Question

cos⁻¹(2x²-1) = 2cos⁻¹x, 0 < x < 1,prove it

Answer 1

In the attachment I have answered this problem.     The calculations are mostly based on  inverse trigonometric functions.   See the attachment for detailed solution

Answer 2

LHS = cos^-1(2x^2 - 1)

Let cos^-1(2x^2 - 1) = A

cosA = (2x² - 1)

cosA + 1 = 2x²


we know by formula, 2cos²P = (1 + cos2P)
so, (1 + cosA) = 2cos²A/2

so, (cosA + 1) = 2cos2A/2 = 2x²

x = ± cosA/2 but x≠ - cosA/2 because 0 < x < 1

hence, x = cosA/2

cos^-1x= A/2

A = 2cos^-1x

hence, cos^-1(2x^2 - 1) = 2cos^-1x