We have to prove that 2*arc cos(x) = arc cos(2x-1)

Let arc cos x = y

=> x= cos y

2x - 1 = 2*cos y - 1

For the given relation to be true for all values of x, 2y = arc cos( 2*cos y - 1). But arc cos(2*cos y - 1) cannot be simplified further to make it of the form 2y.

**Therefore the given relation is not true for all values of x.**

To prove that 2arccos(x)=arccos(2x-1), we'll have to create a function:

f(x) = 2arccos(x)-arccos(2x-1) = 0

Since the function f(x) is a constant, therefore the first derivative must be zero.

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

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

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

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

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

**As we can see, the first derivative is not cancelling, therefore the given expression does not represent an identity.**