Question

Show that sin^-1(2x√(1-x^2))=-2π+2cos^-1x if -1≤x≤-1/√2​

Answer 1

Given:

sin^-1(2x√(1-x^2))=-2π+2cos^-1x

To find:

Show that sin^-1(2x√(1-x^2))=-2π+2cos^-1x if -1≤x≤-1/√2

Solution:

Start with the identity sin^-1(x) = cos^-1(√(1-x^2)) for -1≤x≤1.

Substitute 2x√(1-x^2) for x in the identity above: sin^-1(2x√(1-x^2)) = cos^-1(√(1-(2x√(1-x^2))^2))

Simplify the expression inside the square root: sin^-1(2x√(1-x^2)) = cos^-1(√(1-(4x^2(1-x^2)))

Simplify further: sin^-1(2x√(1-x^2)) = cos^-1(√(1-4x^2+4x^4))

Use the identity cos^-1(x) = -cos^-1(-x) for -1≤x≤1.

Substitute -4x^2+4x^4 for x in the identity above: sin^-1(2x√(1-x^2)) = -cos^-1(4x^2-4x^4)

Use the identity cos^-1(x) = π/2 - sin^-1(√(1-x^2)) for 0≤x≤1

Substitute 4x^2-4x^4 for x in the identity above: sin^-1(2x√(1-x^2)) = -(π/2 - sin^-1(√(1-(4x^2-4x^4)^2))

Simplify the expression inside the square root: sin^-1(2x√(1-x^2)) = -(π/2 - sin^-1(√(1-16x^4+16x^6-16x^8)))

Use the identity sin^-1(x) = π - sin^-1(-x) for -1≤x≤1.

Substitute -16x^4+16x^6-16x^8 for x in the identity above: sin^-1(2x√(1-x^2)) = -(π/2 - (π - sin^-1(16x^4-16x^6+16x^8))

Simplify the expression inside the sin^-1: sin^-1(2x√(1-x^2)) = -(π/2 - (π - (2π - 2cos^-1(x))

Combine like terms and simplify:

sin^-1(2x√(1-x^2)) = -2π + 2cos^-1(x)

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