Inverse Trigonometric Functions

The inverse trigonometric functions arcsin, arccos and arctan "undo" the standard trig functions. Since sin, cos and tan are not one-to-one, we must restrict their domains to define proper inverses. This topic is tested at A-Level (9MA0).

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Why We Need Restricted Domains

sin(x) = 0.5 has infinitely many solutions: x = pi/6, 5pi/6, pi/6 + 2pi, ...

To define a unique inverse, we restrict each trig function to a domain where it is one-to-one (passes the horizontal line test).

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The Three Inverse Functions

arcsin (or sin^(-1))

Property	Value
Domain	-1 <= x <= 1
Range (principal values)	-pi/2 <= y <= pi/2
arcsin(0)	0
arcsin(1/2)	pi/6
arcsin(1)	pi/2
arcsin(-1)	-pi/2

arccos (or cos^(-1))

Property	Value
Domain	-1 <= x <= 1
Range (principal values)	0 <= y <= pi
arccos(1)	0
arccos(1/2)	pi/3
arccos(0)	pi/2
arccos(-1)	pi

arctan (or tan^(-1))

Property	Value
Domain	All real numbers
Range (principal values)	-pi/2 < y < pi/2
arctan(0)	0
arctan(1)	pi/4
arctan(-1)	-pi/4
As x tends to +infinity	y tends to pi/2
As x tends to -infinity	y tends to -pi/2

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Key Relationships

• sin(arcsin(x)) = x for -1 <= x <= 1
• arcsin(sin(x)) = x for -pi/2 <= x <= pi/2

Similar for cos/arccos and tan/arctan, within their principal ranges.

Worked Example 1

Find the exact value of arcsin(sqrt(3)/2).

We need the angle theta in [-pi/2, pi/2] such that sin(theta) = sqrt(3)/2.

theta = pi/3

Worked Example 2

Find arccos(-sqrt(2)/2).

We need theta in [0, pi] such that cos(theta) = -sqrt(2)/2.

cos(3pi/4) = -sqrt(2)/2, so arccos(-sqrt(2)/2) = 3pi/4

Worked Example 3

Find arctan(-sqrt(3)).

We need theta in (-pi/2, pi/2) such that tan(theta) = -sqrt(3).

tan(-pi/3) = -sqrt(3), so arctan(-sqrt(3)) = -pi/3

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Graphs of Inverse Trig Functions

Each inverse trig graph is obtained by reflecting the restricted trig function in the line y = x.

y = arcsin(x)

• Domain: [-1, 1]
• Range: [-pi/2, pi/2]
• Passes through (0, 0)
• Increasing function

y = arccos(x)

• Domain: [-1, 1]
• Range: [0, pi]
• Passes through (0, pi/2)
• Decreasing function

y = arctan(x)

• Domain: all reals
• Range: (-pi/2, pi/2)
• Passes through (0, 0)
• Horizontal asymptotes at y = pi/2 and y = -pi/2
• Increasing function

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Important Identity

arcsin(x) + arccos(x) = pi/2 for all x in [-1, 1]

Worked Example 4

If arcsin(x) = pi/5, find arccos(x).

arccos(x) = pi/2 - pi/5 = 5pi/10 - 2pi/10 = 3pi/10

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Using Inverse Trig to Solve Equations

Worked Example 5

Solve 2sin(x) - 1 = 0 for -pi <= x <= pi.

sin(x) = 1/2

Principal value: x = arcsin(1/2) = pi/6

Second solution in [-pi, pi]: x = pi - pi/6 = 5pi/6

Solutions: x = pi/6 and x = 5pi/6

Worked Example 6

Solve tan(2x) = 3 for 0 <= x <= pi.

2x = arctan(3) + n x pi for integer n

arctan(3) ≈ 1.2490

So 2x = 1.2490, 1.2490 + pi, 1.2490 + 2pi, ...

2x = 1.2490 or 4.3906

x = 0.6245 or 2.1953

Check: both are in [0, pi].

Solutions: x ≈ 0.625 and x ≈ 2.195 (to 3 d.p.)

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Exam Tips

Tip	Detail
Principal values	Always state answers within the correct principal range
Notation	sin^(-1) and arcsin mean the same thing; use whichever the question uses
Exact values	Know arcsin, arccos, arctan for standard angles (0, 1/2, sqrt(2)/2, sqrt(3)/2, 1)
Graphs	Be able to sketch all three inverse functions with correct domains and ranges
Identity	arcsin(x) + arccos(x) = pi/2 is very useful for simplifying

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Practice Problems

1. Find the exact value of arccos(1/2). [Answer: pi/3]
2. Find arctan(1/sqrt(3)). [Answer: pi/6]
3. If arccos(x) = pi/4, find x. [Answer: sqrt(2)/2]
4. Solve cos(x) = -0.6 for 0 <= x <= 2pi. [Answer: x ≈ 2.214, 4.069]
5. Prove that arcsin(x) + arccos(x) = pi/2 for 0 <= x <= 1.

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Summary

• arcsin, arccos, arctan are inverse functions defined on restricted domains.
• arcsin: [-1,1] to [-pi/2, pi/2]; arccos: [-1,1] to [0, pi]; arctan: R to (-pi/2, pi/2).
• arcsin(x) + arccos(x) = pi/2 for all valid x.
• Use inverse functions to find principal values, then find other solutions from graph symmetry.

A-Level Deep Dive: Inverse Trigonometric Functions

Spec mapping

Edexcel 9MA0 Pure Mathematics specification, section 5 (Trigonometry — inverse functions) covers the definitions of arcsin, arccos and arctan; their relationships to sin, cos and tan; understanding of their graphs; their domains and ranges (refer to the official specification document for exact wording). This sub-strand sits inside Paper 2 — Pure Mathematics but is examined synoptically across both Paper 1 and Paper 2. The principal-value conventions are non-negotiable: arcsin: domain $[-1, 1]$[−1,1], range $[-\pi/2, \pi/2]$[−π/2,π/2]; arccos: domain $[-1, 1]$[−1,1], range $[0, \pi]$[0,π]; arctan: domain $\mathbb{R}$R, range $(-\pi/2, \pi/2)$(−π/2,π/2). The Edexcel formula booklet gives compound-angle formulae but does not list these domains and ranges — they must be memorised.

Synoptic reach is significant: differentiation of inverse trig functions ($\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$dxd​arcsinx=1−x2​1​, $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$dxd​arctanx=1+x21​) sits in section 9 — Differentiation (Year 2 Pure); integration giving inverse trig (e.g. $\int \frac{1}{1 + x^2}\,dx = \arctan x + C$∫1+x21​dx=arctanx+C, $\int \frac{1}{\sqrt{1 - x^2}}\,dx = \arcsin x + C$∫1−x2​1​dx=arcsinx+C) sits in section 8 — Integration; the underlying notion of one-to-one restriction so that an inverse exists is section 2 — Functions (composite and inverse functions); and the use of inverse trig to obtain principal solutions of trig equations sits in section 5 — Trigonometric equations. A single 9MA0 question can legitimately draw on three or four of these strands.

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Worked example with full mark scheme

Question (8 marks): Solve, for $x \in \mathbb{R}$x∈R, the equation

$\arctan(2x) + \arctan(x) = \dfrac{\pi}{4}$arctan(2x)+arctan(x)=4π​

giving $x$x in exact form.

Solution with mark scheme:

Step 1 — apply tan to both sides.

$\tan\big(\arctan(2x) + \arctan(x)\big) = \tan\dfrac{\pi}{4} = 1$tan(arctan(2x)+arctan(x))=tan4π​=1

M1 (AO1.1a) — recognising that \arctan\ on its own is hard to manipulate but $\tan(\arctan u) = u$tan(arctanu)=u, so applying \tan\ to both sides converts the equation to algebraic form. Common slip: writing $\arctan(2x) + \arctan(x) = \arctan(3x)$arctan(2x)+arctan(x)=arctan(3x) — this is wrong; arctan is not additive.

Step 2 — apply the addition formula $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$tan(A+B)=1−tanAtanBtanA+tanB​.

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

$\tan(A + B) = \dfrac{2x + x}{1 - (2x)(x)} = \dfrac{3x}{1 - 2x^2}$tan(A+B)=1−(2x)(x)2x+x​=1−2x23x​

M1 (AO1.1b) — correct application of the compound-angle formula with the right substitutions.

A1 (AO1.1b) — correct simplified expression $\dfrac{3x}{1 - 2x^2}$1−2x23x​.

Step 3 — set the expression equal to 1 and form the algebraic equation.

$\dfrac{3x}{1 - 2x^2} = 1$1−2x23x​=1 $3x = 1 - 2x^2$3x=1−2x2 $2x^2 + 3x - 1 = 0$2x2+3x−1=0

M1 (AO1.1b) — clearing the fraction (valid because $1 - 2x^2 \neq 0$1−2x2=0 at the solution; this needs checking later).

A1 (AO2.1) — correct quadratic in standard form.

Step 4 — solve the quadratic using the formula.

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

M1 (AO1.1b) — correct use of the quadratic formula with discriminant $9 + 8 = 17$9+8=17.

Step 5 — reject extraneous solutions.

The two candidate values are $x_1 = \dfrac{-3 + \sqrt{17}}{4} \approx 0.281$x1​=4−3+17​​≈0.281 and $x_2 = \dfrac{-3 - \sqrt{17}}{4} \approx -1.781$x2​=4−3−17​​≈−1.781.

Check the principal-value constraint. The left-hand side $\arctan(2x) + \arctan(x)$arctan(2x)+arctan(x) has range $(-\pi, \pi)$(−π,π), but applying \tan\ folds the answer back modulo $\pi$π. We need the original sum to equal exactly $\pi/4$π/4, not $\pi/4 + \pi = 5\pi/4$π/4+π=5π/4.

For $x_1 \approx 0.281$x1​≈0.281: $\arctan(0.562) + \arctan(0.281) \approx 0.512 + 0.274 \approx 0.786 \approx \pi/4$arctan(0.562)+arctan(0.281)≈0.512+0.274≈0.786≈π/4. Valid.

For $x_2 \approx -1.781$x2​≈−1.781: $\arctan(-3.562) + \arctan(-1.781) \approx -1.297 + -1.058 \approx -2.355 \approx -3\pi/4$arctan(−3.562)+arctan(−1.781)≈−1.297+−1.058≈−2.355≈−3π/4, not $\pi/4$π/4. Reject.

A1 (AO2.4) — final answer $x = \dfrac{-3 + \sqrt{17}}{4}$x=4−3+17​​, with explicit rejection of the other root justified by checking the principal-value range.

Total: 8 marks (M4 A4). The synoptic difficulty is the AO2.4 mark for the rejection — a candidate who writes both roots without checking will lose this mark even if both are mathematically generated by valid algebra.

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks):

(a) Show that $\arcsin x + \arccos x = \dfrac{\pi}{2}$arcsinx+arccosx=2π​ for all $x \in [-1, 1]$x∈[−1,1]. (3)

(b) Hence, or otherwise, solve $2 \arcsin x = \arccos x$2arcsinx=arccosx for $x \in [-1, 1]$x∈[−1,1], giving $x$x exactly. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO2.1) — let $\theta = \arcsin x$θ=arcsinx so that $\sin\theta = x$sinθ=x with $\theta \in [-\pi/2, \pi/2]$θ∈[−π/2,π/2].
• M1 (AO1.1b) — using the co-function identity $\cos(\pi/2 - \theta) = \sin\theta = x$cos(π/2−θ)=sinθ=x, with $\pi/2 - \theta \in [0, \pi]$π/2−θ∈[0,π] (the arccos range).
• A1 (AO2.4) — concluding $\arccos x = \pi/2 - \theta = \pi/2 - \arcsin x$arccosx=π/2−θ=π/2−arcsinx, hence $\arcsin x + \arccos x = \pi/2$arcsinx+arccosx=π/2, with the range justification stated.

(b)

• M1 (AO3.1a) — substituting $\arccos x = \pi/2 - \arcsin x$arccosx=π/2−arcsinx from (a) to get $2\arcsin x = \pi/2 - \arcsin x$2arcsinx=π/2−arcsinx.
• M1 (AO1.1b) — solving $3 \arcsin x = \pi/2$3arcsinx=π/2, giving $\arcsin x = \pi/6$arcsinx=π/6.
• A1 (AO1.1b) — $x = \sin(\pi/6) = \dfrac{1}{2}$x=sin(π/6)=21​.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Edexcel uses inverse-trig identities of this kind as small AO3 problem-solving moments — the "Hence" command tells the candidate that part (a) is the lever that unlocks part (b).

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Synoptic links

Connects to:

1. Section 5 — Compound-angle (addition) formulae: the identity $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$tan(A+B)=1−tanAtanBtanA+tanB​ is the engine that converts equations like $\arctan(2x) + \arctan(x) = \pi/4$arctan(2x)+arctan(x)=π/4 into algebraic ones. Without confident addition-formula manipulation, no inverse-trig equation involving sums or differences of arctans can be tackled.

2. Section 8 — Integration giving arctan/arcsin: $\int \dfrac{1}{1 + x^2}\,dx = \arctan x + C$∫1+x21​dx=arctanx+C and $\int \dfrac{1}{\sqrt{1 - x^2}}\,dx = \arcsin x + C$∫1−x2​1​dx=arcsinx+C are standard results in the Edexcel formula booklet. Recognising integrands as derivatives of inverse trig functions is a Year 2 Pure skill.

3. Section 9 — Differentiation of inverse trig: $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^2}}$dxd​arcsinx=1−x2​1​, $\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^2}$dxd​arctanx=1+x21​, and $\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1 - x^2}}$dxd​arccosx=−1−x2​1​. Derivable by implicit differentiation of $\sin y = x$siny=x (Year 2).

4. Section 2 — Functions (one-to-one restriction): sin, cos and tan are not one-to-one on their full domains, so they have no inverse on those domains. To define arcsin, arccos, arctan we restrict the domain of the original function to a maximal interval on which it is monotonic. This is the prototype example of "restrict to invert" — a recurring theme across A-Level functions.

5. Section 5 — Trigonometric equations: when solving $\sin x = k$sinx=k for general $x$x, the principal value $\arcsin k$arcsink gives one solution; the others come from the symmetries $\sin(\pi - \theta) = \sin\theta$sin(π−θ)=sinθ and $\sin(\theta + 2\pi) = \sin\theta$sin(θ+2π)=sinθ. Inverse functions and general solutions are inseparable.

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Mark-scheme literacy

Inverse-trig questions on 9MA0 split AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Stating principal-value ranges, applying compound-angle formulae, manipulating algebraic forms
AO2 (reasoning / interpretation)	25–35%	Justifying rejection of extraneous roots, using range constraints to validate solutions, deriving identities like $\arcsin x + \arccos x = \pi/2$arcsinx+arccosx=π/2
AO3 (problem-solving)	5–15%	Selecting an appropriate strategy (apply tan to both sides? use the addition formula? use a co-function identity?) when the question doesn't dictate the method

Examiner-rewarded phrasing: "by definition, arcsin gives the value in $[-\pi/2, \pi/2]$[−π/2,π/2]"; "applying tan to both sides, valid because tan is defined at every value in the principal range $(-\pi/2, \pi/2)$(−π/2,π/2)"; "since $\arctan(2x) + \arctan(x)$arctan(2x)+arctan(x) must lie in $(-\pi, \pi)$(−π,π), we reject the candidate root that places the sum outside $\pi/4$π/4 modulo $\pi$π"; "by the co-function identity, $\arccos x = \pi/2 - \arcsin x$arccosx=π/2−arcsinx for $x \in [-1, 1]$x∈[−1,1]".

Phrases that lose marks: writing "arcsin$(\sqrt{2}) = \ldots$(2​)=…" without flagging that $\sqrt{2} \notin [-1, 1]$2​∈/[−1,1] (the answer is undefined); accepting both roots of a quadratic without checking each against the original principal-value constraint; quoting \arcsin\ and \sin^{-1}\ inconsistently within one solution; giving "$\arcsin x = 30$arcsinx=30" when the formula booklet works in radians and the question is in radians.

A specific Edexcel pattern to watch: when a question writes "$\sin^{-1} x$sin−1x" rather than "$\arcsin x$arcsinx", the meaning is identical, but candidates sometimes confuse $\sin^{-1} x$sin−1x with $\dfrac{1}{\sin x} = \csc x$sinx1​=cscx. Read the symbol in context — an exponent of $-1$−1 on a function name almost always means the inverse function, not the reciprocal.

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Grade-band model answers

3-mark question

Question: Find the exact value of $\arcsin\!\left(\dfrac{\sqrt{3}}{2}\right)$arcsin(23​​), giving your answer in radians.

Grade C response (~210 words):

We want the angle whose sine is $\dfrac{\sqrt{3}}{2}$23​​. From the standard exact-value table, $\sin(\pi/3) = \dfrac{\sqrt{3}}{2}$sin(π/3)=23​​, so $\arcsin(\sqrt{3}/2) = \pi/3$arcsin(3​/2)=π/3.

The answer is $\pi/3$π/3 radians.