Inverse Trigonometric Functions

This lesson covers the inverse trigonometric functions — arcsin (sin⁻¹), arccos (cos⁻¹), and arctan (tan⁻¹) — their definitions, domains, ranges, graphs, and their use in solving equations. Understanding these functions is essential for solving trigonometric equations and for integration involving inverse trig functions.

Why Do We Need Inverse Trig Functions?

If sin θ = 0.5, what is θ? You know that θ = 30° is one answer, but θ = 150° also works, as do θ = 390°, θ = −210°, and infinitely many others. The sine function is many-to-one, so it does not have a straightforward inverse unless we restrict its domain.

An inverse function requires the original function to be one-to-one. We achieve this by restricting the domain of sin, cos, and tan to a suitable interval.

Definitions and Restricted Domains

arcsin (sin⁻¹)

We restrict sin to the domain [−π/2, π/2], where it is one-to-one and covers its full range [−1, 1].

arcsin: [−1, 1] → [−π/2, π/2]

Domain of arcsin: [−1, 1]
Range of arcsin: [−π/2, π/2] (i.e., −90° to 90°)

arcsin x gives the unique angle θ in [−π/2, π/2] such that sin θ = x.

Example 1: arcsin(1/2) = π/6, because sin(π/6) = 1/2 and π/6 ∈ [−π/2, π/2].

Example 2: arcsin(−√3/2) = −π/3, because sin(−π/3) = −√3/2 and −π/3 ∈ [−π/2, π/2].

arccos (cos⁻¹)

We restrict cos to the domain [0, π], where it is one-to-one and covers its full range [−1, 1].

arccos: [−1, 1] → [0, π]

Domain of arccos: [−1, 1]
Range of arccos: [0, π] (i.e., 0° to 180°)

arccos x gives the unique angle θ in [0, π] such that cos θ = x.

Example 3: arccos(1/2) = π/3, because cos(π/3) = 1/2 and π/3 ∈ [0, π].

Example 4: arccos(−1) = π, because cos(π) = −1 and π ∈ [0, π].

arctan (tan⁻¹)

We restrict tan to the domain (−π/2, π/2), where it is one-to-one and covers its full range (−∞, ∞).

arctan: (−∞, ∞) → (−π/2, π/2)

Domain of arctan: (−∞, ∞) — all real numbers
Range of arctan: (−π/2, π/2) — note the open interval (the endpoints are excluded)

arctan x gives the unique angle θ in (−π/2, π/2) such that tan θ = x.

Example 5: arctan(1) = π/4, because tan(π/4) = 1 and π/4 ∈ (−π/2, π/2).

Example 6: arctan(−√3) = −π/3, because tan(−π/3) = −√3 and −π/3 ∈ (−π/2, π/2).

Graphs of Inverse Trig Functions

y = arcsin x

Domain: [−1, 1]; Range: [−π/2, π/2]
Passes through (0, 0), (1, π/2), (−1, −π/2)
Increasing function throughout
The graph is the reflection of y = sin x (restricted to [−π/2, π/2]) in the line y = x

y = arccos x

Domain: [−1, 1]; Range: [0, π]
Passes through (1, 0), (0, π/2), (−1, π)
Decreasing function throughout
The graph is the reflection of y = cos x (restricted to [0, π]) in the line y = x

y = arctan x

Domain: (−∞, ∞); Range: (−π/2, π/2)
Passes through (0, 0), (1, π/4), (−1, −π/4)
Increasing function throughout
Horizontal asymptotes at y = π/2 and y = −π/2
The graph is the reflection of y = tan x (restricted to (−π/2, π/2)) in the line y = x

Important Properties

Composition Rules

For x in the appropriate domain:

sin(arcsin x) = x     for −1 ≤ x ≤ 1
cos(arccos x) = x     for −1 ≤ x ≤ 1
tan(arctan x) = x     for all x

But be careful in the other direction:

arcsin(sin θ) = θ     only if −π/2 ≤ θ ≤ π/2
arccos(cos θ) = θ     only if 0 ≤ θ ≤ π
arctan(tan θ) = θ     only if −π/2 < θ < π/2

Example 7: arcsin(sin(5π/6)) ≠ 5π/6.

sin(5π/6) = sin(π/6) = 1/2, so arcsin(sin(5π/6)) = arcsin(1/2) = π/6.

Relationship Between arcsin and arccos

For any x ∈ [−1, 1]:

arcsin x + arccos x = π/2

This follows from the complementary relationship sin θ = cos(π/2 − θ).

Example 8: Verify for x = 1/2: arcsin(1/2) + arccos(1/2) = π/6 + π/3 = π/2. ✓

Solving Equations Using Inverse Trig Functions

The inverse trig functions give us the principal value — the first solution in the restricted range. To find all solutions in a given interval, we use the symmetry properties of sin, cos, and tan.

Example 9: Solve sin x = 0.6 for 0 ≤ x ≤ 2π.

The principal value is x = arcsin(0.6) = 0.6435... radians.

Since sin is positive in the first and second quadrants:

x₁ = 0.644 (3 d.p.)
x₂ = π − 0.6435... = 2.498 (3 d.p.)

Example 10: Solve cos 2x = −0.3 for 0 ≤ x ≤ π.

Let u = 2x, so 0 ≤ u ≤ 2π.

Principal value: u = arccos(−0.3) = 1.8755...

cos is negative in the second and third quadrants:

u₁ = 1.8755... = 1.876 (3 d.p.)
u₂ = 2π − 1.8755... = 4.408 (3 d.p.)

So x₁ = u₁/2 = 0.938 and x₂ = u₂/2 = 2.204 (3 d.p.)

Example 11: Solve tan(x − π/4) = 2 for 0 ≤ x ≤ 2π.

x − π/4 = arctan 2 + nπ
x = arctan 2 + π/4 + nπ
x = 1.1071... + 0.7854... + nπ
x = 1.893 + nπ

For n = 0: x = 1.893 For n = 1: x = 1.893 + π = 5.034

Solutions: x = 1.89 and x = 5.03 (3 s.f.)

Exact Values of Inverse Trig Functions

Expression	Value
arcsin(0)	0
arcsin(1/2)	π/6
arcsin(√2/2)	π/4
arcsin(√3/2)	π/3
arcsin(1)	π/2
arccos(0)	π/2
arccos(1/2)	π/3
arccos(√2/2)	π/4
arccos(√3/2)	π/6
arccos(1)	0
arctan(0)	0
arctan(1/√3)	π/6
arctan(1)	π/4
arctan(√3)	π/3

Summary

arcsin x has domain [−1, 1] and range [−π/2, π/2].
arccos x has domain [−1, 1] and range [0, π].
arctan x has domain (−∞, ∞) and range (−π/2, π/2).
arcsin x + arccos x = π/2 for all x ∈ [−1, 1].
Inverse trig functions give the principal value; use symmetry to find other solutions.
sin⁻¹ x means arcsin x — it does not mean 1/sin x (which is cosec x).

Exam Tip: When solving trigonometric equations, always use the inverse trig function to find the principal value first, then apply the CAST diagram or symmetry to find all solutions in the required interval. State your solutions clearly and check they lie within the given range. Remember that calculators give arcsin, arccos, and arctan — you must find additional solutions yourself.

A-Level Deep Dive: Inverse Trigonometric Functions

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section E (Trigonometry), Year 2 content covers the definitions of arcsin, arccos and arctan; their relationships to sine, cosine and tangent; understanding of their graphs; their ranges and domains (refer to the official specification document for exact wording). Inverse trig is foundational across Section G (Differentiation, where $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ ), Section H (Integration, where $\int \frac{1}{1 + x^2}\,dx = \arctan x + C$ ) and Section J (Numerical methods, where iterative root-finding for $\tan x = k$ requires the principal-value branch of $\arctan$ ). The AQA formula booklet does list the standard inverse-trig derivatives and integrals, but does not state the principal-value ranges — these must be memorised.

The defining ranges (principal values):

$\arcsin: [-1, 1] \to [-\pi/2, \pi/2]$
$\arccos: [-1, 1] \to [0, \pi]$
$\arctan: \mathbb{R} \to (-\pi/2, \pi/2)$

Each is the unique inverse of the corresponding trigonometric function restricted to a monotone branch. Without restriction, $\sin$ , $\cos$ , $\tan$ are not injective and so have no inverse; the chosen branches are conventional.

Worked example with full mark scheme

Question (8 marks): Solve $\arctan(2x) + \arctan(x) = \dfrac{\pi}{4}$ , giving all real solutions exactly.

Solution with mark scheme:

Step 1 — apply tangent to both sides.

$\tan\big(\arctan(2x) + \arctan(x)\big) = \tan\!\left(\dfrac{\pi}{4}\right) = 1$

M1 — recognising that the principal-value identity allows $\tan$ to be applied. Candidates who try to expand the LHS without applying $\tan$ first usually stall.

Step 2 — apply the compound-angle formula for tangent.

Let $A = \arctan(2x)$ and $B = \arctan(x)$ , so $\tan A = 2x$ and $\tan B = x$ . Then:

$\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B} = \dfrac{2x + x}{1 - (2x)(x)} = \dfrac{3x}{1 - 2x^2}$

M1 — correct compound-angle expansion. Sign error in the denominator (writing $1 + 2x^2$ ) is the most frequent slip.

A1 — correct simplified rational expression $\dfrac{3x}{1 - 2x^2}$ .

Step 3 — equate and form a quadratic.

$\dfrac{3x}{1 - 2x^2} = 1 \implies 3x = 1 - 2x^2 \implies 2x^2 + 3x - 1 = 0$

M1 — clearing the fraction (valid provided $1 - 2x^2 \neq 0$ , i.e. $x \neq \pm 1/\sqrt{2}$ — note this for the rejection step).

Step 4 — solve the quadratic.

By the quadratic formula:

$x = \dfrac{-3 \pm \sqrt{9 + 8}}{4} = \dfrac{-3 \pm \sqrt{17}}{4}$

M1 A1 — correct application of the formula with correct discriminant $17$ .

Step 5 — check for extraneous solutions.

Applying $\tan$ to both sides can introduce extraneous solutions because $\tan$ has period $\pi$ . The original equation requires $\arctan(2x) + \arctan(x) = \pi/4$ specifically — not $\pi/4 + n\pi$ .

For $x_1 = \dfrac{-3 + \sqrt{17}}{4} \approx 0.281$ : both $\arctan(2x_1)$ and $\arctan(x_1)$ are positive (since $x_1 > 0$ ) and lie in $(0, \pi/2)$ , so their sum lies in $(0, \pi)$ . Numerically, $\arctan(0.561) + \arctan(0.281) \approx 0.511 + 0.274 = 0.785 \approx \pi/4$ . Valid.

For $x_2 = \dfrac{-3 - \sqrt{17}}{4} \approx -1.781$ : both arctangents are negative and lie in $(-\pi/2, 0)$ , so their sum is negative and cannot equal $\pi/4$ . The "solution" arises from the period- $\pi$ ambiguity introduced in Step 1. Reject.

B1 — explicit rejection of $x_2$ with the principal-value justification.

A1 — final answer $x = \dfrac{-3 + \sqrt{17}}{4}$ .

Total: 8 marks (M5 A2 B1). The B1 for rejecting the extraneous root is the discriminator between a Grade A and a Grade A* response.

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): Let $f(x) = \arcsin x$ for $-1 \leq x \leq 1$ .

(a) Show that $\cos(\arcsin x) = \sqrt{1 - x^2}$ , justifying the choice of sign. (3)

(b) Hence find an exact value for $\sin\big(2\arcsin(\tfrac{3}{5})\big)$ . (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — letting $\theta = \arcsin x$ , so $\sin \theta = x$ and $\theta \in [-\pi/2, \pi/2]$ .
M1 (AO1.1b) — applying $\sin^2 \theta + \cos^2 \theta = 1$ to obtain $\cos^2 \theta = 1 - x^2$ , so $\cos \theta = \pm\sqrt{1 - x^2}$ .
A1 (AO2.4) — selecting the positive root with the explicit reason that $\theta \in [-\pi/2, \pi/2]$ implies $\cos \theta \geq 0$ .

(b)

M1 (AO1.1b) — applying the double-angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$ .
M1 (AO1.1b) — using part (a) with $x = 3/5$ : $\cos(\arcsin(3/5)) = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5$ .
A1 (AO1.1b) — combining: $\sin(2\arcsin(3/5)) = 2 \cdot (3/5) \cdot (4/5) = 24/25$ .

Total: 6 marks split AO1 = 5, AO2 = 1. AQA reserves the AO2 mark for the principal-value justification; many candidates write "since $\cos\theta\geq 0$ " without naming the range, and lose this mark.

Synoptic links

Connects to:

Section E — Compound-angle formulae: evaluating $\arctan A + \arctan B$ uses the identity $\arctan A + \arctan B = \arctan\!\left(\dfrac{A + B}{1 - AB}\right)$ (with a $\pm\pi$ correction when $AB > 1$ ). The correction term comes from the periodicity of tangent and is the principal source of "extraneous" arctangent solutions.
Section G — Differentiation of inverse functions: $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^2}}$ , $\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1 - x^2}}$ , $\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^2}$ . Each follows from implicit differentiation of $\sin y = x$ , $\cos y = x$ , $\tan y = x$ respectively, with the principal-value range fixing the sign on the radical.
Section H — Integration giving inverse trig: $\displaystyle\int \dfrac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\!\left(\dfrac{x}{a}\right) + C$ and $\displaystyle\int \dfrac{1}{a^2 + x^2}\,dx = \dfrac{1}{a}\arctan\!\left(\dfrac{x}{a}\right) + C$ . Recognising these "standard forms" early saves time on Paper 2 — candidates who attempt substitution without recognising the pattern often spend two minutes on a thirty-second integral.
Section B — Functions and inverse functions: an inverse exists only on a domain where the function is one-to-one. $\sin$ , $\cos$ and $\tan$ are restricted to monotone branches before inversion, exactly as a general function $f$ must be restricted before $f^{-1}$ is well-defined. Inverse trig is the canonical worked example of "domain restriction to enable inversion".
Section F — Exponentials and logarithms: the logarithm is the inverse of an exponential, with no need for a principal-value choice because $e^x$ is already injective on $\mathbb{R}$ . Comparing $\ln$ (no branches needed) with $\arcsin$ (branches required) is a fertile prompt for deeper understanding of when inversion is "easy" versus "delicate".

Mark-scheme literacy

Inverse-trig questions on AQA 7357 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Stating principal-value ranges, applying compound-angle and double-angle identities, differentiating and integrating standard forms
AO2 (reasoning / interpretation)	25–35%	Justifying sign choices using the principal-value range, rejecting extraneous solutions, recognising "standard form" integrals
AO3 (problem-solving)	5–15%	Combining inverse trig with implicit differentiation, optimisation via inverse trig, related-rate problems on a triangle

Inverse Trigonometric Functions

Inverse Trigonometric Functions

Why Do We Need Inverse Trig Functions?

Definitions and Restricted Domains

arcsin (sin⁻¹)

arccos (cos⁻¹)

arctan (tan⁻¹)

Graphs of Inverse Trig Functions

y = arcsin x

y = arccos x

y = arctan x

Important Properties

Composition Rules

Relationship Between arcsin and arccos

Solving Equations Using Inverse Trig Functions

Exact Values of Inverse Trig Functions

Summary

A-Level Deep Dive: Inverse Trigonometric Functions

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

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