From cos^(-1)(x) = θ, we have cos(θ) = x. Therefore, sin(θ) = √(1 - cos^2(θ)) = √(1 - x^2).

The equation sin^(-1)(x) + cos^(-1)(x) = π/2 holds for all x in the domain of the functions, which is [-1, 1]. Therefore, x can be any value in this range.

Using the identity sin^(-1)(x) + cos^(-1)(x) = π/2, we can conclude that x can take any value in the range [-1, 1].

If sin^(-1)(x) = π/4, then x = sin(π/4) = √2/2.

tan^(-1)(x) = π/4 implies that x = tan(π/4) = 1.

Since x = cos^(-1)(1/2) = π/3, then sin^(-1)(1/2) = π/6.

Since cos^(-1)(1/2) = π/3, we have sin^(-1)(√(1 - (1/2)^2)) = sin^(-1)(√(3/4)) = π/3.

cos^(-1)(1/2) = π/3.

Using the identity sin(x) = sqrt(1 - cos^2(x)), we have sin(x) = sqrt(1 - (1/2)^2) = √3/2.

sin^(-1)(-1/2) = -π/6, since sin(-π/6) = -1/2.

sin^(-1)(-1/2) = -π/6, since sin(-π/6) = -1/2.

If x = sin^(-1)(1/2), then x = π/6. Therefore, cos(x) = cos(π/6) = √3/2.

Using the identity cos^(-1)(√(1 - sin^2(x))) = π/2 - x, we find that cos^(-1)(√(1 - (1/3)^2)) = π/2 - sin^(-1)(1/3).

If x = sin^(-1)(1/√2), then sin(x) = 1/√2. Therefore, cos(x) = √(1 - sin^2(x)) = √(1 - (1/√2)^2) = √(1/2) = √2/2.

Since x = sin^(-1)(1/√2) = π/4, then cos^(-1)(x) = π/2 - π/4 = π/4.

Using the identity cos^2(x) + sin^2(x) = 1, we find cos(x) = √(1 - (3/5)^2) = 4/5.

Using the identity cos(x) = sqrt(1 - sin^2(x)), we have cos(x) = sqrt(1 - (3/5)^2) = sqrt(16/25) = 4/5.

tan^(-1)(1) = π/4, since tan(π/4) = 1.

x = tan^(-1)(√3) = π/3, thus sin^(-1)(x) = sin^(-1)(√3/2) = π/3.

Since tan^(-1)(√3) = π/3, then 2x = 2π/3 and sin(2x) = sin(2π/3) = √3/2.

cos^(-1)(1/2) = π/3, since cos(π/3) = 1/2.

The derivative of y = cos^(-1)(x) is dy/dx = -1/√(1-x^2).

The derivative of cos^(-1)(x) is dy/dx = -1/√(1-x^2).

The derivative of y = sin^(-1)(x) is dy/dx = 1/sqrt(1-x^2).

sin^(-1)(-1) corresponds to the angle whose sine is -1, which is -π/2.

sin^(-1)(√3/2) corresponds to the angle whose sine is √3/2, which is π/3.

tan^(-1)(√3) corresponds to the angle whose tangent is √3, which is π/3.

The derivative of sin^(-1)(x) is 1/√(1-x^2)

The derivative of y = sin^(-1)(x) is 1/√(1-x^2)

The range of sin^(-1)(x) is [-π/2, π/2].