Justify a principal inverse before using it

Try

Draw the horizontal line y=1/2 through the unit circle. It meets the circle twice, at angles π/6 and 5π/6 in one revolution. Adding full turns produces infinitely many angle inputs with the same sine. An inverse function cannot return all of them as a single output, so a convention must choose a one-to-one portion of the original function.

The notation arcsin, or the calculator key sin⁻¹, means this inverse function. It does not mean 1/sin, which is the reciprocal called cosecant. Keep input and output roles visible: sine accepts an angle; arcsine accepts a coordinate value.

Predict

Which half of the circle contains exactly one point at each vertical coordinate from −1 to 1? Use the right semicircle, including its top and bottom. Predict whether vertical coordinate increases or decreases as its angle moves from −π/2 to π/2. Make the corresponding prediction for horizontal coordinate along the upper semicircle.

Experiment

Trace the chosen arcs and record their coordinate changes. The table is a construction guide, not a replacement for the monotonicity argument.

Function restricted to its chosen interval	Coordinate progression	Inverse output interval
Sine, [−π/2,π/2]	Strictly rises from −1 to 1	arcsin returns [−π/2,π/2]
Cosine, [0,π]	Strictly falls from 1 to −1	arccos returns [0,π]
Tangent, (−π/2,π/2)	Strictly rises through every real slope	arctan returns (−π/2,π/2)

Place 7π/6 on the circle. Its sine is −1/2, but the principal angle with that sine is −π/6. Thus arcsin(sin(7π/6))=−π/6. Circle endpoints agree in vertical coordinate, while the real angles differ. Explain why this is not a malfunction of the inverse key.

Understand

On the right semicircle, each horizontal level between −1 and 1 has exactly one point. Moving counterclockwise from the bottom to the top passes those levels in increasing order. Hence sine is strictly increasing there and takes every value in [−1,1]. Strict increase guarantees that distinct angles cannot share an output; geometric continuity guarantees no values are skipped. These facts justify an inverse, not merely a memorized range.

On the upper semicircle, each vertical line between horizontal coordinates −1 and 1 meets it exactly once. Moving counterclockwise from the right endpoint to the left decreases the horizontal coordinate strictly. Thus cosine is strictly decreasing on [0,π] and covers [−1,1]. An inverse can come from strict decrease just as well as strict increase.

For tangent, intersect the ray at angle x with the vertical line X=1. When −π/2<x<π/2, its height is sin x/cos x. Increasing x turns the ray upward, increasing the intersection height strictly through every real value. Conversely any real height on X=1 determines one such ray. The vertical endpoint rays never meet X=1 and have zero cosine, explaining the open interval. This also makes the unbounded tangent range visible without requiring calculus.

Consequently sin(arcsin y)=y for −1≤y≤1, but arcsin(sin x)=x only on the selected input interval. Similar qualifications apply to cosine. Tangent and arctangent reverse each other on their respective restricted domains. The symbols alone do not remove domain conditions.

To solve sin x=r when −1<r<1, let α=arcsin r. Reflection across the vertical axis gives the other angle π−α, and full turns give the two families α+2kπ and π−α+2kπ for integers k. At r=1 or −1 these families coincide, so avoid double counting. For cosine, β=arccos r yields ±β+2kπ; its endpoint targets ±1 likewise need duplicate handling. The choice of a principal angle is a starting point for all-solution reasoning, not its conclusion.

Construct the three inverse graphs

Restrict the parent before reflecting across y=x. A complete inverse graph is the reflection of the entire restricted curve, not straight segments arbitrarily joining a few landmarks.

Inverse	Domain and range	Exact reflected landmarks	Full direction and boundary
arcsin x	[−1,1] → [−π/2,π/2]	(−1,−π/2), (0,0), (1,π/2)	Increasing; both endpoints included
arccos x	[−1,1] → [0,π]	(−1,π), (0,π/2), (1,0)	Decreasing; both endpoints included
arctan x	R → (−π/2,π/2)	(−1,−π/4), (0,0), (1,π/4)	Increasing; limiting heights ±π/2 are never reached

The parent restrictions are respectively [−π/2,π/2] for sine, [0,π] for cosine, and (−π/2,π/2) for tangent. Reflecting tangent’s vertical asymptotes gives horizontal limiting levels for arctangent as x tends to positive or negative infinity. A calculator returning an angle strictly inside that range is consistent with the whole graph; assigning either boundary angle at a finite input is not. The registered construction task accepts a labeled graph or a full coordinate-and-curve description with domains, ranges, directions and endpoint conventions, so no essential numerical information depends on a picture alone.

Apply

Let β=arccos(1/5), leaving it symbolic. List all roots of cos x=1/5 in [0,4π]: β, 2π−β, 2π+β and 4π−β. Explain why these are complete by counting the two intersections per revolution. Neither interval endpoint has the target cosine, so neither belongs in the list.

Continue

Use inverse-trigonometry practice when the branch conventions need repetition. Next, turn angle solutions into times, filter a physical observation interval and interpret threshold inequalities. A calculator supplies a principal value; the complete answer still needs mathematical reasoning.