# Prove that 3arcsinx-arcsin(3x-4x^3)=0

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To verify if the given identity, we'll use one of Lagrange's consequences.

f(x) = 3arcsin x -[arcsin(3x-4x^3)] = 0

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

We'll calculate the first derivative of the function, both sides:

f'(x) = {3arcsin x -[arcsin(3x-4x^3)]}'

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

[arcsin(3x-4x^3)]' = (3x-4x^3)'/sqrt[1 - (3x-4x^3)^2]

[arcsin(3x-4x^3)]' = (3 - 12x^2)/sqrt[1 - (3x-4x^3)^2] (2)

(3arcsinx)' = 3/sqrt(1-x^2) (1)

{3arcsin x -[arcsin(3x-4x^3)]}' = (1) - (2)

3/sqrt(1-x^2) - (3 - 12x^2)/sqrt[1 - (3x-4x^3)^2]

If (1) - (2) = 0, then the identity is verified.

Another method is to calculate sine function both sides:

sin (3arcsin x) = sin [arcsin(3x-4x^3)]

Let arcsin x = t => sin t = x and we'll use the identity sin (arcsin x) = x.

sin 3t = (3x-4x^3)

We'll apply the triple angle identity;

sin 3t = sin (2t+t) = sin2t*cost + sint*cos2t

sin3t = 2sint*[1-(sin t)^2] + sint*[1-2(sin t)^2]

sin3t = 2sint - 2(sin t)^3 + sint - 2(sin t)^3

sin3t = 3sint- 4(sin t)^3

We'll replace sin t by x:

sin3t = 3x - 4x^3

**We'll notice that calculating the difference 3/sqrt(1-x^2) - (3 - 12x^2)/sqrt[1 - (3x-4x^3)^2] we'll get zero, therefore 3arcsin x -[arcsin(3x-4x^3)] = 0**