# Prove the identity arc sin squareroot(1-x^2)+ arc cos x=pi, if -1=<x=<0.

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### 1 Answer

Since the result of the given sum is a constant, then the value of th derivative of the sum must be zero.

We'll assign a function to the given expression:

f(x) = arcsin sqrt(1-x^2) + arccos x

We'll differentiate with respect to x:

f'(x) = [sqrt(1-x^2)]'/sqrt(1 - 1 + x^2) - 1/sqrt(1 - x^2)

f'(x) = -2x/2*|x|*sqrt(1-x^2) - 1/sqrt(1 - x^2)

Since x is in the interval [-1 , 0] => |x| = -x

f'(x) = -2x/-2x*sqrt(1-x^2) - 1/sqrt(1 - x^2)

f'(x) = 1/sqrt(1 - x^2) - 1/sqrt(1 - x^2)

f'(x) = 0

Since the 1st derivative of f()x is cancelling out, then the function f(x) is a constant.

Now, we'll check if the constant is pi.

Let x = -1. We'll compute the value of f(x) for x =-1.

f(-1) = arcsin sqrt(1 - 1) + arccos(-1)

f(-1) = arcsin 0 + arccos(-1)

f(-1) = 0 + pi

f(-1) = pi

We'll replace x by 0 and we'll get:

f(0) = arcsin sqrt(1 - 0) + arccos(0)

f(0) = arcsin 1 + arccos 0

f(0) = pi/2 + pi/2

f(0) = 2pi/2

f(0) = pi

**The identity arcsin sqrt(1-x^2) + arccos x = pi is verified for any value of x, over the range [-1 , 0].**