Trigonometry Overview – A Level Maths Year 2

1 The Big Idea

Year 2 trigonometry (DfE content area E · Trigonometry) does two things. It switches the unit of angle from degrees to radians, which is the natural unit for calculus. And it gives you a much bigger toolkit of identities, so that almost any trigonometric expression can be rewritten in a form you can solve, differentiate or integrate.

Nearly every question in this topic is the same game: rewrite the expression until it contains only one trig function of one angle, then solve. The skill is choosing which identity gets you there.

What is new compared with Year 1

Radians, arcs and sectors · small angle approximations
$\sec$, $\operatorname{cosec}$, $\cot$ and $\arcsin$, $\arccos$, $\arctan$
Compound and double angle formulae · $R\cos(\theta \pm \alpha)$ · proof and modelling

Calculators are allowed, but check the mode every time: RAD when the interval is in radians, DEG when you see degree signs. When a question asks for exact answers or says "show detailed reasoning", you must use exact values and identities, not decimal answers from a solver.

2 The Language You Need

Word	Meaning	Example
Radian	Angle subtended by an arc equal in length to the radius.	$\pi$ rad $= 180^\circ$
Sector / segment	Region bounded by two radii and an arc / by a chord and an arc.	Sector area $\tfrac12r^2\theta$
Reciprocal function	One over a trig function.	$\sec x = \dfrac{1}{\cos x}$
Inverse function	Gives the angle from the ratio, on a restricted range.	$\arcsin\tfrac12 = \tfrac{\pi}{6}$
Identity	True for every value of the variable; written with $\equiv$.	$1 + \tan^2 x \equiv \sec^2 x$
Compound angle	A sum or difference of angles.	$\sin(A + B)$
Amplitude	The height of a wave above its centre line.	$R$ in $R\sin(\theta + \alpha)$
Period	The length of one full cycle.	$\sin 3x$ has period $\tfrac{2\pi}{3}$

Inverse is not reciprocal. $\sin^{-1}x$ means $\arcsin x$, an angle. The reciprocal $\dfrac{1}{\sin x}$ is $\operatorname{cosec} x$. But $\sin^2 x$ does mean $(\sin x)^2$ — the notation is inconsistent, so read carefully.

3 The Toolkit

1 — Radians, arcs and sectors

$\pi$ rad $= 180^\circ$ · $s = r\theta$ · $A = \tfrac12r^2\theta$ · segment $\tfrac12r^2(\theta - \sin\theta)$

2 — Small angles ($\theta$ in radians)

$\sin\theta \approx \theta$, $\tan\theta \approx \theta$, $\cos\theta \approx 1 - \tfrac12\theta^2$

3 — Pythagorean identities

$\sin^2\theta + \cos^2\theta \equiv 1$ · $1 + \tan^2\theta \equiv \sec^2\theta$ · $1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta$

4 — Compound and double angles

$\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B$ · $\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B$
$\sin 2A \equiv 2\sin A\cos A$ · $\cos 2A \equiv 2\cos^2 A - 1 \equiv 1 - 2\sin^2 A$ · $\tan 2A \equiv \dfrac{2\tan A}{1 - \tan^2 A}$

5 — Harmonic form

$a\cos\theta + b\sin\theta \equiv R\cos(\theta - \alpha)$, $R = \sqrt{a^2 + b^2}$, $\tan\alpha = \tfrac{b}{a}$

Worked Example — Several tools in one question

(a) Prove that $\dfrac{\cos 2\theta}{\cos\theta + \sin\theta} \equiv \cos\theta - \sin\theta$. (b) Hence solve $\dfrac{\cos 2\theta}{\cos\theta + \sin\theta} = 1$ for $0 \leqslant \theta \leqslant 2\pi$.

①(a) $\cos 2\theta = \cos^2\theta - \sin^2\theta = (\cos\theta - \sin\theta)(\cos\theta + \sin\theta)$.

②Cancel the factor $(\cos\theta + \sin\theta)$: LHS $= \cos\theta - \sin\theta$ $\blacksquare$

③(b) Solve $\cos\theta - \sin\theta = 1$. Write it as $R\cos(\theta + \alpha)$: $R\cos\alpha = 1$, $R\sin\alpha = 1$, so $R = \sqrt2$, $\alpha = \tfrac{\pi}{4}$.

④$\cos\left(\theta + \tfrac{\pi}{4}\right) = \tfrac{\sqrt2}{2}$, with $\tfrac{\pi}{4} \leqslant \theta + \tfrac{\pi}{4} \leqslant \tfrac{9\pi}{4}$.

⑤$\theta + \tfrac{\pi}{4} = \tfrac{\pi}{4}, \tfrac{7\pi}{4}, \tfrac{9\pi}{4}$, so $\theta = 0, \tfrac{3\pi}{2}, 2\pi$.

⑥Check none makes the original denominator zero ($\cos\theta + \sin\theta = 0$ only at $\tfrac{3\pi}{4}, \tfrac{7\pi}{4}$) ✓

This one question uses a double angle formula, the difference of two squares, the harmonic form, and an equation in radians. Expect that kind of mixing.

4 The Trigonometry Subtopics

All of the Year 2 content in DfE content area E, and the page that covers each part. The Year 1 pages on trigonometric graphs, identities and equations are assumed throughout.

ERadians, Arcs and SectorsRadian measure, exact values, arc length $r\theta$, sector area $\tfrac12r^2\theta$, segments.
ESmall Angle Approximations$\sin\theta \approx \theta$, $\cos\theta \approx 1 - \tfrac12\theta^2$, $\tan\theta \approx \theta$, and using them.
EReciprocal and Inverse Functions$\sec$, $\operatorname{cosec}$, $\cot$; $\arcsin$, $\arccos$, $\arctan$; the two new Pythagorean identities.
ECompound and Double AnglesThe addition formulae, double angle formulae, exact values and equations.
EThe R cos(x + a) FormWriting $a\cos\theta + b\sin\theta$ as a single wave; maxima, minima and equations.
ETrigonometric Proof and ModellingProving identities, solving equations in radians, tides, Ferris wheels and daylight.

5 Common Mistakes to Avoid

Wrong calculator mode. $\sin 2$ in degree mode is $0.0349$; in radians it is $0.9093$. Any radian question answered in degree mode is wrong throughout.

Using $r\theta$ or $\tfrac12r^2\theta$ with $\theta$ in degrees. These formulae are only true in radians.

Splitting a compound angle. $\cos(A + B) \neq \cos A + \cos B$. Use the formula, and remember that the sign in the middle of $\cos(A \pm B)$ is the opposite one.

Dividing by a trig function. In $\sin 2x = \sin x$, dividing by $\sin x$ loses every solution of $\sin x = 0$. Factorise instead.

Forgetting to transform the interval. For $\cos 2x$ with $0 \leqslant x \leqslant 2\pi$ you need $0 \leqslant 2x \leqslant 4\pi$; for $\sin(x + \alpha)$ you need to shift the interval by $\alpha$.

Dropping the cosine term in small angles. $1 - \cos\theta \approx \tfrac12\theta^2$, not $0$.

6 Quick Reference

Radians

$\pi$ rad $= 180^\circ$; $s = r\theta$, $A = \tfrac12r^2\theta$

Small angles

$\sin\theta \approx \tan\theta \approx \theta$; $\cos\theta \approx 1 - \tfrac12\theta^2$

Reciprocals

$\sec = \tfrac{1}{\cos}$, $\operatorname{cosec} = \tfrac{1}{\sin}$, $\cot = \tfrac{1}{\tan}$

Pythagorean

$1 + \tan^2 \equiv \sec^2$; $1 + \cot^2 \equiv \operatorname{cosec}^2$

Inverse ranges

arcsin, arctan: about $0$, up to $\pm\tfrac{\pi}{2}$; arccos: $0$ to $\pi$.

Addition

$\sin(A\pm B)$, $\cos(A\pm B)$, $\tan(A\pm B)$

Double angle

$\cos 2A$ has three forms — pick the one that leaves one function.

Harmonic form

$R = \sqrt{a^2 + b^2}$; max $R$, min $-R$.

Models

$A + B\sin(\omega t + c)$: period $\tfrac{2\pi}{\omega}$.

7 Practice Questions

Ten questions drawing on the whole of Year 2 Trigonometry. Work each one through before opening the solution.

Question 1

Convert $\dfrac{3\pi}{5}$ to degrees and $210^\circ$ to radians.

▶ Show solution

$\dfrac{3 \times 180^\circ}{5} = 108^\circ$. $210 \times \dfrac{\pi}{180} = \dfrac{7\pi}{6}$.

Question 2

A sector has radius $9$ cm and angle $\dfrac{2\pi}{3}$. Find the exact arc length and area.

▶ Show solution

$s = 9 \times \tfrac{2\pi}{3} = 6\pi$ cm. $A = \tfrac12 \times 81 \times \tfrac{2\pi}{3} = 27\pi$ cm$^2$.

Question 3

For small $\theta$, find an approximation for $\dfrac{1 - \cos\theta}{\theta\sin 2\theta}$.

▶ Show solution

$\dfrac{\frac12\theta^2}{\theta \times 2\theta} = \dfrac{\frac12\theta^2}{2\theta^2} = \dfrac14$.

Question 4

Find the exact values of $\sec\dfrac{7\pi}{6}$ and $\arccos\left(-\dfrac{\sqrt3}{2}\right)$.

▶ Show solution

$\cos\tfrac{7\pi}{6} = -\tfrac{\sqrt3}{2}$, so $\sec\tfrac{7\pi}{6} = -\tfrac{2}{\sqrt3} = -\tfrac{2\sqrt3}{3}$.

$\arccos\left(-\tfrac{\sqrt3}{2}\right) = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$ (in the range $0$ to $\pi$).

Question 5

Find the exact value of $\cos 15^\circ$.

▶ Show solution

$\cos(45^\circ - 30^\circ) = \tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2} + \tfrac{\sqrt2}{2}\cdot\tfrac12 = \dfrac{\sqrt6 + \sqrt2}{4}$.

Question 6

Solve $\sec^2 x - 2\tan x = 4$ for $0^\circ \leqslant x \leqslant 360^\circ$.

▶ Show solution

$1 + \tan^2 x - 2\tan x - 4 = 0 \Rightarrow \tan^2 x - 2\tan x - 3 = 0 \Rightarrow (\tan x - 3)(\tan x + 1) = 0$.

$\tan x = 3$: $x = 71.6^\circ, 251.6^\circ$. $\tan x = -1$: $x = 135^\circ, 315^\circ$.

Question 7

Prove that $\sin(A + B)\sin(A - B) \equiv \sin^2 A - \sin^2 B$.

▶ Show solution

LHS $= (\sin A\cos B + \cos A\sin B)(\sin A\cos B - \cos A\sin B) = \sin^2 A\cos^2 B - \cos^2 A\sin^2 B$

$= \sin^2 A(1 - \sin^2 B) - (1 - \sin^2 A)\sin^2 B = \sin^2 A - \sin^2 B$ $\blacksquare$

Question 8

Solve $\cos 2x + \sin x = 0$ for $0 \leqslant x \leqslant 2\pi$.

▶ Show solution

$1 - 2\sin^2 x + \sin x = 0 \Rightarrow 2\sin^2 x - \sin x - 1 = 0 \Rightarrow (2\sin x + 1)(\sin x - 1) = 0$.

$\sin x = 1$: $x = \tfrac{\pi}{2}$. $\sin x = -\tfrac12$: $x = \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$.

Question 9

Express $5\sin x + 12\cos x$ in the form $R\sin(x + \alpha)$ with $0^\circ < \alpha < 90^\circ$, and state its maximum value and the value of $x$ between $0^\circ$ and $360^\circ$ at which it occurs.

▶ Show solution

$R\cos\alpha = 5$, $R\sin\alpha = 12$: $R = 13$, $\alpha = \arctan\tfrac{12}{5} = 67.38^\circ$.

Maximum $13$ when $x + 67.38^\circ = 90^\circ$, i.e. $x = 22.6^\circ$.

Question 10

An alternating voltage is modelled by $V = 3\sin(100\pi t) + 4\cos(100\pi t)$ volts, $t$ seconds after switching on.

(a) Express $V$ in the form $R\sin(100\pi t + \alpha)$, with $0 < \alpha < \tfrac{\pi}{2}$. (b) State the peak voltage and find the first time it occurs. (c) Find the period and frequency. (d) Find the fraction of each cycle for which $V > 4$.

▶ Show solution

(a) $R\cos\alpha = 3$, $R\sin\alpha = 4$: $R = 5$, $\alpha = \arctan\tfrac43 = 0.9273$. $V = 5\sin(100\pi t + 0.9273)$.

(b) Peak $5$ V when $100\pi t + 0.9273 = \tfrac{\pi}{2}$: $t = \dfrac{1.5708 - 0.9273}{100\pi} = 0.00205$ s (3 s.f.).

(c) Period $= \dfrac{2\pi}{100\pi} = \dfrac{1}{50}$ s, so the frequency is $50$ cycles per second ($50$ Hz).

(d) $\sin\phi > 0.8$ where $\phi = 100\pi t + 0.9273$. Over one cycle this holds for $0.9273 < \phi < \pi - 0.9273$, an interval of length $\pi - 1.8546 = 1.2870$.

Fraction $= \dfrac{1.2870}{2\pi} = 0.205$ (3 s.f.) — about a fifth of each cycle.

Trigonometry Overview · A Level Maths Year 2 · Created with MathJax