Inverse Trigonometric Functions

Builds on: aqa-alevel-maths-pure-1 / trigonometry — this lesson is the A2-level deep dive.

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.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers the inverse functions arcsin, arccos, arctan with restricted domains and ranges content of Section E, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

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Why Do We Need Inverse Trig Functions?

If $\sin\theta = 0.5$sinθ=0.5, what is $\theta$θ? You know that $\theta = 30°$θ=30° is one answer, but $\theta = 150°$θ=150° also works, as do $\theta = 390°$θ=390°, $\theta = -210°$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.

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Definitions and Restricted Domains

$\arcsin$arcsin ($\sin^{-1}$sin−1)

We restrict $\sin$sin to the domain $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$[−2π​,2π​], where it is one-to-one and covers its full range $[-1, 1]$[−1,1].

$$\arcsin\colon [-1, 1] \to \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$$arcsin:[−1,1]→[−2π​,2π​]

• Domain of arcsin: $[-1, 1]$[−1,1]
• Range of arcsin: $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$[−2π​,2π​] (i.e., $-90°$−90° to $90°$90°)

$\arcsin x$arcsinx gives the unique angle $\theta$θ in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$[−2π​,2π​] such that $\sin\theta = x$sinθ=x.

Example 1: $\arcsin(1/2) = \tfrac{\pi}{6}$arcsin(1/2)=6π​, because $\sin(\tfrac{\pi}{6}) = \tfrac{1}{2}$sin(6π​)=21​ and $\tfrac{\pi}{6} \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$6π​∈[−2π​,2π​].

Example 2: $\arcsin(-\tfrac{\sqrt{3}}{2}) = -\tfrac{\pi}{3}$arcsin(−23​​)=−3π​, because $\sin(-\tfrac{\pi}{3}) = -\tfrac{\sqrt{3}}{2}$sin(−3π​)=−23​​ and $-\tfrac{\pi}{3} \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$−3π​∈[−2π​,2π​].

$\arccos$arccos ($\cos^{-1}$cos−1)

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

$$\arccos\colon [-1, 1] \to [0, \pi]$$arccos:[−1,1]→[0,π]

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

$\arccos x$arccosx gives the unique angle $\theta$θ in $[0, \pi]$[0,π] such that $\cos\theta = x$cosθ=x.

Example 3: $\arccos(1/2) = \tfrac{\pi}{3}$arccos(1/2)=3π​, because $\cos(\tfrac{\pi}{3}) = \tfrac{1}{2}$cos(3π​)=21​ and $\tfrac{\pi}{3} \in [0, \pi]$3π​∈[0,π].

Example 4: $\arccos(-1) = \pi$arccos(−1)=π, because $\cos(\pi) = -1$cos(π)=−1 and $\pi \in [0, \pi]$π∈[0,π].

$\arctan$arctan ($\tan^{-1}$tan−1)

We restrict $\tan$tan to the domain $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$(−2π​,2π​), where it is one-to-one and covers its full range $(-\infty, \infty)$(−∞,∞).

$$\arctan\colon (-\infty, \infty) \to \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$$arctan:(−∞,∞)→(−2π​,2π​)

• Domain of arctan: $(-\infty, \infty)$(−∞,∞) — all real numbers
• Range of arctan: $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$(−2π​,2π​) — note the open interval (the endpoints are excluded)

$\arctan x$arctanx gives the unique angle $\theta$θ in $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$(−2π​,2π​) such that $\tan\theta = x$tanθ=x.

Example 5: $\arctan(1) = \tfrac{\pi}{4}$arctan(1)=4π​, because $\tan(\tfrac{\pi}{4}) = 1$tan(4π​)=1 and $\tfrac{\pi}{4} \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$4π​∈(−2π​,2π​).

Example 6: $\arctan(-\sqrt{3}) = -\tfrac{\pi}{3}$arctan(−3​)=−3π​, because $\tan(-\tfrac{\pi}{3}) = -\sqrt{3}$tan(−3π​)=−3​ and $-\tfrac{\pi}{3} \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$−3π​∈(−2π​,2π​).

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Graphs of Inverse Trig Functions

$y = \arcsin x$y=arcsinx

• Domain: $[-1, 1]$[−1,1]; Range: $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$[−2π​,2π​]
• Passes through $(0, 0)$(0,0), $(1, \tfrac{\pi}{2})$(1,2π​), $(-1, -\tfrac{\pi}{2})$(−1,−2π​)
• Increasing function throughout
• The graph is the reflection of $y = \sin x$y=sinx (restricted to $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$[−2π​,2π​]) in the line $y = x$y=x

$y = \arccos x$y=arccosx

• Domain: $[-1, 1]$[−1,1]; Range: $[0, \pi]$[0,π]
• Passes through $(1, 0)$(1,0), $(0, \tfrac{\pi}{2})$(0,2π​), $(-1, \pi)$(−1,π)
• Decreasing function throughout
• The graph is the reflection of $y = \cos x$y=cosx (restricted to $[0, \pi]$[0,π]) in the line $y = x$y=x

$y = \arctan x$y=arctanx

• Domain: $(-\infty, \infty)$(−∞,∞); Range: $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$(−2π​,2π​)
• Passes through $(0, 0)$(0,0), $(1, \tfrac{\pi}{4})$(1,4π​), $(-1, -\tfrac{\pi}{4})$(−1,−4π​)
• Increasing function throughout
• Horizontal asymptotes at $y = \tfrac{\pi}{2}$y=2π​ and $y = -\tfrac{\pi}{2}$y=−2π​
• The graph is the reflection of $y = \tan x$y=tanx (restricted to $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$(−2π​,2π​)) in the line $y = x$y=x

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Important Properties

Composition Rules

For x in the appropriate domain:

$$\begin{aligned} \sin(\arcsin x) &= x \quad \text{for } -1 \leq x \leq 1 \\ \cos(\arccos x) &= x \quad \text{for } -1 \leq x \leq 1 \\ \tan(\arctan x) &= x \quad \text{for all } x \end{aligned}$$sin(arcsinx)cos(arccosx)tan(arctanx)​=xfor −1≤x≤1=xfor −1≤x≤1=xfor all x​

But be careful in the other direction:

$$\begin{aligned} \arcsin(\sin\theta) &= \theta \quad \text{only if } -\pi/2 \leq \theta \leq \pi/2 \\ \arccos(\cos\theta) &= \theta \quad \text{only if } 0 \leq \theta \leq \pi \\ \arctan(\tan\theta) &= \theta \quad \text{only if } -\pi/2 < \theta < \pi/2 \end{aligned}$$arcsin(sinθ)arccos(cosθ)arctan(tanθ)​=θonly if −π/2≤θ≤π/2=θonly if 0≤θ≤π=θonly if −π/2<θ<π/2​

Example 7: $\arcsin(\sin(\tfrac{5\pi}{6})) \neq \tfrac{5\pi}{6}$arcsin(sin(65π​))=65π​.

$\sin(\tfrac{5\pi}{6}) = \sin(\tfrac{\pi}{6}) = \tfrac{1}{2}$sin(65π​)=sin(6π​)=21​, so $\arcsin(\sin(\tfrac{5\pi}{6})) = \arcsin(\tfrac{1}{2}) = \tfrac{\pi}{6}$arcsin(sin(65π​))=arcsin(21​)=6π​.

Relationship Between arcsin and arccos

For any $x \in [-1, 1]$x∈[−1,1]:

$$\arcsin x + \arccos x = \frac{\pi}{2}$$arcsinx+arccosx=2π​

This follows from the complementary relationship $\sin\theta = \cos(\tfrac{\pi}{2} - \theta)$sinθ=cos(2π​−θ).

Example 8: Verify for $x = \tfrac{1}{2}$x=21​: $\arcsin(\tfrac{1}{2}) + \arccos(\tfrac{1}{2}) = \tfrac{\pi}{6} + \tfrac{\pi}{3} = \tfrac{\pi}{2}$arcsin(21​)+arccos(21​)=6π​+3π​=2π​. ✓

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Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

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 $\sin x = 0.6$sinx=0.6 for $0 \leq x \leq 2\pi$0≤x≤2π.

The principal value is $x = \arcsin(0.6) = 0.6435\ldots$x=arcsin(0.6)=0.6435… radians.

Since sin is positive in the first and second quadrants:

$$\begin{aligned} x_{1} = 0.644 (3 d.p.) \\ x_{2} = \pi - 0.6435... = 2.498 (3 d.p.) \end{aligned}$$x1​=0.644(3d.p.)x2​=π−0.6435...=2.498(3d.p.)​

Example 10: Solve $\cos 2x = -0.3$cos2x=−0.3 for $0 \leq x \leq \pi$0≤x≤π.

Let $u = 2x$u=2x, so $0 \leq u \leq 2\pi$0≤u≤2π.

Principal value: $u = \arccos(-0.3) = 1.8755\ldots$u=arccos(−0.3)=1.8755…

$\cos$cos is negative in the second and third quadrants:

$$\begin{aligned} u_{1} = 1.8755... = 1.876 (3 d.p.) \\ u_{2} = 2 \pi - 1.8755... = 4.408 (3 d.p.) \end{aligned}$$u1​=1.8755...=1.876(3d.p.)u2​=2π−1.8755...=4.408(3d.p.)​

So $x_1 = u_1/2 = 0.938$x1​=u1​/2=0.938 and $x_2 = u_2/2 = 2.204$x2​=u2​/2=2.204 (3 d.p.)

Example 11: Solve $\tan(x - \tfrac{\pi}{4}) = 2$tan(x−4π​)=2 for $0 \leq x \leq 2\pi$0≤x≤2π.

$$\begin{aligned} x - \frac{\pi}{4} &= \arctan 2 + n\pi \\ x &= \arctan 2 + \frac{\pi}{4} + n\pi \\ x &= 1.1071\ldots + 0.7854\ldots + n\pi \\ x &= 1.893 + n\pi \end{aligned}$$x−4π​xxx​=arctan2+nπ=arctan2+4π​+nπ=1.1071…+0.7854…+nπ=1.893+nπ​

For $n = 0$n=0: $x = 1.893$x=1.893 For $n = 1$n=1: $x = 1.893 + \pi = 5.034$x=1.893+π=5.034

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

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Exact Values of Inverse Trig Functions

Expression	Value
$\arcsin(0)$arcsin(0)	$0$0
$\arcsin(\tfrac{1}{2})$arcsin(21​)	$\tfrac{\pi}{6}$6π​
$\arcsin(\tfrac{\sqrt{2}}{2})$arcsin(22​​)	$\tfrac{\pi}{4}$4π​
$\arcsin(\tfrac{\sqrt{3}}{2})$arcsin(23​​)	$\tfrac{\pi}{3}$3π​
$\arcsin(1)$arcsin(1)	$\tfrac{\pi}{2}$2π​
$\arccos(0)$arccos(0)	$\tfrac{\pi}{2}$2π​
$\arccos(\tfrac{1}{2})$arccos(21​)	$\tfrac{\pi}{3}$3π​
$\arccos(\tfrac{\sqrt{2}}{2})$arccos(22​​)	$\tfrac{\pi}{4}$4π​
$\arccos(\tfrac{\sqrt{3}}{2})$arccos(23​​)	$\tfrac{\pi}{6}$6π​
$\arccos(1)$arccos(1)	$0$0
$\arctan(0)$arctan(0)	$0$0
$\arctan(\tfrac{1}{\sqrt{3}})$arctan(3​1​)	$\tfrac{\pi}{6}$6π​
$\arctan(1)$arctan(1)	$\tfrac{\pi}{4}$4π​
$\arctan(\sqrt{3})$arctan(3​)	$\tfrac{\pi}{3}$3π​

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Synoptic Links

This topic connects to:

• Solving trig equations (aqa-alevel-maths-trigonometry-depth / solving-trigonometric-equations) — inverse trig delivers principal values from which all solutions follow.
• Integration giving arctan (aqa-alevel-maths-calculus-applications / integration-by-substitution) — $\int \tfrac{1}{1+x^2}\, dx = \arctan x + C$∫1+x21​dx=arctanx+C.
• Functions and inverses (aqa-alevel-maths-pure-2 / functions) — restricted domains are required because $\sin$sin, $\cos$cos, $\tan$tan are not one-to-one.

Summary

• $\arcsin x$arcsinx has domain $[-1, 1]$[−1,1] and range $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$[−2π​,2π​].
• $\arccos x$arccosx has domain $[-1, 1]$[−1,1] and range $[0, \pi]$[0,π].
• $\arctan x$arctanx has domain $(-\infty, \infty)$(−∞,∞) and range $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$(−2π​,2π​).
• $\arcsin x + \arccos x = \tfrac{\pi}{2}$arcsinx+arccosx=2π​ for all $x \in [-1, 1]$x∈[−1,1].
• Inverse trig functions give the principal value; use symmetry to find other solutions.
• $\sin^{-1} x$sin−1x means $\arcsin x$arcsinx — it does not mean $\tfrac{1}{\sin x}$sinx1​ (which is $\cosec x$cosecx).

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 $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$dxd​arcsinx=1−x2​1​), Section H (Integration, where $\int \frac{1}{1 + x^2}\,dx = \arctan x + C$∫1+x21​dx=arctanx+C) and Section J (Numerical methods, where iterative root-finding for $\tan x = k$tanx=k requires the principal-value branch of $\arctan$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):

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

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

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Worked example with full mark scheme

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

Solution with mark scheme:

Step 1 — apply tangent to both sides.

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

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

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

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

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

M1 — correct compound-angle expansion. Sign error in the denominator (writing $1 + 2x^2$1+2x2) is the most frequent slip.

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

Step 3 — equate and form a quadratic.

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

M1 — clearing the fraction (valid provided $1 - 2x^2 \neq 0$1−2x2=0, i.e. $x \neq \pm 1/\sqrt{2}$x=±1/2​ — note this for the rejection step).

Step 4 — solve the quadratic.

By the quadratic formula:

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

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

Step 5 — check for extraneous solutions.

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

For $x_1 = \dfrac{-3 + \sqrt{17}}{4} \approx 0.281$x1​=4−3+17​​≈0.281: both $\arctan(2x_1)$arctan(2x1​) and $\arctan(x_1)$arctan(x1​) are positive (since $x_1 > 0$x1​>0) and lie in $(0, \pi/2)$(0,π/2), so their sum lies in $(0, \pi)$(0,π). Numerically, $\arctan(0.561) + \arctan(0.281) \approx 0.511 + 0.274 = 0.785 \approx \pi/4$arctan(0.561)+arctan(0.281)≈0.511+0.274=0.785≈π/4. Valid.

For $x_2 = \dfrac{-3 - \sqrt{17}}{4} \approx -1.781$x2​=4−3−17​​≈−1.781: both arctangents are negative and lie in $(-\pi/2, 0)$(−π/2,0), so their sum is negative and cannot equal $\pi/4$π/4. The "solution" arises from the period-$\pi$π ambiguity introduced in Step 1. Reject.

B1 — explicit rejection of $x_2$x2​ with the principal-value justification.

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

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

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Specimen question modelled on the AQA 7357 Paper 2 format

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

(a) Show that $\cos(\arcsin x) = \sqrt{1 - x^2}$cos(arcsinx)=1−x2​, justifying the choice of sign. (3)

(b) Hence find an exact value for $\sin\big(2\arcsin(\tfrac{3}{5})\big)$sin(2arcsin(53​)). (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — letting $\theta = \arcsin x$θ=arcsinx, so $\sin \theta = x$sinθ=x and $\theta \in [-\pi/2, \pi/2]$θ∈[−π/2,π/2].
• M1 (AO1.1b) — applying $\sin^2 \theta + \cos^2 \theta = 1$sin2θ+cos2θ=1 to obtain $\cos^2 \theta = 1 - x^2$cos2θ=1−x2, so $\cos \theta = \pm\sqrt{1 - x^2}$cosθ=±1−x2​.
• A1 (AO2.4) — selecting the positive root with the explicit reason that $\theta \in [-\pi/2, \pi/2]$θ∈[−π/2,π/2] implies $\cos \theta \geq 0$cosθ≥0.

(b)

• M1 (AO1.1b) — applying the double-angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$sin(2θ)=2sinθcosθ.
• M1 (AO1.1b) — using part (a) with $x = 3/5$x=3/5: $\cos(\arcsin(3/5)) = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5$cos(arcsin(3/5))=1−9/25​=16/25​=4/5.
• A1 (AO1.1b) — combining: $\sin(2\arcsin(3/5)) = 2 \cdot (3/5) \cdot (4/5) = 24/25$sin(2arcsin(3/5))=2⋅(3/5)⋅(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 $\cos\theta\geq 0$cosθ≥0" without naming the range, and lose this mark.

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Synoptic links

Connects to:

1. Section E — Compound-angle formulae: evaluating $\arctan A + \arctan B$arctanA+arctanB uses the identity $\arctan A + \arctan B = \arctan\!\left(\dfrac{A + B}{1 - AB}\right)$arctanA+arctanB=arctan(1−ABA+B​) (with a $\pm\pi$±π correction when $AB > 1$AB>1). The correction term comes from the periodicity of tangent and is the principal source of "extraneous" arctangent solutions.

2. Section G — Differentiation of inverse functions: $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^2}}$dxd​arcsinx=1−x2​1​, $\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1 - x^2}}$dxd​arccosx=−1−x2​1​, $\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^2}$dxd​arctanx=1+x21​. Each follows from implicit differentiation of $\sin y = x$siny=x, $\cos y = x$cosy=x, $\tan y = x$tany=x respectively, with the principal-value range fixing the sign on the radical.

3. Section H — Integration giving inverse trig: $\displaystyle\int \dfrac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\!\left(\dfrac{x}{a}\right) + C$∫a2−x2​1​dx=arcsin(ax​)+C and $\displaystyle\int \dfrac{1}{a^2 + x^2}\,dx = \dfrac{1}{a}\arctan\!\left(\dfrac{x}{a}\right) + C$∫a2+x21​dx=a1​arctan(ax​)+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.

4. Section B — Functions and inverse functions: an inverse exists only on a domain where the function is one-to-one. $\sin$sin, $\cos$cos and $\tan$tan are restricted to monotone branches before inversion, exactly as a general function $f$f must be restricted before $f^{-1}$f−1 is well-defined. Inverse trig is the canonical worked example of "domain restriction to enable inversion".

5. Section F — Exponentials and logarithms: the logarithm is the inverse of an exponential, with no need for a principal-value choice because $e^x$ex is already injective on $\mathbb{R}$R. Comparing $\ln$ln (no branches needed) with $\arcsin$arcsin (branches required) is a fertile prompt for deeper understanding of when inversion is "easy" versus "delicate".

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Mark-scheme literacy

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