# Trigonometry

> CIE A-Level Mathematics · Pure Mathematics 2 & 3
> Source: https://www.owlsprep.com/study/cie-9709-u2-trigonometry/

This sub-topic covers the trigonometry that is common to CIE A-Level Pure Mathematics 2 and Pure Mathematics 3: the reciprocal functions sec, cosec and cot with their identities, compound angles, double angles, and the R-form method for combining sine and cosine terms. You will learn to apply these to prove identities and solve equations.

**Prerequisites:** [Basic trigonometric ratios and Pythagorean identities](https://www.owlsprep.com/study/cie-9709-u1-trigonometry/); [Solving simple trigonometric equations over a given interval](https://www.owlsprep.com/study/cie-9709-u1-trigonometry/)

## Learning objectives

- Understand the reciprocal functions $\sec\theta$, $\csc\theta$ and $\cot\theta$, sketch their graphs, and apply the identities $\sec^2\theta \equiv 1 + \tan^2\theta$ and $\csc^2\theta \equiv 1 + \cot^2\theta$
- Recall and apply compound angle identities for sine, cosine and tangent
- Derive and use double-angle identities (and power-reduction forms) to prove identities and solve equations
- Express a linear combination of sine and cosine as a single trigonometric function in R-form
- Apply these identities to solve problems involving maximum/minimum values

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P2 past papers (2016–2025): **Trigonometry** appears **168 times** in the last 10 years — **20%** of all P2 questions (168 of 455).

Most-tested forms: Double angle and R sin(θ+α) form (73), Solving trig equations using compound/double-angle identities (71), Compound angle formulae sin(A±B), cos(A±B), tan(A±B) (36), sec, cosec, cot — graphs and identities (36).

## Reciprocal Trigonometric Functions (sec, cosec, cot)

**Reciprocal Trigonometric Functions** — Each of the three basic ratios has a reciprocal: the cosecant (cosec), secant (sec) and cotangent (cot). They are defined by $\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$ and $\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$.

*Example:* $\sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2$

$$\csc\theta = \frac{1}{\sin\theta}, \qquad \sec\theta = \frac{1}{\cos\theta}, \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$$

> **info**
>
> Pairing tip: it is the third letter that names the partner - coSecant goes with Sine, seCant goes with Cosine, coTangent goes with Tangent. So $\sec\theta = 1/\cos\theta$, never $1/\sin\theta$.

**Graphs of all six functions (angles of any magnitude)** — A reciprocal function is undefined - and its graph has a vertical asymptote - wherever the original ratio is zero. $y = \csc\theta$ has asymptotes where $\sin\theta = 0$ (at $\theta = 0^\circ, 180^\circ, 360^\circ, \ldots$); $y = \sec\theta$ has asymptotes where $\cos\theta = 0$ (at $\theta = 90^\circ, 270^\circ, \ldots$); and $y = \cot\theta$ has asymptotes where $\sin\theta = 0$ (at $\theta = 0^\circ, 180^\circ, \ldots$). Because $|\sin\theta| \le 1$ and $|\cos\theta| \le 1$, the graphs of $\sec$ and $\csc$ never take values strictly between $-1$ and $1$.

*Example:* Near $\theta = 0^\circ$, $\sin\theta \to 0$, so $\csc\theta = 1/\sin\theta \to \pm\infty$.

> **info**
>
> Dividing the Pythagorean identity $\sin^2\theta + \cos^2\theta \equiv 1$ by $\cos^2\theta$ gives $\tan^2\theta + 1 \equiv \sec^2\theta$; dividing instead by $\sin^2\theta$ gives $1 + \cot^2\theta \equiv \csc^2\theta$.

$$\sec^2\theta \equiv 1 + \tan^2\theta \qquad\qquad \csc^2\theta \equiv 1 + \cot^2\theta$$

**Worked example:** Solve $2\sec^2\theta - \tan\theta = 5$ for $0^\circ \leq \theta \leq 360^\circ$.

1. Use $\sec^2\theta \equiv 1 + \tan^2\theta$ to write the equation entirely in terms of $\tan\theta$:
2. $$2(1 + \tan^2\theta) - \tan\theta = 5$$
3. Expand and collect every term on one side to form a quadratic in $\tan\theta$:
4. $$2\tan^2\theta - \tan\theta - 3 = 0$$
5. Factorise the quadratic:
6. $$(2\tan\theta - 3)(\tan\theta + 1) = 0$$
7. Solve each factor for $\tan\theta$:
8. $$\tan\theta = \frac{3}{2} \quad \text{or} \quad \tan\theta = -1$$
9. Give every solution in the interval $0^\circ \leq \theta \leq 360^\circ$:
10. $$\theta = 56.3^\circ,\ 236.3^\circ \quad \text{or} \quad \theta = 135^\circ,\ 315^\circ$$

**Check your understanding**

Which identity turns $\sec^2\theta$ into an expression in $\tan\theta$?

1. Which identity rewrites $\sec^2\theta$ in terms of $\tan\theta$?

   - $\sec^2\theta = 1 + \tan^2\theta$
   - $\sec^2\theta = 1 - \tan^2\theta$
   - $\sec^2\theta = \tan^2\theta - 1$
   - $\sec^2\theta = 1 + \cot^2\theta$

   *Answer:* $\sec^2\theta = 1 + \tan^2\theta$

   *Why:* Correct: dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ gives $\tan^2\theta + 1 = \sec^2\theta$.

> **Exam tip:** $\sec$, $\csc$ and $\cot$ are reciprocals, not inverse functions. Rewrite them with $\sec^2\theta \equiv 1 + \tan^2\theta$ or $\csc^2\theta \equiv 1 + \cot^2\theta$ so the equation becomes a quadratic in a single ratio.

## Compound Angle Identities

**Compound Angle Identity** — Formulae that express trigonometric functions of the sum or difference of two angles in terms of functions of the individual angles

*Example:* Expanding $\sin(A+B)$ using the sine addition rule

$$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$

$$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$

$$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$

**Worked example:** Find the exact value of $\sin 15^\circ$

1. Write $15^\circ$ as the difference of two angles with known trigonometric values:
2. $$15^\circ = 45^\circ - 30^\circ$$
3. Apply the sine compound angle identity for $\sin(A-B)$:
4. $$\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$$
5. Substitute known exact values:
6. $$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$$
7. Final simplified result:
8. $$\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$$

**Worked example:** Simplify $\cos(x - 60^\circ) - \sqrt{3}\,\sin(x - 60^\circ)$, giving your answer as a single term.

1. Expand each compound angle using the addition formulae:
2. $$\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ = \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x$$
3. $$\sin(x - 60^\circ) = \sin x \cos 60^\circ - \cos x \sin 60^\circ = \frac{1}{2}\sin x - \frac{\sqrt{3}}{2}\cos x$$
4. Substitute both expansions and multiply out the $\sqrt{3}$:
5. $$\cos(x - 60^\circ) - \sqrt{3}\sin(x - 60^\circ) = \left(\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x\right) - \sqrt{3}\left(\frac{1}{2}\sin x - \frac{\sqrt{3}}{2}\cos x\right)$$
6. The $\sin x$ terms cancel and the $\cos x$ terms combine:
7. $$= \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x - \frac{\sqrt{3}}{2}\sin x + \frac{3}{2}\cos x = 2\cos x$$

> **Exam tip:** Always check the sign of the middle term: for cosine compound angles, the sign flips relative to the angle's sign.

## Double-Angle Identities (and power-reduction forms)

> **info**
>
> Double angle identities are just a special case of compound angle identities, where $A = B$.

**Double Angle Identity** — Identities derived from compound angle identities when both angles are equal, relating trigonometric functions of $2\theta$ to functions of $\theta$

$$\sin 2\theta = 2 \sin \theta \cos \theta$$

$$\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta$$

$$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$$

**Worked example:** Prove the identity $\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta$

1. Start with the left-hand side (LHS) and substitute double angle identities:
2. Replace $1 - \cos 2\theta$ with $2\sin^2 \theta$ (from the identity $\cos 2\theta = 1 - 2\sin^2 \theta$) and $\sin 2\theta$ with $2\sin \theta \cos \theta$:
3. $$\text{LHS} = \frac{2 \sin^2 \theta}{2 \sin \theta \cos \theta}$$
4. Cancel common factors $2$ and $\sin \theta$ (for $\sin \theta \neq 0$):
5. $$\text{LHS} = \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{RHS}$$

**Check your understanding**

Which expression is equivalent to $\cos^2 x$?

1. Which expression is equivalent to $\cos^2 x$?

   - $\frac{1 + \cos 2x}{2}$
   - $\frac{1 - \cos 2x}{2}$
   - $1 + \cos 2x$
   - $\cos 2x - 1$

   *Answer:* $\frac{1 + \cos 2x}{2}$

   *Why:* Correct: Rearranging $\cos 2x = 2\cos^2 x - 1$ gives this result.

## R-form for Linear Combinations of Sine and Cosine

**R-form (Amplitude-Phase Form)** — A method to rewrite a linear combination $a \sin \theta + b \cos \theta$ (same angle $\theta$) as a single trigonometric function, with $R > 0$ and $0 < \alpha < 90^\circ$ (or $\frac{\pi}{2}$ radians)

*Notation:* R \sin(\theta \pm \alpha), R \cos(\theta \pm \alpha)

*Example:* $3\sin x + 4\cos x = 5\sin(x + 53.1^\circ)$

**Worked example:** Express $3 \sin x + 4 \cos x$ in the form $R \sin(x + \alpha)$, where $R > 0$ and $0 < \alpha < 90^\circ$.

1. Expand the target form using the compound angle identity for sine:
2. $$R \sin(x + \alpha) = R \sin x \cos \alpha + R \cos x \sin \alpha$$
3. Equate coefficients with $3 \sin x + 4 \cos x$:
4. $$R \cos \alpha = 3 \quad \text{and} \quad R \sin \alpha = 4$$
5. Calculate $R$ by squaring and adding both equations, using $\cos^2 \alpha + \sin^2 \alpha = 1$:
6. $$R^2 (\cos^2 \alpha + \sin^2 \alpha) = 3^2 + 4^2 = 25 \implies R = 5$$
7. Calculate $\alpha$ by dividing the two equations to eliminate $R$:
8. $$\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{4}{3} \implies \alpha \approx 53.1^\circ$$
9. Final result:
10. $$3 \sin x + 4 \cos x = 5 \sin(x + 53.1^\circ)$$

> **Exam tip:** Always check what form the question asks for: $R \cos(x - \alpha)$ will have different coefficient arrangements, so expand first before solving.

## Common pitfalls

- **Wrong:** Writing $\cos(A - B) = \cos A \cos B - \sin A \sin B$
  - Why it fails: The sign of the sine term flips for cosine compound angle identities, opposite to the angle's sign
  - Correct: Use $\cos(A - B) = \cos A \cos B + \sin A \sin B$
- **Wrong:** Cancelling $\sin \theta$ from both sides of an equation when solving
  - Why it fails: This removes solutions where $\sin \theta = 0$, which are valid in most intervals
  - Correct: Bring all terms to one side, factor out $\sin \theta$, and solve each factor separately
- **Wrong:** Writing $\tan \alpha = \frac{a}{b}$ for $a \sin x + b \cos x = R \sin(x + \alpha)$
  - Why it fails: Coefficients are swapped when equating, leading to an incorrect value for $\alpha$
  - Correct: Always expand the R-form expression first to equate coefficients before solving for $\alpha$
- **Wrong:** Keeping the original interval when solving for $2\theta$, e.g., $0 < 2\theta < \pi$ for $0 < \theta < \pi$
  - Why it fails: The interval scales with the angle coefficient, leading to missing solutions
  - Correct: Multiply the interval bounds by the angle coefficient, so $0 < 2\theta < 2\pi$ for $0 < \theta < \pi$

## Cheatsheet

| Identity Type | Key Formulae |
| --- | --- |
| Reciprocal ratios & identities | $\csc\theta = \frac{1}{\sin\theta},\ \sec\theta = \frac{1}{\cos\theta},\ \cot\theta = \frac{\cos\theta}{\sin\theta}$ <br> $\sec^2\theta \equiv 1 + \tan^2\theta$ <br> $\csc^2\theta \equiv 1 + \cot^2\theta$ |
| Compound Angles | $\sin(A\pm B) = \sin A \cos B \pm \cos A \sin B$ <br> $\cos(A\pm B) = \cos A \cos B \mp \sin A \sin B$ <br> $\tan(A\pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ |
| Double Angles | $\sin 2\theta = 2 \sin \theta \cos \theta$ <br> $\cos 2\theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta$ <br> $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$ |
| R-form: $a\sin x + b\cos x$ | $R = \sqrt{a^2 + b^2}, \tan \alpha = \frac{b}{a}$ <br> $a\sin x + b\cos x = R\sin(x + \alpha)$ |

## What's next

These core trigonometric identities are foundational for almost all further topics in Pure Mathematics 2 and 3, and also appear frequently in mechanics problems involving harmonic motion. You will use them regularly to solve complex trigonometric equations, differentiate and integrate trigonometric functions, simplify expressions for inverse trigonometry, and find maximum and minimum values of combined periodic functions. Mastery of these identities is essential for scoring high marks on Paper 2 and Paper 3, as they appear in multiple questions every exam series.

- [Differentiation](https://www.owlsprep.com/study/cie-9709-u2-differentiation/)
- [Integration](https://www.owlsprep.com/study/cie-9709-u2-integration/)
- [Numerical methods](https://www.owlsprep.com/study/cie-9709-u2-numerical-methods/)

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