What are the primary trigonometric functions covered in Edexcel International A Levels Pure Mathematics 3?

In Edexcel International A Levels Pure Mathematics 3, the primary trigonometric functions include sine (sin), cosine (cos), and tangent (tan). These functions are fundamental in understanding the relationships between the angles and sides of triangles, particularly in right-angled triangles.

How do you solve trigonometric equations in Pure Mathematics 3?

To solve trigonometric equations in Pure Mathematics 3, you typically isolate the trigonometric function and use identities or inverse functions to find the angle solutions. It's important to consider the periodic nature of trigonometric functions and provide solutions within the specified domain, often using the unit circle or reference angles.

What are trigonometric identities and how are they used in Edexcel International A Levels?

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables involved. In Edexcel International A Levels, identities such as Pythagorean identities, angle sum and difference identities, and double angle identities are used to simplify expressions and solve equations. They are essential tools for proving mathematical statements and solving complex trigonometric problems.

How is the unit circle used to understand trigonometric functions in Pure Mathematics 3?

The unit circle is a fundamental concept in Pure Mathematics 3 for understanding trigonometric functions. It is a circle with a radius of one centered at the origin of a coordinate plane. The unit circle helps visualize the values of sine, cosine, and tangent for different angles, and it illustrates the periodic nature of these functions. It is also used to derive the values of trigonometric functions for common angles.

What is the significance of the graphs of trigonometric functions in Edexcel International A Levels?

The graphs of trigonometric functions, such as sine, cosine, and tangent, are significant in Edexcel International A Levels as they provide a visual representation of these functions' behavior over different intervals. Understanding these graphs helps students analyze the amplitude, period, phase shift, and vertical shift of trigonometric functions, which are crucial for solving real-world problems involving periodic phenomena.

Why is "Writing $\sec\theta = 1/\sin\theta$" a common mistake in Trigonometric functions?

Why it happens: The names suggest secant connects with sine. How to avoid it: Mnemonic: 'co-co-co' — only the cosecant has 'co' (matches sine) and the cotangent ('co' matches tangent's reciprocal of cosine over sine). has no 'co' and pairs with .

Why is "Reading $\sin^{-1}(x)$ as $\dfrac{1}{\sin x}$" a common mistake in Trigonometric functions?

Why it happens: The -1 superscript usually denotes reciprocal in algebra. How to avoid it: On a calculator and in -style notation, ^-1(x) ALWAYS means the inverse function (i.e. x). For the reciprocal, write x or ( x)^-1. Source: WMA13/01 examiner reports — flagged annually

Why is "Stating $\arcsin(1/2) = 5\pi/6$ (a valid sine solution but outside the principal range)" a common mistake in Trigonometric functions?

Why it happens: Treating as 'any angle with = 1/2'. How to avoid it: Always quote the PRINCIPAL VALUE: x lies in [-/2, /2]. For (1/2) = /6 (NOT 5/6).

Why is "Quoting $1 + \tan^{2}\theta = \sec^{2}\theta$ without noting $\cos\theta \neq 0$" a common mistake in Trigonometric functions?

Why it happens: Forgetting that the identity has a domain restriction. How to avoid it: If the question's answer set could include = /2 etc., always state the exclusion. Mark schemes don't always penalise but quoting it shows mathematical maturity.

Why is "Drawing $y = \sec x$ passing through the origin" a common mistake in Trigonometric functions?

Why it happens: Confusing it with y = x. How to avoid it: 0 = 1, NOT 0. The minimum on the central interval is (0, 1). Plot (, -1) on the outer intervals. The graph never crosses the x-axis.

Edexcel International A Levels Mathematics (XMA01-YMA01)

Trigonometric functions

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Short Notes - Trigonometric functions

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Detailed Study Notes

Detailed notes on Pure Mathematics 3 for Edexcel International A Levels Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Trigonometric Functions Study Notes — Edexcel IAL Mathematics P3 (WMA13/01, 2018 spec — 2026 onwards)

Reciprocal trig functions $\sec$ , $\csc$ , $\cot$ — their definitions, graphs, asymptotes, and the Pythagorean identities $1 + \tan^{2}\theta = \sec^{2}\theta$ and $1 + \cot^{2}\theta = \csc^{2}\theta$ . Inverse trig functions $\arcsin$ , $\arccos$ , $\arctan$ with restricted principal-value ranges. Identities-based equation solving.

At a glance

$\sec\theta = 1/\cos\theta$ , $\csc\theta = 1/\sin\theta$ , $\cot\theta = \cos\theta/\sin\theta$ .
Pythagorean identities: $1 + \tan^{2}\theta = \sec^{2}\theta$ (divide by $\cos^{2}$ ), $1 + \cot^{2}\theta = \csc^{2}\theta$ (divide by $\sin^{2}$ ).
$\arcsin: [-1, 1] \to [-\pi/2, \pi/2]$ (closed range).
$\arccos: [-1, 1] \to [0, \pi]$ (closed range).
$\arctan: \mathbb{R} \to (-\pi/2, \pi/2)$ (open range).
Sketches: $\sec$ has asymptotes where $\cos = 0$ ; $\csc$ where $\sin = 0$ ; $\cot$ where $\sin = 0$ .
Identity-substitution converts mixed trig equations to single-variable quadratics.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XMA01-YMA01 syllabus (2026 onwards).

P3 3.1 — Define and use the reciprocal trig functions $\sec\theta$ , $\csc\theta$ , $\cot\theta$ .
P3 3.2 — Sketch graphs of reciprocal trig functions, identifying asymptotes and key points.
P3 3.3 — Derive and use $1 + \tan^{2}\theta = \sec^{2}\theta$ and $1 + \cot^{2}\theta = \csc^{2}\theta$ .
P3 3.4 — Solve trigonometric equations involving reciprocal functions.
P3 3.5 — Define inverse trigonometric functions $\arcsin$ , $\arccos$ , $\arctan$ and sketch their graphs with domain and range.

Reciprocal trig functions: definitions

Three new functions — secant, cosecant, cotangent.

Definitions. $\sec\theta = \dfrac{1}{\cos\theta}, \quad \csc\theta = \dfrac{1}{\sin\theta}, \quad \cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}.$

Mnemonic for pairing.

$\sec$ pairs with $\cos$ (no 'co' in $\sec$ , matches the 'co' in $\cos$ ).
$\csc$ pairs with $\sin$ ('co' in $\csc$ matches no-'co' in $\sin$ ).
$\cot$ pairs with $\tan$ .

This anti-symmetry is the source of much confusion — write it down.

Domains (each undefined where its denominator is zero):

Function	Undefined when	At $\theta =$
$\sec\theta$	$\cos\theta = 0$	$\theta = \pi/2 + k\pi$
$\csc\theta$	$\sin\theta = 0$	$\theta = k\pi$
$\cot\theta$	$\sin\theta = 0$	$\theta = k\pi$

Ranges.

$|\sec\theta| \geq 1$ (so $\sec\theta \leq -1$ or $\sec\theta \geq 1$ ).
$|\csc\theta| \geq 1$ (so $\csc\theta \leq -1$ or $\csc\theta \geq 1$ ).
$\cot\theta \in \mathbb{R}$ (all real values).

Periodicity. $\sec$ and $\csc$ have period $2\pi$ (same as $\cos$ , $\sin$ ). $\cot$ has period $\pi$ (same as $\tan$ ).

Sec x = 1/cos x has vertical asymptotes where cos x = 0 and touches y = ±1 at the maxima of |cos x|.

$\sec = 1/\cos$ , $\csc = 1/\sin$ , $\cot = 1/\tan$ .
Asymptotes where the reciprocal's denominator is zero.
$|\sec|, |\csc| \geq 1$ ; $\cot$ takes all real values.
Periods: $2\pi$ ( $\sec, \csc$ ), $\pi$ ( $\cot$ ).

Graphs of reciprocal trig functions

Vertical asymptotes where denominator $= 0$ ; touch $\pm 1$ where denominator $= \pm 1$ .

$y = \sec\theta$ . Vertical asymptotes at $\theta = \pi/2 + k\pi$ . Touches $y = 1$ at $\theta = 2k\pi$ (where $\cos = 1$ ). Touches $y = -1$ at $\theta = (2k+1)\pi$ (where $\cos = -1$ ). Shape: U-curves opening upward where $\cos > 0$ ; inverted U-curves (downward) where $\cos < 0$ .

$y = \csc\theta$ . Same shape as $\sec$ but shifted by $\pi/2$ . Asymptotes at $\theta = k\pi$ . Touches $y = 1$ at $\theta = \pi/2 + 2k\pi$ ; touches $y = -1$ at $\theta = 3\pi/2 + 2k\pi$ .

$y = \cot\theta$ . Asymptotes at $\theta = k\pi$ . Decreasing within each period. $\cot(\pi/4) = 1$ , $\cot(\pi/2) = 0$ (the only zero of $\cot$ per period), $\cot(3\pi/4) = -1$ .

Symmetries.

$\sec(-\theta) = \sec\theta$ (even).
$\csc(-\theta) = -\csc\theta$ (odd).
$\cot(-\theta) = -\cot\theta$ (odd).

Sketching tips. Always lightly sketch $\cos$ or $\sin$ first (depending which is the denominator), mark zeros (asymptotes) and turning points ( $\pm 1$ ), and INVERT to get $\sec$ or $\csc$ .

Asymptotes at zeros of $\cos$ ( $\sec$ ) or $\sin$ ( $\csc$ , $\cot$ ).
$\sec, \csc$ never enter $(-1, 1)$ strip.
$\cot$ decreases continuously between asymptotes.
$\sec$ is even; $\csc, \cot$ are odd.

Pythagorean identities (sec / csc forms)

Divide the basic identity by $\cos^{2}\theta$ or $\sin^{2}\theta$ .

Foundation. $\sin^{2}\theta + \cos^{2}\theta = 1$ .

Divide by $\cos^{2}\theta$ (where $\cos\theta \neq 0$ ): $\dfrac{\sin^{2}\theta}{\cos^{2}\theta} + 1 = \dfrac{1}{\cos^{2}\theta} \;\Rightarrow\; \tan^{2}\theta + 1 = \sec^{2}\theta.$

Divide by $\sin^{2}\theta$ (where $\sin\theta \neq 0$ ): $1 + \dfrac{\cos^{2}\theta}{\sin^{2}\theta} = \dfrac{1}{\sin^{2}\theta} \;\Rightarrow\; 1 + \cot^{2}\theta = \csc^{2}\theta.$

Use case: solving equations. If an equation mixes $\tan\theta$ with $\sec\theta$ , substitute $\sec^{2}\theta = 1 + \tan^{2}\theta$ to reduce to a single variable.

Worked example. Solve $\sec^{2}\theta = 3\tan\theta - 1$ for $0 \leq \theta \leq 2\pi$ .

Substitute: $1 + \tan^{2}\theta = 3\tan\theta - 1 \Rightarrow \tan^{2}\theta - 3\tan\theta + 2 = 0$ .

Factorise: $(\tan\theta - 1)(\tan\theta - 2) = 0$ . So $\tan\theta = 1$ or $\tan\theta = 2$ .

$\tan\theta = 1$ : $\theta = \pi/4, 5\pi/4$ (period $\pi$ ).
$\tan\theta = 2$ : $\theta = \arctan(2) \approx 1.107, 4.249$ .

Four solutions in $[0, 2\pi]$ .

Identity-proving questions. When asked to 'show that' an identity holds, manipulate one side until it equals the other. Common toolkit: substitute reciprocals, apply Pythagoras, factor differences of squares.

$1 + \tan^{2}\theta = \sec^{2}\theta$ — divide Pythagoras by $\cos^{2}$ .
$1 + \cot^{2}\theta = \csc^{2}\theta$ — divide by $\sin^{2}$ .
Use to substitute and reduce mixed-trig equations.
Both identities in the IAL formula booklet.

Inverse trig functions

Restricted ranges so each is a function. Principal values.

Why restrict? $\sin$ , $\cos$ , $\tan$ are NOT one-to-one over their natural domains — they repeat. To define inverses we restrict the domain to a principal-value interval.

$\arcsin$ (or $\sin^{-1}$ ).

Defined as the inverse of $\sin x$ restricted to $\bigl[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\bigr]$ .
Domain: $-1 \leq x \leq 1$ . Range: $-\tfrac{\pi}{2} \leq y \leq \tfrac{\pi}{2}$ .
Graph: S-shape, passing through $(-1, -\tfrac{\pi}{2})$ , $(0, 0)$ , $(1, \tfrac{\pi}{2})$ . Increasing.

$\arccos$ (or $\cos^{-1}$ ).

Inverse of $\cos x$ restricted to $[0, \pi]$ .
Domain: $-1 \leq x \leq 1$ . Range: $0 \leq y \leq \pi$ .
Graph: decreasing, through $(-1, \pi)$ , $(0, \tfrac{\pi}{2})$ , $(1, 0)$ .

$\arctan$ (or $\tan^{-1}$ ).

Inverse of $\tan x$ restricted to $\bigl(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\bigr)$ .
Domain: all real numbers. Range: $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ .
Graph: increasing S-shape with horizontal asymptotes $y = \pm\tfrac{\pi}{2}$ , through $(0, 0)$ , $(1, \tfrac{\pi}{4})$ , $(-1, -\tfrac{\pi}{4})$ .

Worked example. $\arcsin(\sin(5\pi/6)) = ?$

Note $\sin(5\pi/6) = 1/2$ . Then $\arcsin(1/2) = \pi/6$ (the PRINCIPAL value in $[-\pi/2, \pi/2]$ ).

So $\arcsin(\sin(5\pi/6)) = \pi/6$ , NOT $5\pi/6$ .

Useful identities.

$\arcsin x + \arccos x = \tfrac{\pi}{2}$ .
$\arctan(1) = \tfrac{\pi}{4}$ , $\arctan(\sqrt{3}) = \tfrac{\pi}{3}$ , $\arctan(1/\sqrt{3}) = \tfrac{\pi}{6}$ .

Notation warning. $\sin^{-1}(x)$ ALWAYS means $\arcsin(x)$ in P3/calculator context, NOT $1/\sin(x)$ . The reciprocal is $\csc(x)$ or $(\sin x)^{-1}$ .

Principal-value ranges: $\arcsin, \arccos$ closed; $\arctan$ open.
$\arcsin, \arctan$ increasing; $\arccos$ decreasing.
$\arcsin x + \arccos x = \pi/2$ .
$\sin^{-1}$ means inverse function — NOT reciprocal.

Quick recap

$\sec = 1/\cos$ ; $\csc = 1/\sin$ ; $\cot = 1/\tan$ .
Pythagorean identities: $1 + \tan^{2} = \sec^{2}$ ; $1 + \cot^{2} = \csc^{2}$ .
Reciprocal graphs have asymptotes where the original = 0.
$\arcsin, \arccos$ : domain $[-1, 1]$ . $\arctan$ : domain $\mathbb{R}$ .
$\arctan$ range is OPEN $(-\pi/2, \pi/2)$ ; the others CLOSED.
Always quote principal values for inverse trig.
Substitution-via-identity converts mixed equations to single-variable.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Reciprocals: $\sec = 1/\cos$ , $\csc = 1/\sin$ , $\cot = \cos/\sin$ .
Pythagorean (sec): $1 + \tan^{2}\theta = \sec^{2}\theta$ .
Pythagorean (csc): $1 + \cot^{2}\theta = \csc^{2}\theta$ .
$\arcsin$ : domain $[-1, 1]$ , range $[-\pi/2, \pi/2]$ .
$\arccos$ : domain $[-1, 1]$ , range $[0, \pi]$ .
$\arctan$ : domain $\mathbb{R}$ , range $(-\pi/2, \pi/2)$ .
$\arcsin x + \arccos x = \pi/2$ .

How it’s examined

Trigonometric Functions is examined in every WMA13/01 sitting — typically as a $6$ - $10$ -mark mid-paper question OR as part of a longer trig problem. Common setups: 'Show that ...' an identity ( $4$ - $5$ marks); solve $\sec^{2}\theta + a\tan\theta = b$ via substitution ( $6$ - $7$ marks); state domain / range / draw graph of $\arcsin x$ ( $4$ marks); evaluate $\arcsin(\sin(\text{ang}))$ paying attention to the principal-value range. Examiner reports consistently flag: principal-value mistakes and missing $\sec^{2} = 1 + \tan^{2}$ substitution. Memorise the identities — they're checkable.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA13/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Trigonometric functions

Step-by-step solutions to past-paper-style questions on trigonometric functions, written exactly the way a tutor would explain them at the board.

1Prove $\tan^{2}\theta + 1 = \sec^{2}\theta$ from first principles (3 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q6(a)• identity, proof
Question
Starting from $\sin^{2}\theta + \cos^{2}\theta = 1$ , show that $\tan^{2}\theta + 1 \equiv \sec^{2}\theta$ .
Step-by-step solution
1. Step 1
  Start with the Pythagorean identity and divide every term by $\cos^{2}\theta$ (assumed non-zero, i.e. $\theta \neq \tfrac{\pi}{2} + k\pi$ ).
  $\dfrac{\sin^{2}\theta}{\cos^{2}\theta} + \dfrac{\cos^{2}\theta}{\cos^{2}\theta} = \dfrac{1}{\cos^{2}\theta}$
2. Step 2
  Rewrite each fraction using $\tan = \sin/\cos$ and $\sec = 1/\cos$ :
  $\tan^{2}\theta + 1 = \sec^{2}\theta$
3. Step 3
  Identity stated. (Identity, not equation — true for all $\theta$ in the common domain.)
Answer
$\tan^{2}\theta + 1 \equiv \sec^{2}\theta$ , provided $\cos\theta \neq 0$ .
Examiner tip
M1 dividing through by $\cos^{2}\theta$ . A1 recognising the resulting ratios as $\tan^{2}$ and $\sec^{2}$ . A1 stating the identity with the $\equiv$ sign and domain condition. 'Show that' demands every algebraic step. The analogous identity $1 + \cot^{2}\theta = \csc^{2}\theta$ comes from dividing by $\sin^{2}\theta$ instead.
2Solve $\sec^{2}\theta = 3 \tan\theta - 1$ for $0 \leq \theta \leq 2\pi$ (6 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q5• trig equation, secant, identity-substitution
Question
Solve $\sec^{2}\theta = 3\tan\theta - 1$ for $0 \leq \theta \leq 2\pi$ , giving answers to 3 s.f. where exact form is not available.
Step-by-step solution
1. Step 1
  Use the identity $\sec^{2}\theta = 1 + \tan^{2}\theta$ to convert to a quadratic in $\tan\theta$ .
  $1 + \tan^{2}\theta = 3\tan\theta - 1$
2. Step 2
  Rearrange:
  $\tan^{2}\theta - 3\tan\theta + 2 = 0$
3. Step 3
  Factorise: $(\tan\theta - 1)(\tan\theta - 2) = 0$ . So $\tan\theta = 1$ or $\tan\theta = 2$ .
4. Step 4
  $\tan\theta = 1$ : principal value $\theta = \tfrac{\pi}{4}$ ; $\tan$ has period $\pi$ , so within $[0, 2\pi]$ : $\theta = \tfrac{\pi}{4}, \tfrac{5\pi}{4}$ .
5. Step 5
  $\tan\theta = 2$ : principal value $\theta = \arctan(2) \approx 1.107$ rad. Within $[0, 2\pi]$ : $\theta \approx 1.107$ and $\theta \approx 1.107 + \pi \approx 4.249$ .
Answer
$\theta = \tfrac{\pi}{4}, \tfrac{5\pi}{4}$ (exact); $\theta \approx 1.11, 4.25$ (3 s.f.).
Examiner tip
M1 for using the identity to convert. A1 for the quadratic in $\tan\theta$ . M1 for factorising. A1 for both values of $\tan\theta$ . A1 for the two exact solutions from $\tan\theta = 1$ . A1 for the two decimal solutions. Exact forms are required where available (mark scheme awards A0 for $\tan\theta = 1$ giving only $0.785$ rad).
3Prove the identity $\cot\theta + \tan\theta \equiv 2\csc(2\theta)$ (5 marks, P3)
Extended• Adapted from WMA13/01 January 2023 Q7• identity, proof, double angle
Question
Show that $\cot\theta + \tan\theta \equiv 2 \csc(2\theta)$ , $\theta \neq \tfrac{k\pi}{2}$ , $k \in \mathbb{Z}$ .
Step-by-step solution
1. Step 1
  LHS = $\dfrac{\cos\theta}{\sin\theta} + \dfrac{\sin\theta}{\cos\theta}$ .
2. Step 2
  Common denominator $\sin\theta \cos\theta$ :
  $\dfrac{\cos^{2}\theta + \sin^{2}\theta}{\sin\theta \cos\theta} = \dfrac{1}{\sin\theta \cos\theta}$
3. Step 3
  Use the double-angle identity $\sin(2\theta) = 2 \sin\theta \cos\theta$ , so $\sin\theta \cos\theta = \tfrac{1}{2} \sin(2\theta)$ .
  $\dfrac{1}{\tfrac{1}{2} \sin(2\theta)} = \dfrac{2}{\sin(2\theta)} = 2\csc(2\theta)$
4. Step 4
  So $\cot\theta + \tan\theta \equiv 2\csc(2\theta)$ . ∎
Answer
$\cot\theta + \tan\theta = \dfrac{1}{\sin\theta\cos\theta} = \dfrac{2}{\sin 2\theta} = 2\csc(2\theta)$ .
Examiner tip
M1 expressing both terms with common denominator. A1 collapsing the numerator via Pythagoras. M1 using $\sin 2\theta = 2\sin\theta\cos\theta$ . A1 reaching the final form. B1 stating the identity and (implicitly or explicitly) the domain. 'Show that' demands strict working from LHS to RHS; mixing both sides (working backwards) is acceptable in a strict 'show' but cleaner to go LHS → RHS.
4Range of $\arcsin$ and an inverse-trig evaluation (4 marks, P3)
Core• Adapted from WMA13/01 June 2023 Q2• arcsin, inverse trig, principal value
Question
(a) State the domain and range of $y = \arcsin x$ . (b) Find the exact value of $\arcsin(-\tfrac{1}{2}) + \arccos(\tfrac{1}{2})$ .
Step-by-step solution
1. Step 1
  (a) Domain of $\arcsin$ : $-1 \leq x \leq 1$ . Range: $-\tfrac{\pi}{2} \leq y \leq \tfrac{\pi}{2}$ .
2. Step 2
  (b) $\arcsin(-\tfrac{1}{2}) = -\tfrac{\pi}{6}$ (principal value, third-quadrant-style answer rejected because of range).
3. Step 3
  $\arccos(\tfrac{1}{2}) = \tfrac{\pi}{3}$ (principal value).
4. Step 4
  Sum: $-\tfrac{\pi}{6} + \tfrac{\pi}{3} = -\tfrac{\pi}{6} + \tfrac{2\pi}{6} = \tfrac{\pi}{6}$ .
Answer
(a) Domain $[-1, 1]$ , range $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ . (b) $\tfrac{\pi}{6}$ .
Examiner tip
(a) B1 domain. B1 range. (b) B1 each correct exact value. B1 final answer. Note the range of $\arcsin$ is closed (includes $\pm \tfrac{\pi}{2}$ ); the range of $\arctan$ is open ( $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ ). Distinguish carefully.
5Sketch $y = \sec x$ and identify asymptotes (5 marks, P3)
Extended• Adapted from WMA13/01 January 2024 Q4• sketch, secant, asymptotes
Question
Sketch the graph of $y = \sec x$ for $-\pi \leq x \leq \pi$ . State the equations of any asymptotes and the coordinates of any turning points.
Step-by-step solution
1. Step 1
  $\sec x = 1/\cos x$ . Asymptotes where $\cos x = 0$ , i.e. $x = \pm \tfrac{\pi}{2}$ .
2. Step 2
  Sign analysis: $\cos x > 0$ for $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ , so $\sec x > 0$ there; $\cos x < 0$ for $\tfrac{\pi}{2} < x < \pi$ and $-\pi < x < -\tfrac{\pi}{2}$ , so $\sec x < 0$ there.
3. Step 3
  Minimum of $\sec x$ on $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ is at $x = 0$ , where $\sec 0 = 1$ . Maxima on the outer intervals at $x = \pm \pi$ , where $\sec(\pm\pi) = -1$ .
4. Step 4
  Sketch: U-shape opening upward in the middle interval with minimum $(0, 1)$ ; inverted U-shapes opening downward in the outer intervals with maxima $(\pm\pi, -1)$ . Vertical asymptotes $x = \pm\tfrac{\pi}{2}$ .
Answer
Asymptotes $x = \pm\tfrac{\pi}{2}$ . Minimum $(0, 1)$ in the middle; maxima $(\pm\pi, -1)$ on the outer intervals.
Examiner tip
B1 asymptotes. B1 minimum point. B1 maxima coordinates. M1 correct shape on the middle interval. M1 correct shape on outer intervals. The graph of $\sec x$ touches $y = \pm 1$ where $\cos x = \pm 1$ — $y = 1$ at $x = 0$ and $y = -1$ at $x = \pm\pi$ .

Key Formulae — Trigonometric functions

The formulae you need to memorise for trigonometric functions on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Reciprocal trig functions
$\sec\theta = \dfrac{1}{\cos\theta}, \quad \csc\theta = \dfrac{1}{\sin\theta}, \quad \cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$
$\sec\theta$
secant of $\theta$ , undefined where $\cos\theta = 0$
$\csc\theta$
cosecant of $\theta$ , undefined where $\sin\theta = 0$
$\cot\theta$
cotangent of $\theta$ , undefined where $\sin\theta = 0$
When to use
Whenever a question involves $\sec$ , $\csc$ , $\cot$ — re-express in terms of $\sin$ and $\cos$ if you get stuck. Definitions are in the IAL formula booklet but the relationships should be instinctive.
Example
$\sec(\pi/3) = 1/\cos(\pi/3) = 1/(1/2) = 2$ .
Pythagorean identity (sec form)
$1 + \tan^{2}\theta \equiv \sec^{2}\theta$
$\theta$
angle with $\cos\theta \neq 0$
When to use
Solving equations that mix $\tan$ and $\sec$ . Substitute to reduce to a single-variable polynomial. IN the IAL formula booklet (listed under trig identities).
Example
$\sec^{2}\theta = 3\tan\theta - 1 \Rightarrow 1 + \tan^{2}\theta = 3\tan\theta - 1 \Rightarrow \tan^{2}\theta - 3\tan\theta + 2 = 0$ .
Pythagorean identity (csc form)
$1 + \cot^{2}\theta \equiv \csc^{2}\theta$
$\theta$
angle with $\sin\theta \neq 0$
When to use
Solving equations that mix $\cot$ and $\csc$ . Substitute to reduce to a single-variable polynomial. IN the IAL formula booklet.
Example
$\csc^{2}\theta + \cot\theta = 7 \Rightarrow 1 + \cot^{2}\theta + \cot\theta = 7$ .
Principal values of inverse trig functions
$\arcsin: [-1, 1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}], \quad \arccos: [-1, 1] \to [0, \pi], \quad \arctan: \mathbb{R} \to (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$
$\arcsin x$
inverse of $\sin x$ restricted to $[-\pi/2, \pi/2]$
$\arccos x$
inverse of $\cos x$ restricted to $[0, \pi]$
$\arctan x$
inverse of $\tan x$ restricted to $(-\pi/2, \pi/2)$
When to use
Whenever quoting an inverse-trig value. NOT in the IAL formula booklet — memorise the ranges. Note $\arctan$ 's range is OPEN; the others are CLOSED.
Example
$\arcsin(\sin(5\pi/6)) = \pi - 5\pi/6 = \pi/6$ (NOT $5\pi/6$ , because the principal range excludes it).

Key Definitions and Keywords — Trigonometric functions

Definitions to memorise and the exact keywords mark schemes credit for trigonometric functions answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Secant ( $\sec\theta$ )
Examiner keyword
$\sec\theta = \dfrac{1}{\cos\theta}$ . Undefined where $\cos\theta = 0$ , i.e. $\theta = \tfrac{\pi}{2} + k\pi$ . Range: $\sec\theta \leq -1$ or $\sec\theta \geq 1$ .
Cosecant ( $\csc\theta$ )
Examiner keyword
$\csc\theta = \dfrac{1}{\sin\theta}$ . Undefined where $\sin\theta = 0$ , i.e. $\theta = k\pi$ . Range: $\csc\theta \leq -1$ or $\csc\theta \geq 1$ .
Cotangent ( $\cot\theta$ )
Examiner keyword
$\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$ . Undefined where $\sin\theta = 0$ , i.e. $\theta = k\pi$ . Range: all real values.
Arcsine ( $\arcsin$ , sometimes $\sin^{-1}$ )
Examiner keyword
The inverse of $\sin x$ restricted to $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ . Domain: $[-1, 1]$ . Range: $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ . $\sin^{-1}$ here does NOT mean reciprocal.
Arccosine ( $\arccos$ , sometimes $\cos^{-1}$ )
Examiner keyword
The inverse of $\cos x$ restricted to $[0, \pi]$ . Domain: $[-1, 1]$ . Range: $[0, \pi]$ .
Arctangent ( $\arctan$ , sometimes $\tan^{-1}$ )
Examiner keyword
The inverse of $\tan x$ restricted to $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ . Domain: all reals. Range: $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ (open, because $\tan$ tends to $\pm\infty$ ).

Common Mistakes and Misconceptions — Trigonometric functions

The traps other students keep falling into on trigonometric functions questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Writing $\sec\theta = 1/\sin\theta$
Why it happens
The names suggest secant connects with sine.
How to avoid it
Mnemonic: 'co-co-co' — only the cosecant has 'co' (matches sine) and the cotangent ('co' matches tangent's reciprocal of cosine over sine). $\sec$ has no 'co' and pairs with $\cos$ .
✕Reading $\sin^{-1}(x)$ as $\dfrac{1}{\sin x}$
WMA13/01 examiner reports — flagged annually
Why it happens
The $-1$ superscript usually denotes reciprocal in algebra.
How to avoid it
On a calculator and in $\arcsin$ -style notation, $\sin^{-1}(x)$ ALWAYS means the inverse function (i.e. $\arcsin x$ ). For the reciprocal, write $\csc x$ or $(\sin x)^{-1}$ .
✕Stating $\arcsin(1/2) = 5\pi/6$ (a valid sine solution but outside the principal range)
Why it happens
Treating $\arcsin$ as 'any angle with $\sin = 1/2$ '.
How to avoid it
Always quote the PRINCIPAL VALUE: $\arcsin x$ lies in $[-\pi/2, \pi/2]$ . For $\arcsin(1/2) = \pi/6$ (NOT $5\pi/6$ ).
✕Quoting $1 + \tan^{2}\theta = \sec^{2}\theta$ without noting $\cos\theta \neq 0$
Why it happens
Forgetting that the identity has a domain restriction.
How to avoid it
If the question's answer set could include $\theta = \pi/2$ etc., always state the exclusion. Mark schemes don't always penalise but quoting it shows mathematical maturity.
✕Drawing $y = \sec x$ passing through the origin
Why it happens
Confusing it with $y = \cos x$ .
How to avoid it
$\sec 0 = 1$ , NOT $0$ . The minimum on the central interval is $(0, 1)$ . Plot $(\pm\pi, -1)$ on the outer intervals. The graph never crosses the $x$ -axis.

Trigonometric functions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Trigonometric functions

Short Notes - Trigonometric functions

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Prove tan⁡2θ+1=sec⁡2θ\tan^{2}\theta + 1 = \sec^{2}\thetatan2θ+1=sec2θ from first principles (3 marks, P3)

2Solve sec⁡2θ=3tan⁡θ−1\sec^{2}\theta = 3 \tan\theta - 1sec2θ=3tanθ−1 for 0≤θ≤2π0 \leq \theta \leq 2\pi0≤θ≤2π (6 marks, P3)

3Prove the identity cot⁡θ+tan⁡θ≡2csc⁡(2θ)\cot\theta + \tan\theta \equiv 2\csc(2\theta)cotθ+tanθ≡2csc(2θ) (5 marks, P3)

4Range of arcsin⁡\arcsinarcsin and an inverse-trig evaluation (4 marks, P3)

5Sketch y=sec⁡xy = \sec xy=secx and identify asymptotes (5 marks, P3)

Reciprocal trig functions

Pythagorean identity (sec form)

Pythagorean identity (csc form)

Principal values of inverse trig functions

Secant (sec⁡θ\sec\thetasecθ)

Cosecant (csc⁡θ\csc\thetacscθ)

Cotangent (cot⁡θ\cot\thetacotθ)

Arcsine (arcsin⁡\arcsinarcsin, sometimes sin⁡−1\sin^{-1}sin−1)

Arccosine (arccos⁡\arccosarccos, sometimes cos⁡−1\cos^{-1}cos−1)

Arctangent (arctan⁡\arctanarctan, sometimes tan⁡−1\tan^{-1}tan−1)

✕Writing sec⁡θ=1/sin⁡θ\sec\theta = 1/\sin\thetasecθ=1/sinθ

✕Reading sin⁡−1(x)\sin^{-1}(x)sin−1(x) as 1sin⁡x\dfrac{1}{\sin x}sinx1​

✕Stating arcsin⁡(1/2)=5π/6\arcsin(1/2) = 5\pi/6arcsin(1/2)=5π/6 (a valid sine solution but outside the principal range)

✕Quoting 1+tan⁡2θ=sec⁡2θ1 + \tan^{2}\theta = \sec^{2}\theta1+tan2θ=sec2θ without noting cos⁡θ≠0\cos\theta \neq 0cosθ=0

✕Drawing y=sec⁡xy = \sec xy=secx passing through the origin

Trigonometric functions — frequently asked questions

1Prove $\tan^{2}\theta + 1 = \sec^{2}\theta$ from first principles (3 marks, P3)

2Solve $\sec^{2}\theta = 3 \tan\theta - 1$ for $0 \leq \theta \leq 2\pi$ (6 marks, P3)

3Prove the identity $\cot\theta + \tan\theta \equiv 2\csc(2\theta)$ (5 marks, P3)

4Range of $\arcsin$ and an inverse-trig evaluation (4 marks, P3)

5Sketch $y = \sec x$ and identify asymptotes (5 marks, P3)

Secant ( $\sec\theta$ )

Cosecant ( $\csc\theta$ )

Cotangent ( $\cot\theta$ )

Arcsine ( $\arcsin$ , sometimes $\sin^{-1}$ )

Arccosine ( $\arccos$ , sometimes $\cos^{-1}$ )

Arctangent ( $\arctan$ , sometimes $\tan^{-1}$ )

✕Writing $\sec\theta = 1/\sin\theta$

✕Reading $\sin^{-1}(x)$ as $\dfrac{1}{\sin x}$

✕Stating $\arcsin(1/2) = 5\pi/6$ (a valid sine solution but outside the principal range)

✕Quoting $1 + \tan^{2}\theta = \sec^{2}\theta$ without noting $\cos\theta \neq 0$

✕Drawing $y = \sec x$ passing through the origin