Trigonometry

CIE A-Level Mathematics· 40 min read

1. Reciprocal Trigonometric Functions (sec, cosec, cot)★★☆☆☆⏱ 10 min

📘 Definition

Reciprocal Trigonometric Functions

Each of the three basic ratios has a reciprocal: the cosecant (cosec), secant (sec) and cotangent (cot). They are defined by $csc θ = \frac{1}{s i n θ}$ , $sec θ = \frac{1}{c o s θ}$ and $cot θ = \frac{1}{t a n θ} = \frac{c o s θ}{s i n θ}$ .

Example:

$sec 6 0^{\circ} = \frac{1}{c o s 6 0 ^{\circ}} = \frac{1}{\frac{1}{2}} = 2$

\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}

📘 Definition

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. $y = csc θ$ has asymptotes where $sin θ = 0$ (at $θ = 0^{\circ}, 18 0^{\circ}, 36 0^{\circ}, \dots$ ); $y = sec θ$ has asymptotes where $cos θ = 0$ (at $θ = 9 0^{\circ}, 27 0^{\circ}, \dots$ ); and $y = cot θ$ has asymptotes where $sin θ = 0$ (at $θ = 0^{\circ}, 18 0^{\circ}, \dots$ ). Because $∣ sin θ ∣ \leq 1$ and $∣ cos θ ∣ \leq 1$ , the graphs of $sec$ and $csc$ never take values strictly between $- 1$ and $1$ .

Example:

Near $θ = 0^{\circ}$ , $sin θ \to 0$ , so $csc θ = 1/ sin θ \to \pm \infty$ .

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

📐 Worked Example

Solve $2 sec^{2} θ - tan θ = 5$ for $0^{\circ} \leq θ \leq 36 0^{\circ}$ .

1
Use $sec^{2} θ \equiv 1 + tan^{2} θ$ to write the equation entirely in terms of $tan θ$ :
2
$2(1 + \tan^2\theta) - \tan\theta = 5$
3
Expand and collect every term on one side to form a quadratic in $tan θ$ :
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 $tan θ$ :
8
$\tan\theta = \frac{3}{2} \quad \text{or} \quad \tan\theta = -1$
9
Give every solution in the interval $0^{\circ} \leq θ \leq 36 0^{\circ}$ :
10
$\theta = 56.3^\circ,\ 236.3^\circ \quad \text{or} \quad \theta = 135^\circ,\ 315^\circ$

✓ Quick check

Which identity turns $sec^{2} θ$ into an expression in $tan θ$ ?

Which identity rewrites $sec^{2} θ$ in terms of $tan θ$ ?
- $sec^{2} θ = 1 + tan^{2} θ$
- $sec^{2} θ = 1 - tan^{2} θ$
- $sec^{2} θ = tan^{2} θ - 1$
- $sec^{2} θ = 1 + cot^{2} θ$

Exam tip:

$sec$ , $csc$ and $cot$ are reciprocals, not inverse functions. Rewrite them with $sec^{2} θ \equiv 1 + tan^{2} θ$ or $csc^{2} θ \equiv 1 + cot^{2} θ$ so the equation becomes a quadratic in a single ratio.

2. Compound Angle Identities★★☆☆☆⏱ 10 min

📘 Definition

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 $sin 1 5^{\circ}$

1
Write $1 5^{\circ}$ 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 $cos (x - 6 0^{\circ}) - 3 sin (x - 6 0^{\circ})$ , 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 $3$ :
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.

3. Double-Angle Identities (and power-reduction forms)★★★☆☆⏱ 10 min

📘 Definition

Double Angle Identity

Identities derived from compound angle identities when both angles are equal, relating trigonometric functions of $2 θ$ to functions of $θ$

\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 $\frac{1 - c o s 2 θ}{s i n 2 θ} = tan θ$

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

✓ Quick check

Which expression is equivalent to $cos^{2} x$ ?

Which expression is equivalent to $cos^{2} x$ ?
- $\frac{1 + c o s 2 x}{2}$
- $\frac{1 - c o s 2 x}{2}$
- $1 + cos 2 x$
- $cos 2 x - 1$

4. R-form for Linear Combinations of Sine and Cosine★★★☆☆⏱ 10 min

📘 Definition

R-form (Amplitude-Phase Form)

R \sin(\theta \pm \alpha), R \cos(\theta \pm \alpha)

A method to rewrite a linear combination $a sin θ + b cos θ$ (same angle $θ$ ) as a single trigonometric function, with $R > 0$ and $0 < α < 9 0^{\circ}$ (or $\frac{π}{2}$ radians)

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 $R sin (x + α)$ , where $R > 0$ and $0 < α < 9 0^{\circ}$ .

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 $cos^{2} α + sin^{2} α = 1$ :
6
$R^2 (\cos^2 \alpha + \sin^2 \alpha) = 3^2 + 4^2 = 25 \implies R = 5$
7
Calculate $α$ 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: $R cos (x - α)$ will have different coefficient arrangements, so expand first before solving.

5. Common Pitfalls

Wrong move:

Writing $cos (A - B) = cos A cos B - sin A sin B$

Why:

The sign of the sine term flips for cosine compound angle identities, opposite to the angle's sign

Correct move:

Use $cos (A - B) = cos A cos B + sin A sin B$

Wrong move:

Cancelling $sin θ$ from both sides of an equation when solving

Why:

This removes solutions where $sin θ = 0$ , which are valid in most intervals

Correct move:

Bring all terms to one side, factor out $sin θ$ , and solve each factor separately

Wrong move:

Writing $tan α = \frac{a}{b}$ for $a sin x + b cos x = R sin (x + α)$

Why:

Coefficients are swapped when equating, leading to an incorrect value for $α$

Correct move:

Always expand the R-form expression first to equate coefficients before solving for $α$

Wrong move:

Keeping the original interval when solving for $2 θ$ , e.g., $0 < 2 θ < π$ for $0 < θ < π$

Why:

The interval scales with the angle coefficient, leading to missing solutions

Correct move:

Multiply the interval bounds by the angle coefficient, so $0 < 2 θ < 2 π$ for $0 < θ < π$

6. Quick Reference Cheatsheet

Identity Type	Key Formulae
Reciprocal ratios & identities	$csc θ = \frac{1}{s i n θ}, sec θ = \frac{1}{c o s θ}, cot θ = \frac{c o s θ}{s i n θ}$ $sec^{2} θ \equiv 1 + tan^{2} θ$ $csc^{2} θ \equiv 1 + cot^{2} θ$
Compound Angles	$sin (A \pm B) = sin A cos B \pm cos A sin B$ $cos (A \pm B) = cos A cos B \mp sin A sin B$ $tan (A \pm B) = \frac{t a n A \pm t a n B}{1 \mp t a n A t a n B}$
Double Angles	$sin 2 θ = 2 sin θ cos θ$ $cos 2 θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$ $tan 2 θ = \frac{2 t a n θ}{1 - t a n ^{2} θ}$
R-form: $a sin x + b cos x$	$R = a^{2} + b^{2}, tan α = \frac{b}{a}$ $a sin x + b cos x = R sin (x + α)$

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.