# Prove that the function f(x)=arcsinx+3arccosx+arcsin(2x*square root(1-x^2)) is constant.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We could use derivative of the function to prove that f(x) is constant. If we'll prove that the derivative of f(x) is cancelling, therefore the function is constant, since the derivative of any constant function is 0.

We'll differentiate the function with respect to x:

f'(x) = 1/sqrt(1-x^2) - 3/sqrt(1-x^2) + [2x*sqrt(1-x)^2]'/sqrt[1-4x^2*(1-x^2)]

f'(x) = -2/sqrt(1-x^2) + [2sqrt(1-x^2) - 2x*2x/2sqrt(1-x^2)]/sqrt(1-4x^2 + 4x^4)

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

f'(x) = -2/sqrt(1-x^2) + (2-4x^2)/(1-2x^2)sqrt(1-x^2)

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

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

f'(x) = 0

Since the 1st derivative of the function is zero, therefore the function is a constant.

For x = 0 => f(0) = arcsin 0 + 3arccos 0 + arcsin 0

f(0) = 0 + 3*`pi` /2 + 0

f(0) = 3`pi` /2

The given function f(x) is constant and for x = 0 is f(0) = 3`pi` ` ` /2.