arcsin(x) + arccos(x) = pi/2 ?

sin⁻¹x + cos⁻¹x = π/2

sin⁻¹x = π/2 - cos⁻¹x

sin⁻¹x = sin⁻¹x 

sin^-1(x)+cos^-1(x)=pi/2

Let A=sin^-1(x); B=cos^-1(x)

=>

A+B=pi/2

sin(A+B)=1

sinAcosB+cosAsinB=1

=>

x^2+1-x^2=1

is an identity

=>

|x|<=1 or -1=<x<=1

e.g.

x=0.5=>sin^-1(0.5)=30*; cos^-1(0.5)=60*=>sin^-1(0.5)+cos^-1(0.5)=90*.

x=-0.5=>sin^-1(-0.5)=-30*; cos^-1(-0.5)=120*=>sin^-1(-0.5)+cos^-1(-0.5)=90*

This is true of all numbers "x" in the interval [0,1].  If you use negative numbers, you could run into some problems.

Noting that pi/2 = pi/4 + pi/4,  how about  x = sqrt(2)/2?

Hint: Its an identity. Think of a right triangle with hypotenuse 1 and the two other sides of equal length. 

cos(arcsin(x) + arccos(x)) = cos(pi/2)

cos(arcsin(x))cos(arccos(x)) - sin(arcsin(x))sin(arccos(x)) = 0

sqrt(1 - sin(arcsin(x))^2) * x - x * sqrt(1 - cos(arccos(x))^2) = 0

sqrt(1 - x^2) * x - x * sqrt(1 - x^2) = 0

0 = 0