Principal Value Equations Quiz: Using Principal Values to Solve Equations

Let theta = arccos(-0.2), which is the unique angle in the principal range from 0 to pi whose cosine equals -0.2. Applying cos to that angle immediately recovers -0.2 by definition. The composition cos(arccos(x)) always returns x for any x in the domain of arccos. Option A drops the negative sign without justification. Option C gives 1 minus 0.2, which has no basis in this composition. Option D negates the result of option C.

The answer is True. A strictly decreasing function is one-to-one, meaning no two different inputs produce the same output. On the interval from 0 to pi, cosine decreases from 1 to -1 without repeating any value. This one-to-one behavior guarantees that for every x in the domain from -1 to 1, there is exactly one angle in the interval from 0 to pi whose cosine equals x, making arccos a well-defined function with a unique output for each input.

The principal value for an equation of the form sin(theta) = x is found by applying arcsin directly. arcsin(0.8) returns the unique angle in the principal range from -pi/2 to pi/2 whose sine equals 0.8. Option A gives the supplementary angle, which is a valid solution to sin(theta) = 0.8 but is not the principal value because it lies outside the principal range. Option B uses arctan, which is unrelated. Option D negates the result, giving a negative angle whose sine equals -0.8.

arcsin always returns the principal value from the interval from -pi/2 to pi/2, regardless of the quadrant where the original angle was found. sin(pi/6) = 1/2 and pi/6 lies within the principal range, so arcsin(1/2) = pi/6. Options A and C both give 5pi/6, which lies in the second quadrant outside the principal range. Option B gives 11pi/6, which is also outside the principal range entirely.

arcsin(1/2) = pi/6 because sin(pi/6) = 1/2 and pi/6 is in the principal range, confirming A. arccos(-1) = pi because cos(pi) = -1 and pi is the endpoint of the principal range of arccos, confirming B. Option C gives 5pi/6, which lies outside the principal range of arcsin and is not a principal value. arctan(-1) = -pi/4 because tan(-pi/4) = -1 and -pi/4 is in the principal range, confirming D.

cos(2pi/3) = -1/2 and 2pi/3 lies within the principal range from 0 to pi, so arccos(-1/2) = 2pi/3. Option A gives pi/3, whose cosine is positive 1/2, not negative. Option B gives -pi/3, which lies outside the principal range of arccos since that range only includes non-negative angles. Option D gives pi/6, whose cosine is sqrt(3)/2, not -1/2. Only 2pi/3 satisfies both conditions of the definition.

Option A is false — arctan(0) = 0, not pi. The angle pi lies outside the open principal range of arctan entirely. cos(pi) = -1 and pi is in the principal range of arccos, confirming B. tan(pi/4) = 1 and pi/4 is in the principal range of arctan, confirming C. sin(pi/4) = sqrt(2)/2 and pi/4 is in the principal range of arcsin, confirming D.

The answer is True. arcsin returns the unique angle theta satisfying sin(theta) = x with theta in the closed interval from -pi/2 to pi/2. The endpoints are included because sin(-pi/2) = -1 and sin(pi/2) = 1, meaning the extreme values of the arcsin domain actually correspond to attained boundary angles. This closed range distinguishes arcsin from arctan, whose range is open because tangent is undefined at the boundary angles.

sin(pi/4) = sqrt(2)/2 and pi/4 is in the principal range of arcsin. cos(pi) = -1 and pi is the endpoint of the principal range of arccos. tan(pi/4) = 1 and pi/4 is in the principal range of arctan. sin(-pi/3) = -sqrt(3)/2 and -pi/3 is in the principal range of arcsin. 

tan(theta) = 1 occurs at theta = pi/4 plus any integer multiple of pi. The principal range of arctan is the open interval from -pi/2 to pi/2, and pi/4 lies within this range, making it the unique principal value. Options A, B, and D all lie outside the principal range and are therefore not valid outputs of arctan regardless of their tangent values.

The principal interval of arctan is the open interval from -pi/2 to pi/2. The endpoints are excluded because tan(-pi/2) and tan(pi/2) are undefined — cosine equals zero at those angles, making tangent undefined. Since arctan reverses the tangent function, it can never return an angle where tangent does not exist. Options A and B describe intervals used for other inverse functions. Option C incorrectly uses a closed interval with endpoints that arctan never attains.

tan(-pi/3) = -sqrt(3) and -pi/3 lies within the open principal range from -pi/2 to pi/2, so arctan(-sqrt(3)) = -pi/3. Option A gives -pi/6, whose tangent is -1/sqrt(3), not -sqrt(3). Option C gives -pi/4, whose tangent is -1, not -sqrt(3). Option D gives pi/3, which produces a positive tangent value. Only -pi/3 satisfies the definition of arctan applied to -sqrt(3).

Option A is false — arccos(cos(theta)) = theta only for theta in the closed interval from 0 to pi, not for all real theta. arcsin composed with sin recovers theta only when theta is in the principal range, confirming B. tan(arctan(x)) = x holds for every real x, confirming C. sin composed with arcsin recovers x on the full domain, confirming D.

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

Option A is false — arcsin never returns second quadrant angles, its range from -pi/2 to pi/2 does not extend beyond pi/2. The range of arccos is from 0 to pi covering first and second quadrant angles, confirming B. The range of arctan is the open interval from -pi/2 to pi/2 covering first and fourth quadrant angles, confirming C. The range of arcsin is from -pi/2 to pi/2 covering fourth and first quadrant angles, confirming D.

The answer is False. sin(5pi/6) = 1/2. arcsin then returns the unique angle in the principal range from -pi/2 to pi/2 whose sine equals 1/2, which is pi/6, not 5pi/6. The angle 5pi/6 lies in the second quadrant, outside the principal range of arcsin. arcsin always returns the principal value, not the original angle, when that angle lies outside the interval from -pi/2 to pi/2.

sin(pi/4) = sqrt(2)/2 and pi/4 lies within the principal range from -pi/2 to pi/2, so arcsin(sqrt(2)/2) = pi/4. Option A gives pi/3, whose sine is sqrt(3)/2, not sqrt(2)/2. Option B gives pi/2, whose sine is 1, not sqrt(2)/2. Option C gives pi/6, whose sine is 1/2, not sqrt(2)/2. Only pi/4 satisfies both the sine condition and the principal range requirement.

The principal range of arccos is the closed interval from 0 to pi. cos(5pi/6) = -sqrt(3)/2 and 5pi/6 lies within this range, so the principal value is 5pi/6. Option A gives pi/6, whose cosine is positive sqrt(3)/2, not negative. Options C and D both lie outside the principal range of arccos entirely and are therefore never valid outputs of the inverse cosine function.

The answer is True. The tangent function is odd and strictly increasing on its principal interval, and these properties carry over to its inverse. Since arctan(-x) = -arctan(x) for all x, a positive input always produces a positive output and a negative input always produces a negative output. arctan(0) = 0 confirms the function passes through the origin. The strictly increasing nature means no sign reversal occurs anywhere on the domain.

tan(-pi/4) = -1 and -pi/4 lies within the open principal range from -pi/2 to pi/2, so arctan(-1) = -pi/4. Option A gives pi/4, which corresponds to arctan(1), a positive input. Option B gives -pi/3, whose tangent is -sqrt(3), not -1. Option C gives -pi/6, whose tangent is -1/sqrt(3), not -1. This result also follows from the odd function property since arctan(1) = pi/4 implies arctan(-1) = -pi/4.