arcsin x + arccosx = pi/2 if -1<x<1
Let f(x) = arcsinx + arccosx
If f(x) = 1 , then it is a constant , then f'(x) should be 0:
f'(x) = (arcsinx + arccosx)'
= 1/sqrt(1-x^2) -1/sqrt(1-x^2) = 0
Then we proved that f(x) is a constant number
==> f(x) = C
Now let us substitutw with any value within the interval given.
f(1) = arcsin1 + arccos1 = pi/2 + 0 = pi/2
==> arcsinx + arccosx = pi/2
Draw a right angled triangle : with sides BC = x , AC = 1 and AB sqrt(1-x^2)., with B right angle.
Then angle SinA = x/1 =x Or A = arc sin x....(1).
cos C = x/1 . Or C = Arc cosx ...................(2).
From (1) and (2) A+C = arc sinx + arc cosx. But A+ C = 180 -B deg = 180 - 90 = 90. deg or pi/2 rad.
So arcsinx +arc cosx = pi/2.
We'll associate a function f(x) to the expression (arcsin x + arccos x).
If we want to prove that the function is a constant function, we'll have to do the first derivative test. If the first derivative is cancelling, that means that f(x) is a constant function, knowing the fact that a derivative of a constant function is 0.
f'(x) = (arcsin x + arccos x)'
f'(x) = 1/sqrt(1-x^2) - 1/sqrt(1-x^2)
We'll eliminate like terms:
f'(x)=0, so f(x)=constant
To verify if the constant is pi/2, we'll put x = 1:
f(1)=arcsin 1 + arccos 1 = pi/2 + 0=pi/2