A-Level Mathematics – Chapter 4: Trigonometry
Trigonometry is one of the most substantial chapters in A-Level Mathematics, appearing in both pure mathematics and mechanics. This chapter builds from GCSE foundations through exact values and identities, then extends to radians and arc length, and finally to the Year 2 topics of compound angles, the harmonic form, and reciprocal/inverse functions.
Specification Note
Content labelled Year 2 is A2-level only. All other content applies to both AS and full A-Level.
1. Trigonometric Ratios and Exact Values
SOHCAHTOA
For a right-angled triangle with angle $\theta$, hypotenuse $H$, opposite side $O$, and adjacent side $A$:
$$\sin\theta = \frac{O}{H}, \qquad \cos\theta = \frac{A}{H}, \qquad \tan\theta = \frac{O}{A}$$Exact Values
You must know these exact values without a calculator:
| Angle | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\tfrac{1}{2}$ | $\tfrac{\sqrt{2}}{2}$ | $\tfrac{\sqrt{3}}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\tfrac{\sqrt{3}}{2}$ | $\tfrac{\sqrt{2}}{2}$ | $\tfrac{1}{2}$ | $0$ |
| $\tan\theta$ | $0$ | $\tfrac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | undefined |
Graphs of sin, cos, and tan
The three trigonometric functions have the following key features:
- $y = \sin x$: period $360°$, amplitude $1$, passes through $(0,0)$, $(90°,1)$, $(180°,0)$, $(270°,-1)$.
- $y = \cos x$: period $360°$, amplitude $1$, passes through $(0,1)$, $(90°,0)$, $(180°,-1)$, $(270°,0)$.
- $y = \tan x$: period $180°$, no amplitude, vertical asymptotes at $x = 90° + 180°n$ for integer $n$.
Figure 4.1 — Graphs of $y = \sin x$ (blue), $y = \cos x$ (green), and $y = \tan x$ (red) for $x \in [-360°, 360°]$.
Worked Example 4.1 — Using Exact Values
Without a calculator, find the exact value of $\sin 30° \times \tan 60°$.
Step 1 Write down the exact values: $\sin 30° = \tfrac{1}{2}$ and $\tan 60° = \sqrt{3}$.
Step 2 Multiply: $\tfrac{1}{2} \times \sqrt{3} = \dfrac{\sqrt{3}}{2}$.
Worked Example 4.2 — Sketch a Transformed Graph
Describe how $y = 3\sin(2x) + 1$ differs from $y = \sin x$.
Step 1 The factor $3$ stretches the graph vertically by scale factor $3$ (amplitude becomes $3$).
Step 2 The factor $2$ inside the argument compresses the graph horizontally by factor $\tfrac{1}{2}$, giving period $\tfrac{360°}{2} = 180°$.
Step 3 The $+1$ translates the graph $1$ unit upwards, so the range becomes $[-2, 4]$.
Practice 4.1
Without a calculator, evaluate $\cos^2 45° + \sin^2 60°$. Give your answer in simplified surd form.
Show Solution
$\sin^2 60° = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}$
$\cos^2 45° + \sin^2 60° = \dfrac{1}{2} + \dfrac{3}{4} = \dfrac{2}{4} + \dfrac{3}{4} = \dfrac{5}{4}$
2. Trigonometric Identities
Fundamental Pythagorean Identity
For all values of $\theta$:
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing through by $\cos^2\theta$ gives: $\tan^2\theta + 1 = \sec^2\theta$ Year 2
Dividing through by $\sin^2\theta$ gives: $1 + \cot^2\theta = \csc^2\theta$ Year 2
Quotient Identity
$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$
This follows directly from the definitions of sin, cos, and tan in terms of a right-angled triangle.
Simplifying Expressions
Use the identities to rewrite expressions in simpler or equivalent forms. The most common strategy is to express everything in terms of $\sin$ and $\cos$, then apply the Pythagorean identity.
Worked Example 4.3 — Simplifying
Simplify $\dfrac{1 - \cos^2\theta}{\sin\theta}$.
Step 1 Use $\sin^2\theta + \cos^2\theta = 1$, so $1 - \cos^2\theta = \sin^2\theta$.
Step 2 Substitute: $\dfrac{\sin^2\theta}{\sin\theta} = \sin\theta$.
Proving Identities
To prove an identity, work on one side only (usually the more complex side) and transform it until it equals the other side. Never cross-multiply or assume the identity is true.
Worked Example 4.4 — Proving an Identity
Prove that $\dfrac{\sin\theta}{1 - \cos\theta} - \dfrac{1 + \cos\theta}{\sin\theta} = 0$.
Step 1 Write the left-hand side with a common denominator: $\sin\theta(1-\cos\theta)$.
Step 2 Numerator becomes $\sin^2\theta - (1-\cos\theta)(1+\cos\theta) = \sin^2\theta - (1 - \cos^2\theta)$.
Step 3 Since $1 - \cos^2\theta = \sin^2\theta$, the numerator becomes $\sin^2\theta - \sin^2\theta = 0$.
Step 4 Therefore the left-hand side equals $0$ = right-hand side. QED.
Practice 4.2
Prove that $(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta$.
Show Solution
$(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$
Using $\sin^2\theta + \cos^2\theta = 1$:
$= 1 + 2\sin\theta\cos\theta$ = right-hand side. QED.
3. Solving Trigonometric Equations
The CAST Diagram
The CAST diagram helps identify all angles in $[0°, 360°]$ with a given trig value:
- Cosine is positive in the 4th quadrant ($270°$–$360°$).
- All are positive in the 1st quadrant ($0°$–$90°$).
- Sine is positive in the 2nd quadrant ($90°$–$180°$).
- Tangent is positive in the 3rd quadrant ($180°$–$270°$).
General Solutions
For equations valid for all $\theta \in \mathbb{R}$:
- $\sin\theta = k$: principal value $\alpha = \arcsin k$; general solution $\theta = \alpha + 360°n$ or $\theta = (180° - \alpha) + 360°n$.
- $\cos\theta = k$: principal value $\alpha = \arccos k$; general solution $\theta = \pm\alpha + 360°n$.
- $\tan\theta = k$: principal value $\alpha = \arctan k$; general solution $\theta = \alpha + 180°n$.
Worked Example 4.5 — Solving $\sin\theta = k$
Solve $\sin\theta = \tfrac{\sqrt{3}}{2}$ for $0° \leq \theta \leq 360°$.
Step 1 Find the principal value: $\theta = \arcsin\!\left(\tfrac{\sqrt{3}}{2}\right) = 60°$.
Step 2 Sine is positive in the 1st and 2nd quadrants, so the second solution is $180° - 60° = 120°$.
Answer $\theta = 60°$ or $\theta = 120°$.
Worked Example 4.6 — Equation with Transformation
Solve $2\cos(2\theta - 30°) = \sqrt{2}$ for $0° \leq \theta \leq 360°$.
Step 1 Rearrange: $\cos(2\theta - 30°) = \dfrac{\sqrt{2}}{2}$.
Step 2 Let $u = 2\theta - 30°$. The range for $u$ is $-30° \leq u \leq 690°$.
Step 3 Principal value: $u = \arccos\!\left(\tfrac{\sqrt{2}}{2}\right) = 45°$. Cosine is also positive in the 4th quadrant: $u = 360° - 45° = 315°$. Adding $360°$: $u = 45° + 360° = 405°$ and $u = 315° + 360° = 675°$.
Step 4 Solve for $\theta$: $2\theta - 30° \in \{45°, 315°, 405°, 675°\}$, so $2\theta \in \{75°, 345°, 435°, 705°\}$, thus $\theta \in \{37.5°, 172.5°, 217.5°, 352.5°\}$.
Practice 4.3
Solve $\tan\theta = -1$ for $0° \leq \theta \leq 360°$.
Show Solution
Tangent has period $180°$, so the second solution is $135° + 180° = 315°$.
$\theta = 135°$ or $\theta = 315°$.
4. Radians, Arc Length and Sector Area
Definition: Radian
One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. Thus $2\pi$ radians $= 360°$, giving:
$$1 \text{ rad} = \frac{180°}{\pi} \approx 57.3°, \qquad 1° = \frac{\pi}{180} \text{ rad}$$
Key Conversions
| Degrees | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ | $180°$ | $270°$ | $360°$ |
|---|---|---|---|---|---|---|---|---|
| Radians | $0$ | $\tfrac{\pi}{6}$ | $\tfrac{\pi}{4}$ | $\tfrac{\pi}{3}$ | $\tfrac{\pi}{2}$ | $\pi$ | $\tfrac{3\pi}{2}$ | $2\pi$ |
Arc Length and Sector Area
For a sector of a circle with radius $r$ and central angle $\theta$ in radians:
$$\text{Arc length:} \quad s = r\theta$$
$$\text{Sector area:} \quad A = \tfrac{1}{2}r^2\theta$$
Small Angle Approximations
When $\theta$ is small and measured in radians:
$$\sin\theta \approx \theta, \qquad \cos\theta \approx 1 - \frac{\theta^2}{2}, \qquad \tan\theta \approx \theta$$Exam Tip
Small angle approximations are only valid when $\theta$ is measured in radians. Always state this assumption in your working.
Worked Example 4.7 — Arc Length and Sector Area
A sector has radius $8$ cm and central angle $\tfrac{5\pi}{6}$ radians. Find the arc length and sector area.
Step 1 Arc length: $s = r\theta = 8 \times \tfrac{5\pi}{6} = \dfrac{40\pi}{6} = \dfrac{20\pi}{3} \approx 20.9$ cm.
Step 2 Sector area: $A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2} \times 64 \times \tfrac{5\pi}{6} = \dfrac{320\pi}{12} = \dfrac{80\pi}{3} \approx 83.8$ cm².
Worked Example 4.8 — Small Angle Approximation
Using small angle approximations, simplify $\dfrac{\sin 3\theta}{1 - \cos 2\theta}$ for small $\theta$.
Step 1 Approximate: $\sin 3\theta \approx 3\theta$ and $\cos 2\theta \approx 1 - \tfrac{(2\theta)^2}{2} = 1 - 2\theta^2$.
Step 2 Substitute: $\dfrac{3\theta}{1 - (1 - 2\theta^2)} = \dfrac{3\theta}{2\theta^2} = \dfrac{3}{2\theta}$.
Practice 4.4
A sector of a circle has arc length $12$ cm and area $36$ cm². Find the radius and the angle in radians.
Show Solution
Substitute into $A = \tfrac{1}{2}r^2\theta$:
$36 = \tfrac{1}{2}r^2 \cdot \dfrac{12}{r} = 6r$, so $r = 6$ cm.
Then $\theta = \dfrac{12}{6} = 2$ radians.
5. Compound and Double Angle Formulae Year 2
Compound Angle Formulae
$$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$
$$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$
$$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$
Double Angle Formulae
Setting $B = A$ in the compound angle formulae:
$$\sin 2A = 2\sin A \cos A$$
$$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$$
$$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$$
Worked Example 4.9 — Exact Value using Compound Angles
Find the exact value of $\cos 75°$.
Step 1 Write $75° = 45° + 30°$.
Step 2 Apply $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$\cos 75° = \cos 45°\cos 30° - \sin 45°\sin 30° = \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}$
$= \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$.
Worked Example 4.10 — Using Double Angle Formula
Given $\sin\theta = \tfrac{3}{5}$ and $\theta$ is acute, find $\cos 2\theta$.
Step 1 Find $\cos\theta$: using $\sin^2\theta + \cos^2\theta = 1$, $\cos\theta = \tfrac{4}{5}$ (positive since acute).
Step 2 Apply $\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2 \times \tfrac{9}{25} = 1 - \tfrac{18}{25} = \dfrac{7}{25}$.
Practice 4.5
Show that $\cos 2\theta \equiv 1 - 2\sin^2\theta$ by expanding $\cos(\theta + \theta)$.
Show Solution
Using $\cos^2\theta = 1 - \sin^2\theta$:
$= (1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta$. QED.
6. The Harmonic Form Year 2
Any expression of the form $a\sin\theta + b\cos\theta$ can be written as a single sinusoidal function. This is particularly useful for finding maxima and minima, and for solving equations.
Harmonic Form Theorem
$$a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)$$
where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \dfrac{b}{a}$ (with $a > 0$).
Alternatively: $a\sin\theta + b\cos\theta \equiv R\cos(\theta - \beta)$ where $\tan\beta = \dfrac{a}{b}$.
Figure 4.2 — The harmonic form $R\sin(x+\alpha)$ shown alongside $a\sin x + b\cos x$; both are identical curves.
Worked Example 4.11 — Converting to Harmonic Form
Write $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$, where $R > 0$ and $0 < \alpha < 90°$.
Step 1 Find $R$: $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Step 2 Find $\alpha$: $\tan\alpha = \dfrac{4}{3}$, so $\alpha = \arctan\!\left(\tfrac{4}{3}\right) \approx 53.1°$.
Step 3 Write: $3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.1°)$.
Check Maximum value is $R = 5$, occurring when $\theta + \alpha = 90°$, i.e. $\theta \approx 36.9°$.
Worked Example 4.12 — Solving with Harmonic Form
Solve $3\sin\theta + 4\cos\theta = 2$ for $0° \leq \theta \leq 360°$.
Step 1 From Example 4.11: $5\sin(\theta + 53.1°) = 2$, so $\sin(\theta + 53.1°) = 0.4$.
Step 2 Let $u = \theta + 53.1°$; range is $53.1° \leq u \leq 413.1°$.
Step 3 $u = \arcsin(0.4) \approx 23.6°$. This is outside the range. Second solution: $u = 180° - 23.6° = 156.4°$. Also $u = 23.6° + 360° = 383.6°$.
Step 4 $\theta = 156.4° - 53.1° = 103.3°$ and $\theta = 383.6° - 53.1° = 330.5°$.
Practice 4.6
Write $\sin\theta - \sqrt{3}\cos\theta$ in the form $R\sin(\theta - \alpha)$, stating $R$ and $\alpha$ exactly.
Show Solution
Comparing: $R\cos\alpha = 1$ and $R\sin\alpha = \sqrt{3}$.
$R = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$.
$\tan\alpha = \dfrac{\sqrt{3}}{1} = \sqrt{3}$, so $\alpha = 60°$.
Therefore: $\sin\theta - \sqrt{3}\cos\theta = 2\sin(\theta - 60°)$.
7. Reciprocal and Inverse Trig Functions Year 2
Reciprocal Trigonometric Functions
$$\sec\theta = \frac{1}{\cos\theta}, \qquad \csc\theta = \frac{1}{\sin\theta}, \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$$
Note: $\sec\theta$ is undefined when $\cos\theta = 0$; $\csc\theta$ is undefined when $\sin\theta = 0$.
Further Pythagorean Identities
Dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$:
$$\tan^2\theta + 1 = \sec^2\theta \quad \Longleftrightarrow \quad \sec^2\theta - \tan^2\theta = 1$$
Dividing $\sin^2\theta + \cos^2\theta = 1$ by $\sin^2\theta$:
$$1 + \cot^2\theta = \csc^2\theta \quad \Longleftrightarrow \quad \csc^2\theta - \cot^2\theta = 1$$
Inverse Trigonometric Functions
The inverse functions undo the trig operation but require restricted domains to be well-defined:
- $y = \arcsin x$: domain $[-1, 1]$, range $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$.
- $y = \arccos x$: domain $[-1, 1]$, range $[0, \pi]$.
- $y = \arctan x$: domain $\mathbb{R}$, range $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$.
Worked Example 4.13 — Using the sec Identity
Solve $\sec^2\theta - \tan\theta = 1$ for $0 \leq \theta \leq 2\pi$.
Step 1 Replace $\sec^2\theta$ using $\sec^2\theta = 1 + \tan^2\theta$:
$1 + \tan^2\theta - \tan\theta = 1$, i.e. $\tan^2\theta - \tan\theta = 0$.
Step 2 Factorise: $\tan\theta(\tan\theta - 1) = 0$, so $\tan\theta = 0$ or $\tan\theta = 1$.
Step 3 For $\tan\theta = 0$: $\theta = 0, \pi, 2\pi$. For $\tan\theta = 1$: $\theta = \tfrac{\pi}{4}, \tfrac{5\pi}{4}$.
Answer $\theta = 0, \tfrac{\pi}{4}, \pi, \tfrac{5\pi}{4}, 2\pi$.
Practice 4.7
Find the exact value of $\arcsin\!\left(\sin\tfrac{5\pi}{6}\right)$.
Show Solution
The range of $\arcsin$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, so:
$\arcsin\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}$.
8. Practice Problems
Question 1
Find the exact value of $\sin^2 30° + \tan 45° - \cos 60°$.
Show Solution
$\tfrac{1}{4} + 1 - \tfrac{1}{2} = \tfrac{1}{4} + \tfrac{4}{4} - \tfrac{2}{4} = \dfrac{3}{4}$.
Question 2
Prove that $\dfrac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin\theta\cos\theta} \equiv 1 - \tan\theta$.
Show Solution
Question 3
Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $0° \leq \theta \leq 360°$.
Show Solution
$\sin\theta = -\tfrac{1}{2}$: $\theta = 210°$ or $\theta = 330°$.
$\sin\theta = 1$: $\theta = 90°$.
$\theta = 90°, 210°, 330°$.
Question 4
A sector OAB has radius $10$ cm and area $65$ cm². Find the arc length AB and the perimeter of the sector.
Show Solution
Arc length: $s = r\theta = 10 \times 1.3 = 13$ cm.
Perimeter = arc + 2 radii $= 13 + 10 + 10 = 33$ cm.
Question 5
Find the exact value of $\sin 15°$ using $\sin(45° - 30°)$.
Show Solution
$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$.
Question 6
Given $\cos\theta = -\tfrac{5}{13}$ and $\theta$ is obtuse, find $\sin 2\theta$.
Show Solution
$\sin\theta = \sqrt{1 - \tfrac{25}{169}} = \sqrt{\tfrac{144}{169}} = \tfrac{12}{13}$.
$\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \tfrac{12}{13} \cdot \left(-\tfrac{5}{13}\right) = -\dfrac{120}{169}$.
Question 7
Write $5\cos\theta + 12\sin\theta$ in the form $R\cos(\theta - \alpha)$, where $R > 0$ and $0° < \alpha < 90°$. Hence state the maximum value and the value of $\theta$ at which it occurs.
Show Solution
$R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$, so $R\cos\alpha = 5$ and $R\sin\alpha = 12$.
$\tan\alpha = \tfrac{12}{5}$, so $\alpha = \arctan(2.4) \approx 67.4°$.
Maximum value is $13$, occurring when $\theta - \alpha = 0$, i.e. $\theta \approx 67.4°$.
Question 8
Solve $\csc^2\theta - \cot\theta - 3 = 0$ for $0 \leq \theta \leq 2\pi$. Give answers to 3 significant figures.
Show Solution
$1 + \cot^2\theta - \cot\theta - 3 = 0 \Rightarrow \cot^2\theta - \cot\theta - 2 = 0$.
Factorise: $(\cot\theta - 2)(\cot\theta + 1) = 0$.
$\cot\theta = 2 \Rightarrow \tan\theta = \tfrac{1}{2}$: $\theta = 0.464$ rad or $\theta = \pi + 0.464 = 3.61$ rad.
$\cot\theta = -1 \Rightarrow \tan\theta = -1$: $\theta = \pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4} \approx 2.36$ rad or $\theta = 2\pi - \tfrac{\pi}{4} = \tfrac{7\pi}{4} \approx 5.50$ rad.
$\theta \approx 0.464, 2.36, 3.61, 5.50$ rad.
Question 9
Using small angle approximations, show that $\dfrac{\sin\theta\tan\theta}{1 - \cos\theta} \approx 2$ for small $\theta$.
Show Solution
Numerator: $\sin\theta\tan\theta \approx \theta \cdot \theta = \theta^2$.
Denominator: $1 - \cos\theta \approx 1 - \left(1 - \tfrac{\theta^2}{2}\right) = \tfrac{\theta^2}{2}$.
$\dfrac{\theta^2}{\theta^2/2} = 2$. QED.
Question 10
Find the exact value of $\arctan(\tan\tfrac{7\pi}{6})$.
Show Solution
The range of $\arctan$ is $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, so:
$\arctan\!\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$.