Principal Value Quiz: Principal Value and Restricted Domains

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 input values -1 and 1 correspond to actual attained angles at the boundary. This is why the range of arcsin is closed, unlike the open range of arctan where the boundary angles are never attained.

The answer is True. The sine function is odd, meaning sin(-x) = -sin(x), and the principal interval from -pi/2 to pi/2 is symmetric about 0. These two properties together ensure that the inverse inherits the odd symmetry. For any x in the domain of arcsin, arcsin(-x) = -arcsin(x). This can be verified by noting that if arcsin(x) = theta, then sin(theta) = x, and sin(-theta) = -x, so arcsin(-x) = -theta = -arcsin(x).

On the interval from 0 to pi, cosine is strictly decreasing from 1 to -1, making it one-to-one throughout. Every value in the output range from -1 to 1 is attained exactly once, which allows a true inverse to be defined. Option A is false because cosine is decreasing, not increasing, on this interval. Option C is false because cosine and sine are not equal across the entire interval. Option D is false because cosine is an even function, not an odd one.

An inverse function requires that every output corresponds to exactly one input. Since cosine is periodic with period 2pi, the same output value is produced by infinitely many different inputs. For example, cos(0) = 1 and cos(2pi) = 1 are two distinct inputs sharing the same output, which violates the one-to-one requirement. Without restricting the domain to an interval where this repetition does not occur, a well-defined inverse cannot exist.

A principal value is defined by choosing one specific interval where the trigonometric function is one-to-one. That restricted interval is selected deliberately so that each output of the original function appears exactly once, enabling a true inverse to be defined. The principal value is then the unique angle from that interval corresponding to a given input. Without this restriction, the periodic nature of trigonometric functions would make any inverse ambiguous and non-unique.

As theta approaches plus or minus pi/2, tan(theta) goes to infinity so those angles are never outputs of arctan, confirming A. sin(-pi/2) = -1 and sin(pi/2) = 1 are actual attained values confirming B. Since sin attains those boundary values, arcsin includes them in its range, confirming D. Option C is false — arctan does not include -pi/2 and pi/2, which is precisely why its range is open, not closed.

The answer is True. Sine is strictly increasing on the interval from -pi/2 to pi/2, which means no two distinct angles in this interval share the same sine value. For every x between -1 and 1, the strictly increasing nature of sine on this interval guarantees a unique angle theta that satisfies sin(theta) = x. This uniqueness is exactly what makes arcsin a well-defined function with a single output for each valid input.

The horizontal line test determines whether a function is one-to-one. If any horizontal line crosses the graph more than once, the function is not injective and cannot have a well-defined inverse. Since sine repeats every 2pi, infinitely many horizontal lines intersect its graph multiple times across the full real number line. Restricting the domain to from -pi/2 to pi/2 ensures every horizontal line crosses the graph at most once, making the inverse well-defined.

sin(pi/4) = sqrt(2)/2 and pi/4 is in the principal range of arcsin, confirming A. 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. Option D is false — arctan(pi/2) does not equal 1. arctan(1) = pi/4, not arctan(pi/2).

The principal interval for 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 through division by zero. Since arctan reverses tangent, it can never return an angle where tangent does not exist. Options A and B include endpoints that are never valid outputs. Option C is the principal interval of arccos.

On the interval from -pi/2 to pi/2, sin(x) is strictly increasing and passes through the origin, making it one-to-one on that interval. The interval is also symmetric about 0, which gives arcsin its odd function property. Option B covers the range of arccos, where sine increases then decreases. Options C and D do not center on the origin and do not produce the symmetric, increasing behavior required.

Sine and cosine only produce outputs between -1 and 1, so arcsin and arccos only accept inputs in that closed interval, confirming A and B. Tangent produces every real number as output, so arctan accepts any real number as input, confirming C. Option D is false — arcsin is undefined for inputs outside the interval from -1 to 1, so its domain is not all real numbers.

On the interval from 0 to pi, sine increases from 0 to 1 and then decreases back to 0. This means two different angles in that interval share the same sine value, so the function is not one-to-one there. An inverse requires the original function to be injective, and using from 0 to pi violates that requirement. The inverse would then fail to be a well-defined function because a single input could correspond to two different angles.

The answer is True. The entire purpose of restricting a trigonometric function to its principal interval is to ensure each input maps to exactly one output. Without this restriction, a periodic function like sine or cosine would produce infinitely many angles for any given input value. By selecting one specific interval where the function is one-to-one, the inverse is guaranteed to return a single, unambiguous angle for every valid input.

The principal range of arccos is the closed interval from 0 to pi. The endpoints are included because cos(0) = 1 and cos(pi) = -1, meaning the extreme values of the domain of arccos are actually attained at those boundary angles. This distinguishes arccos from arctan, whose range is open because tangent never attains the values at its boundary angles. arcsin also uses a closed interval for the same reason as arccos.

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

The answer is True. Strict monotonicity means tangent produces every real number exactly once on the open interval from -pi/2 to pi/2. Because no two different angles in this interval share the same tangent value, the inverse is uniquely defined. For every real number x, there is precisely one angle theta in the principal range satisfying tan(theta) = x, which is exactly what is needed for arctan to be a proper function.

Trigonometric functions are periodic, meaning they repeat the same output values infinitely many times. This prevents them from having a well-defined inverse on their full domain. By restricting the domain to a specific interval where each output occurs exactly once, the function becomes one-to-one and a true inverse can be defined. The principal value is the unique angle selected from that restricted interval for each valid input.

For an inverse function to be well-defined, every output of the original function must correspond to exactly one input. This property is called being one-to-one or injective. Trigonometric functions are periodic and repeat values across their full domain, so they fail this requirement unless restricted to an interval where each output value is produced exactly once. Continuity alone does not guarantee a unique inverse, and even symmetry actually prevents injectivity.

The answer is True. The tangent function is undefined at -pi/2 and pi/2 because cosine equals zero at those angles, causing division by zero. Since arctan reverses the tangent function, it can never return an angle where tangent is undefined. The values -pi/2 and pi/2 are therefore excluded from the range, making the principal range an open interval that approaches but never includes those boundary values.