Inverse trig functions (Math AA HL)

An inverse must give ONE answer: sin x = 1/2 has infinitely many answers (π/6, 5π/6, π/6 + 2π, …). So sine has no inverse over all reals — it fails the one-input-one-output rule.

The fix: chop sine down to a stretch where it is one-to-one (only rises, never repeats). On that stretch we can run it backwards. arcsin x (also written sin⁻¹x) is the angle in that chosen stretch whose sine is x.

The three chosen ranges (memorise these): arcsin x: domain −1 ≤ x ≤ 1, range −π/2 ≤ y ≤ π/2 (right half: Q4→Q1).

arccos x: domain −1 ≤ x ≤ 1, range 0 ≤ y ≤ π (top half: Q1→Q2).

arctan x: domain all real x, range −π/2 < y < π/2 (open — never reaches ±π/2).

Each graph is the reflection in y = x: An inverse function's graph is the original reflected in the line y = x (swap the x and y axes).

So y = arcsin x is the rising piece of sine reflected over y = x: it climbs from (−1, −π/2) up to (1, π/2).

y = arccos x falls from (−1, π) down to (1, 0).

y = arctan x is an S-curve through the origin with two horizontal asymptotes y = ±π/2 (it never quite reaches them).

Reading the three graphs: y = arcsin x: an increasing S-shaped curve from (−1, −π/2) through (0, 0) up to (1, π/2). It lives only between x = −1 and x = 1.

y = arccos x: a decreasing curve from (−1, π) through (0, π/2) down to (1, 0). Same width, but it falls instead of rises.

y = arctan x: an increasing S through the origin that flattens towards the horizontal asymptotes y = π/2 and y = −π/2, but never touches them. Its width is the whole x-axis.

Composing: build a right triangle: To simplify something like cos(arcsin x), let θ = arcsin x. Then sin θ = x, and θ is in [−π/2, π/2] where cosine is never negative.

Use sin²θ + cos²θ = 1: cos θ = √(1 − x²). So cos(arcsin x) = √(1 − x²).

A handy picture: a right triangle with opposite = x, hypotenuse = 1, so the adjacent side = √(1 − x²).