PROVE 2arccos(x)=arccos(2x-1)Thanks for any help

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.