IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read

Inverse Trigonometric Functions

arcsin, arccos, and arctan undo sin, cos, and tan — but only on a restricted domain, because the trig functions aren’t one-to-one. Master the restricted ranges and you’ll handle every “find the exact value” or “state the range” question without falling into the classic traps.

📘 What you need to know

Notation: arcsin x = sin⁻¹ x; arccos x = cos⁻¹ x; arctan x = tan⁻¹ x.
arcsin x: domain [−1, 1]; range [−π/2, π/2].
arccos x: domain [−1, 1]; range [0, π].
arctan x: domain ℝ; range (−π/2, π/2). Horizontal asymptotes at y = ±π/2.
Always returns a value in the restricted range — even if the original angle was outside it.
arcsin(sin θ) ≠ θ in general — only when θ is already in [−π/2, π/2].
Composition trick: cos(arcsin x) and sin(arccos x) can be found from a right triangle plus the Pythagorean identity.

Why restricted domains?

The trig functions repeat — sin x = 1/2 has infinitely many solutions, so “the” inverse can’t pick all of them. The restriction takes one slice where each y-value appears just once, then inverts that slice.

Function	Restricted domain (input to sin/cos/tan)	Range of inverse (output of arc-)
arcsin	[−π/2, π/2]	[−π/2, π/2]
arccos	[0, π]	[0, π]
arctan	(−π/2, π/2)	(−π/2, π/2)

Visual: each inverse graph is the original trig graph (cut to its restricted domain) reflected in the line y = x.

The three graphs

y = arcsin x

domain [−1, 1]
range [−π/2, π/2]

passes through (0, 0); endpoints (±1, ±π/2)

y = arccos x

domain [−1, 1]
range [0, π]

passes through (0, π/2); endpoints (1, 0) and (−1, π)

y = arctan x domain ℝ range (−π/2, π/2) horizontal asymptotes y = ±π/2

arctan is the only one with an unrestricted domain — you can take arctan of any real number. The others require −1 ≤ x ≤ 1, because that’s the range of sin and cos.

When the angle is outside the range

arcsin only gives values in [−π/2, π/2]. So if sin θ = 1/2 with θ = 5π/6, then arcsin(1/2) = π/6 — not 5π/6.

Use trig symmetry first sin(5π/6) = sin(π/6) → arcsin(sin(5π/6)) = π/6

The trick is to rewrite the trig value using a symmetry (sin θ = sin(π − θ), etc.) so the angle inside sin/cos/tan ends up in the restricted range, then take the inverse.

🧭 Recipe — composition like cos(arcsin x) or sin(arccos x)

Let θ = the inner inverse function. So sin θ = x (or cos θ = x).
State the range of θ from the restricted domain: arcsin → [−π/2, π/2]; arccos → [0, π].
Use the Pythagorean identity: cos²θ = 1 − sin²θ (or sin²θ = 1 − cos²θ).
Take the right sign based on the range from step 2.
Done — the answer is the value found in step 4.

Worked examples

WE 1

Find exact values of inverse trig

Without using a calculator, find the exact values of (a) arcsin(√3/2) and (b) arctan(−1).

(a) Find the angle in [−π/2, π/2] whose sin is √3/2 sin(π/3) = √3/2; π/3 ∈ [−π/2, π/2] ✓ (a) arcsin(√3/2) = π3 (b) Find the angle in (−π/2, π/2) whose tan is −1 tan(−π/4) = −1; −π/4 ∈ (−π/2, π/2) ✓ (b) arctan(−1) = −π4 arctan of a negative number gives a negative angle

WE 2

Find arccos of a negative number

Without using a calculator, find the exact value of arccos(−1/2).

Step 1: Need the angle in [0, π] with cos = −1/2 cos(π/3) = 1/2 → cos(π − π/3) = −1/2 cos(2π/3) = −1/2; 2π/3 ∈ [0, π] ✓ arccos(−1/2) = 2π3 arccos of negative → angle in (π/2, π], the “left half” of the unit circle

WE 3

Composition: cos(arcsin(3/5))

Find the exact value of cos(arcsin(3/5)).

Step 1: Let θ = arcsin(3/5) sin θ = 3/5, θ ∈ [−π/2, π/2] Step 2: Use cos²θ = 1 − sin²θ cos²θ = 1 − 9/25 = 16/25 cos θ = ±4/5 Step 3: θ in [−π/2, π/2] → cos θ ≥ 0 cos θ = 4/5 cos(arcsin(3/5)) = 45 classic 3-4-5 triangle ✓

WE 4

Show arcsin(sin(5π/6)) ≠ 5π/6

Find the exact value of arcsin(sin(5π/6)). Briefly explain why the answer is not 5π/6.

Step 1: 5π/6 is NOT in [−π/2, π/2], so we can’t simplify directly Step 2: Use sin(π − x) = sin x to rewrite sin(5π/6) = sin(π − π/6) = sin(π/6) = 1/2 Step 3: Now apply arcsin — π/6 IS in [−π/2, π/2] arcsin(1/2) = π/6 arcsin(sin(5π/6)) = π6 arcsin always returns a value in [−π/2, π/2]; 5π/6 is outside that range so it gets “folded back” to π/6

WE 5

Find the range of x given arcsin x = k

Given that x satisfies arcsin x = k where 0 < k < π/2, state the range of possible values of x.

Step 1: Rearrange — apply sin to both sides x = sin k Step 2: Sin is increasing on [0, π/2] k = 0 → x = sin 0 = 0 k = π/2 → x = sin(π/2) = 1 Step 3: Strict inequality preserved 0 < x < 1 since arcsin is increasing, the order is preserved → reading off the endpoints gives the range

WE 6

Solve arcsin(2x) = π/3

Solve the equation arcsin(2x) = π/3 for x. Give your answer in exact form.

Step 1: Apply sin to both sides 2x = sin(π/3) = √3/2 Step 2: Solve for x x = √3/4 x = √34 Step 3: Check 2x is in [−1, 1] 2x = √3/2 ≈ 0.866 ✓ π/3 is in the arcsin range [−π/2, π/2], so taking sin and back-substituting works directly

💡 Top tips

Always state the range when computing an inverse — it stops you accepting an out-of-range answer.
For composition like cos(arcsin x), set up a right triangle: opp = x, hyp = 1 → adj = √(1 − x²). Quick and clean.
arccos of negative gives an angle in (π/2, π]; arcsin of negative gives a negative angle in [−π/2, 0).
For arcsin(sin θ) when θ is outside [−π/2, π/2]: rewrite using sin(π − θ) = sin θ to fold it back into range.
Inverse trig is in radians by convention in IB papers — answers like π/6 are exact and preferred.

⚠ Common mistakes

Assuming arcsin(sin θ) = θ. True only when θ ∈ [−π/2, π/2]. Otherwise it folds back.
Confusing inverse with reciprocal. sin⁻¹x = arcsin x, NOT 1/sin x (that’s cosec x).
Wrong sign in composition. cos(arcsin x) is always non-negative because arcsin’s range puts θ in [−π/2, π/2].
Forgetting the domain check. arcsin(2x) needs 2x ∈ [−1, 1] — the domain of arcsin.
Using arccos range for arcsin or vice versa. Memorise: arcsin → [−π/2, π/2]; arccos → [0, π].

That closes Inverse & Reciprocal Trigonometric Functions. Next section: Trigonometric Proof & Equation Strategies. The how-to guide for choosing the right identity, structuring proofs, and tackling tricky exam-style trig equations from start to finish.

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