Your question: show the sin^-1 (cosx)= x + pi/2
answer, you can't because it is not true, at least it is not true for any positive value of x because arcsin (anything) must be between -pi/2 and p/2 and this would give an answer above this range.
It may be true for x values between -pi and 0, I'll have to think about that some more.
I've only really thought this through for acute values of x (0 is less than x is less than pi/2 radians)
but
sin^-1 (cosx)= pi/2 - x (in radians)
Method 1 (0 is less than x is less than pi/2 radians)
You can show this by drawing a right angled triangle. One acute angle is x so the other one has to be pi/2-x
Let the side adjacent to x be p and the one opposite be q and the hypotenuse be h.
cos x = p/h
sine of what angle equals p/h? Sine of the angle at the other vertex equals p/h and the angle at the other vertex is pi/2-x
Try drawing the triangle - it should make it clearer.
Method 2
You can also consider it algebraically (0 is less than x is less than pi/2 radians)
arcsin[cos(x)]
= arcsin[sin(π/2 -x)]
= π/2 - x
It you want me to think about it more let me know.
+118654 
