer
yumean1119
Sep 28, 2017

xx=sqrt(1-k^2)√1−k2 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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