Inverse Sine Quiz: Inverse Sine Fundamentals

The principal range of arcsin is the interval from -pi/2 to pi/2. This interval is chosen because sine is one-to-one on it, allowing a true inverse to be defined. Option A is too wide and includes angles where sine is not one-to-one. Option C describes the principal range of arccos, not arcsin. Option D spans a full circle and includes far more angles than the arcsin function ever returns.

Let theta = arcsin(-0.8), which is the unique angle in the principal range whose sine equals -0.8. Applying sin to that angle immediately recovers -0.8 by definition. Option B gives the positive version, ignoring the negative sign. Option C incorrectly squares the value. Option D is the value of cos(theta) using the Pythagorean identity, which gives sqrt(1 - 0.64) = 0.6, but that is cosine, not sine.

sin(-pi/3) = -sqrt(3)/2 and -pi/3 lies within the principal range from -pi/2 to pi/2, so arcsin(-sqrt(3)/2) = -pi/3. Option A gives -pi/6, whose sine is -1/2, not -sqrt(3)/2. Option C gives pi/3, which is positive and would give a positive sine value. Option D gives 5pi/3, which lies far outside the principal range and is never a valid arcsin output.

Since theta = arcsin(x), we have sin(theta) = x. Using the Pythagorean identity, cos squared theta = 1 - sin squared theta = 1 - x squared. Because theta lies in the principal range from -pi/2 to pi/2, cosine is non-negative there, so cos(theta) = sqrt(1 - x squared). Option A adds instead of subtracts under the radical. Option C is a linear expression with no trigonometric basis. Option D confuses cosine with sine.

Options A and D both capture the full definition of arcsin: the output is the unique angle in the principal range whose sine equals x. Option C correctly identifies that x must belong to the domain of arcsin. Option B is false because it allows infinitely many angles, removing the uniqueness that defines arcsin. 

Let theta = arcsin(0.3), which is the angle in the principal range whose sine equals 0.3. Applying sin to that angle immediately gives back 0.3 by definition of the arcsin function. Option B incorrectly squares the value. Option C subtracts from 1, which has no basis in this composition. Option D negates the value without justification. The composition sin(arcsin(x)) always returns x for any x in the domain of arcsin.

The answer is True. The derivative of arcsin(x) equals 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 confirms the function is strictly increasing throughout the interior of its domain. The function also increases up to and including the endpoints -1 and 1, so arcsin is strictly increasing across its entire domain.

The answer is True. The range of arcsin is the closed interval from -pi/2 to pi/2, which is approximately from -1.571 to 1.571. The value 2 lies above this upper bound and therefore cannot be an output of arcsin for any real input x. Since no value of x in the domain of arcsin produces an output of 2, the equation has no real solution.

sin(0) = 0 and 0 lies within the principal range from -pi/2 to pi/2, so arcsin(0) = 0. Option B gives pi/2, which is the output of arcsin(1), not arcsin(0). Option C gives pi, which lies outside the principal range entirely and is never a valid arcsin output. Option D gives -pi/2, which is the output of arcsin(-1). Only 0 satisfies both conditions required by the definition.

sin composed with arcsin recovers x on the full domain of arcsin. arcsin composed with sin recovers theta only on the principal range. For theta in the interval from pi/2 to 3pi/2, sin(theta) = sin(pi - theta) and pi - theta lies in the principal range, so arcsin returns pi - theta. arcsin is an odd function, confirming option D. 

The answer is False. sin(5pi/6) equals 1/2. The arcsin function always returns the principal value, which is the unique angle in the interval from -pi/2 to pi/2 whose sine equals the input. Since arcsin(1/2) = pi/6, not 5pi/6, the statement is incorrect. The angle 5pi/6 lies outside the principal range of arcsin, so it is never a valid output of the function.

arcsin maps the interval from -1 to 1 onto the interval from -pi/2 to pi/2. The values arcsin(0) = 0, arcsin(1) = pi/2, and arcsin(-1) = -pi/2 are standard results within this range. Every output of arcsin belongs to the interval from -pi/2 to pi/2. Option D is false because arcsin is undefined for any input whose absolute value exceeds 1.

Within the principal interval from -pi/2 to pi/2, the angle whose sine equals -1/2 is -pi/6. This is because sin(-pi/6) = -1/2 and -pi/6 lies within the required interval. Option A gives -pi/3, whose sine is -sqrt(3)/2, not -1/2. Option B gives pi/6, whose sine is positive 1/2. Option D gives pi/3, whose sine is positive sqrt(3)/2. Only -pi/6 is correct.

The arcsin function requires its input to be a value that sine can actually produce. Since the sine function only outputs values between -1 and 1, arcsin is only defined for inputs in that same closed interval. Option A is incorrect because inputs outside this range produce no real output. Options C and D each describe only half the domain, omitting the other half entirely.

Every listed value is a standard result within the principal range from -pi/2 to pi/2. sin(-pi/2) = -1, sin(pi/4) = sqrt(2)/2, sin(pi/3) = sqrt(3)/2, sin(-pi/3) = -sqrt(3)/2, and sin(pi/2) = 1. Each angle lies within the principal range, confirming all five evaluations are correct. These are foundational reference values for the arcsin function.

The answer is False. The identity arcsin(sin(theta)) = theta holds only when theta belongs to the principal range from -pi/2 to pi/2. Outside this interval, arcsin returns the principal angle that shares the same sine value as theta, which is generally different from theta itself. For example, arcsin(sin(5pi/6)) = pi/6, not 5pi/6, because 5pi/6 lies outside the principal range.

sin(pi/6) = 1/2 and pi/6 lies within the principal range from -pi/2 to pi/2, so arcsin(1/2) = pi/6. Option A is incorrect because sin(pi/3) = sqrt(3)/2, not 1/2. Option B is incorrect because sin(pi/4) = sqrt(2)/2, not 1/2. Option D is incorrect because sin(pi/2) = 1, not 1/2. Only pi/6 satisfies both conditions required by the definition of arcsin.

The arcsin function is defined to return exactly one output for each valid input. That output is the unique angle theta restricted to the principal range from -pi/2 to pi/2 satisfying sin(theta) = x. Option A is wrong because infinitely many angles share the same sine value. Option C is wrong because arcsin can return negative angles. Option D describes the range of arccos, not arcsin.

The domain of arcsin is the closed interval from -1 to 1. Only inputs within this interval produce a real output. x = -1, x = 0, and x = 1 all fall within this interval and are valid inputs. x = -3/2 is less than -1 and x = 5/4 is greater than 1, so both fall outside the domain and arcsin is undefined for those values.

The answer is True. By definition, arcsin(x) returns the angle theta in the interval from -pi/2 to pi/2 such that sin(theta) = x. Applying sin to that angle immediately recovers x. This composition works for every x in the domain of arcsin, which is the closed interval from -1 to 1. The result holds without exception throughout that entire domain.