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**