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

sin^(-1)(-1) corresponds to the angle -π/2.

sin^(-1)(-1) corresponds to the angle -π/2.

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

Using the identity cos(x) = √(1 - sin^2(x)), we find cos(sin^(-1)(1/3)) = √(1 - (1/3)^2) = √(8)/3.

tan^(-1)(1) = π/4.

tan^(-1)(1/√3) = π/6, since tan(π/6) = 1/√3.

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

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

By definition, if y = sin^(-1)(x), then x = sin(y).

The second derivative d^2y/dx^2 = -1/√(1-x^2)^3.

The range of y = tan^(-1)(x) is (-π/2, π/2).

The first derivative dy/dx = 1/(1+x^2). The second derivative d^2y/dx^2 = -2/(1+x^2)^2.

The derivative of \( y = \cot^{-1}(x) \) is \( -\frac{1}{1+x^2} \).

The derivative of \( y = \sec^{-1}(x) \) is \( \frac{1}{|x|\sqrt{x^2-1}} \).

Since \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) for all \( x \) in the domain of \( \sin^{-1} \) and \( \cos^{-1} \), the answer is \( \frac{\pi}{2} \).

When \( x = 1 \), \( y = \tan^{-1}(1) + \tan^{-1}(y) \) leads to \( y = \frac{\pi}{4} \).

The pairs are (1,2), (1,3), (1,4), (2,3), (2,4), (3,4). Thus, there are 6 ordered pairs.

The ordered pairs are (1,2), (1,3), (1,4), (2,3), (2,4), (3,4). Thus, there are 6 ordered pairs.

R is reflexive, antisymmetric, and transitive, thus it is a partial order.

The function has a minimum at x = 0, hence it is neither always increasing nor decreasing.

f(x) = (x - 2)^2 is the completed square form.

The function |x - 3| is continuous everywhere, including at x = 3.

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

The absolute value function has a minimum value of 0, hence the range is [0, ∞).

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

Using the right triangle definition, cos(tan^(-1)(x)) = adjacent/hypotenuse = 1/√(1+x^2).

Using the right triangle definition, sin(tan^(-1)(x)) = opposite/hypotenuse = x/√(1+x^2).

Since π/4 is in the range of sin^(-1), sin^(-1)(sin(π/4)) = π/4.

f(g(x)) = f(2x) = 2x + 1.