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.
How arcsin is built
- Sine 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).
IB-style question — exact inverse values
Find the exact value of arcsin(1/2), arccos(−1/2) and arctan(√3).
Step by step
- arcsin(1/2): which angle in [−π/2, π/2] has sine 1/2?
- arccos(−1/2): which angle in [0, π] has cosine −1/2? Cosine is negative in Q2.
- arctan(√3): which angle in (−π/2, π/2) has tangent √3?
Final answer
arcsin(1/2) = π/6, arccos(−1/2) = 2π/3, arctan(√3) = π/3.
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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).
The three reflections
- 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: Each graph has a shape, key points and a width (the x-values it lives on).
| Graph | Shape | Passes through | Width |
|---|---|---|---|
| y = arcsin x | increasing S-shaped curve | (−1, −π/2), (0, 0), (1, π/2) | only between x = −1 and x = 1 |
| y = arccos x | decreasing curve: same width, but it falls instead of rises | (−1, π), (0, π/2), (1, 0) | between x = −1 and x = 1 |
| y = arctan x | increasing S through the origin; flattens towards the horizontal asymptotes y = π/2 and y = −π/2, but never touches them | (0, 0) | 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, and a handy picture: cos(arcsin x) = √(1 − x²). Picture a right triangle with opposite = x, hypotenuse = 1, so the adjacent side = √(1 − x²).
IB-style question — compose to a number
Find the exact value of tan(arcsin(3/5)).
Step by step
- Let θ = arcsin(3/5), so sin θ = 3/5 with θ in [−π/2, π/2] (here Q1).
- Find cos θ using sin²+cos²=1 (positive, since θ is acute).
- Then tan θ = sin θ / cos θ.
Final answer
tan(arcsin(3/5)) = 3/4 (the 3–4–5 triangle).