Arcsin Values Quiz: Evaluating Common Arcsin Values

The answer is False. While sin(3pi/4) = sqrt(2)/2, the angle 3pi/4 lies outside the principal range from -pi/2 to pi/2 and is never returned by arcsin. The arcsin function always selects the unique angle within the principal range, which is pi/4. arcsin(sqrt(2)/2) = pi/4, not 3pi/4.

Let theta = arcsin(1/2), so sin(theta) = 1/2 and theta lies in the principal range where cosine is non-negative. Using the Pythagorean identity: cos squared theta = 1 minus 1/4 = 3/4. Taking the positive square root gives cos(theta) = sqrt(3)/2. Option A gives 1/2, the sine value not cosine. Option C gives sqrt(2)/2, requiring sin(theta) = sqrt(2)/2. Option D gives sqrt(5)/2, which exceeds 1 and is impossible for cosine.

sin(-pi/3) = -sqrt(3)/2 and -pi/3 lies within the principal range, so arcsin(-sqrt(3)/2) = -pi/3. Option A gives -pi/6, whose sine is -1/2. Option B gives pi/3, which produces a positive sine value. Option D gives 2pi/3, which lies outside the principal range.

sin(-pi/6) = -1/2 and -pi/6 lies within the principal range, so arcsin(-1/2) = -pi/6. Option B gives pi/6, which produces a positive sine value. Option C gives -pi/3, whose sine is -sqrt(3)/2. Option D gives 5pi/6, which lies outside the principal range entirely.

Option B gives x = 5/4, which exceeds 1 and falls outside the domain of arcsin, making it undefined there. Options A, C, and D all fall within the closed interval from -1 to 1 and produce valid real outputs.

sin(-pi/2) = -1 and -pi/2 lies within the principal range, so arcsin(-1) = -pi/2. Option A gives 0, the output of arcsin(0). Option B gives pi/2, the output of arcsin(1). Option D gives -pi, which lies outside the principal range and is never a valid arcsin output.

Option D is false because arcsin(0) = 0, not pi. The angle pi lies outside the principal range entirely and is never a valid arcsin output. Options A, B, and C are all correct with each angle lying within the principal range and satisfying the sine condition.

sin(pi/2) = 1 and pi/2 lies within the principal range, so arcsin(1) = pi/2. Option A gives 0, the output of arcsin(0). Option C gives pi, which lies outside the principal range. Option D gives -pi/2, the output of arcsin(-1). Only pi/2 satisfies the definition.

sin(5pi/6) = 1/2. arcsin returns the unique angle in the principal range whose sine equals 1/2, which is pi/6. The angle 5pi/6 lies outside the principal range. Option B is negative and requires a negative sine input. Option D gives -pi/6, whose sine is -1/2, not 1/2.

The answer is True. sin(0) = 0 and 0 lies within the principal range from -pi/2 to pi/2. Both conditions of the arcsin definition are satisfied, so arcsin(0) = 0. There is no ambiguity because 0 is the only angle in the principal range whose sine equals 0.

By definition, arcsin maps every input to a unique angle in the closed interval from -pi/2 to pi/2. Sine is strictly increasing and one-to-one on this interval, making a true inverse possible. Option A describes the range of arccos. Option B is too wide. Option D spans a full circle and includes far more angles than arcsin ever returns.

The answer is True. The domain of arcsin is the closed interval from -1 to 1. The input 1.2 exceeds the upper bound, meaning no real angle exists whose sine equals 1.2. Since sine only produces outputs between -1 and 1, there is no real solution and arcsin(1.2) has no real value.

sin(-pi/4) = -sqrt(2)/2 and -pi/4 belongs to the principal range, so arcsin(-sqrt(2)/2) = -pi/4. Option A gives pi/4, which produces a positive sine value. Option B gives -pi/6, whose sine is -1/2. Option C gives -pi/3, whose sine is -sqrt(3)/2. Only -pi/4 is correct.

sin(pi/4) = sqrt(2)/2 and pi/4 lies within the principal range, so arcsin(sqrt(2)/2) = pi/4. Option A gives -pi/4, which has a negative sine value. Option B gives pi/6, whose sine is 1/2, not sqrt(2)/2. Option D gives 3pi/4, which lies outside the principal range.

sin(0) = 0 and 0 lies within the principal range from -pi/2 to pi/2, so arcsin(0) = 0. Option A gives pi/2, the output of arcsin(1). Option C gives -pi/2, the output of arcsin(-1). Option D gives pi, which lies outside the principal range and is never a valid arcsin output.

Option A gives x = 1/2, producing arcsin(1/2) = pi/6, a positive angle, so it is excluded. Options B, C, and D are all strictly negative inputs producing negative angles: arcsin(-1/2) = -pi/6, arcsin(-sqrt(2)/2) = -pi/4, and arcsin(-sqrt(3)/2) = -pi/3.

The answer is True. The derivative of arcsin(x) is 1 divided by the square root of (1 minus x squared), which is positive for all x strictly between -1 and 1. A positive derivative means the function is strictly increasing throughout the interior of its domain. This increasing behavior extends to the endpoints, so arcsin is strictly increasing across its entire closed domain.

Let theta = arcsin(sqrt(3)/2), the unique angle in the principal range whose sine equals sqrt(3)/2. Applying sin immediately returns sqrt(3)/2 by definition. Option A gives 1/2, which is sin(pi/6). Option B gives sqrt(2)/2, the sine of pi/4. Option D gives 1, the sine of pi/2. The composition sin(arcsin(x)) always returns x for any x in the domain.

The answer is True. The sine function is odd, and arcsin is its inverse defined on a symmetric interval centered at 0. Because the domain from -1 to 1 and the range from -pi/2 to pi/2 are both symmetric about 0, the odd symmetry of sine carries over to its inverse. This means arcsin(-x) equals -arcsin(x) for every x in the domain.

Option C is false because 2 lies outside the domain of arcsin, making arcsin(2) undefined over the real numbers. Options A, B, and D are all correct standard identities confirmed by the principal range definition.