Notes · MathematicsUK · A-Levels

Trigonometry

Trigonometry at A-Level covers radian measure with arc length and sector area, the graphs and exact values of the trigonometric functions, the standard identities, the reciprocal and inverse functions, the compound- and double-angle formulae with the harmonic (R) form, and the solution of trigonometric equations. It supplies the periodic-function toolkit used across calculus, mechanics and modelling.

5 sections~11 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 2

T·0666 / 20

Exam profile

AO1 · Use trigonometric ratios, identities and formulae to solve equationsAO2 · Prove trigonometric identities and reason about periodic functionsAO3 · Model periodic phenomena using trigonometric functions

Operators:solveproveshow thatsketchexpressfind

basic level

AS-Level covers radians, arc and sector, the three ratios, their graphs and exact values, and the identities $\sin^2 + \cos^2 = 1$ and $\tan = \sin/\cos$ .

higher level

The full A-Level adds reciprocal and inverse functions, compound- and double-angle formulae, the $R\sin(x + \alpha)$ form and small-angle approximations.

Depth

Text

Contents · 5 sections

Trigonometry

Radians, arcs and sectors#

FoundationDfE GCE Mathematics E1AQA 7357 E

A circular sector

Fig. 1The sector has arc length

r\theta

and area

\tfrac{1}{2}r^2\theta

when

\theta

is in radians.

A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. Because the full circumference is

2\pi r

, a full turn is

2\pi

radians, giving the conversion

180^\circ = \pi

radians. Radians are the natural unit for calculus: the derivative of

\sin x

\cos x

only when

x

is in radians, so from A-Level onwards angles default to radians unless degrees are stated.

With the angle

\theta

in radians, the two circle-measurement formulae become strikingly simple. The arc length is

s = r\theta

and the area of the sector is

A = \tfrac{1}{2}r^2\theta

. These replace the clumsy 'fraction of 360' versions and follow from them: an arc is the fraction

\dfrac{\theta}{2\pi}

of the circumference

2\pi r

, which simplifies to

r\theta

You should know the exact radian values of the common angles:

30^\circ = \tfrac{\pi}{6}

45^\circ = \tfrac{\pi}{4}

60^\circ = \tfrac{\pi}{3}

90^\circ = \tfrac{\pi}{2}

, and be able to convert both ways fluently. Mixing radian formulae with an angle left in degrees is a frequent and heavily penalised error, so always check the mode.

Sector and segment problems combine these formulae with triangle work. The area of a segment is the sector area minus the area of the triangle,

\tfrac{1}{2}r^2\theta - \tfrac{1}{2}r^2\sin\theta = \tfrac{1}{2}r^2(\theta - \sin\theta)

, and the perimeter of a sector is two radii plus the arc,

2r + r\theta

. Reading whether a question wants an arc, a perimeter, a sector area or a segment area is half the battle.

s = r\theta, \qquad A = \tfrac{1}{2}r^2\theta

Arc length and sector area (theta in radians)

Both hold only when $\theta$ is measured in radians.

180^\circ = \pi \ \text{radians}

Degree-radian conversion

Multiply degrees by $\pi/180$ to get radians, or radians by $180/\pi$ to get degrees.

Worked example

Arc, area and segment

A sector has radius 6 cm and angle 1.2 radians. Find the arc length, the sector area, and the area of the segment cut off by the chord.

Arc length
$s = r\theta = 6 \times 1.2 = 7.2$ cm.
Sector area
$A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(36)(1.2) = 21.6$ cm $^2$ .
Segment area
$\tfrac{1}{2}r^2(\theta - \sin\theta) = \tfrac{1}{2}(36)(1.2 - \sin 1.2) = 18(1.2 - 0.932) = 4.82$ cm $^2$ (3 s.f.).
$\tfrac{1}{2}(36)(1.2 - \sin 1.2) \approx 4.82$

Result: Arc $7.2$ cm, sector area $21.6$ cm $^2$ , segment area $\approx 4.82$ cm $^2$ .

Typical mistakes

Applying $s = r\theta$ or $A = \tfrac{1}{2}r^2\theta$ with $\theta$ in degrees.
Confusing arc length with sector perimeter (which also includes the two radii).

Active revision

A sector of a circle of radius 8 cm has angle $\tfrac{3\pi}{4}$ radians. Find its arc length, perimeter and area.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Trigonometric graphs and exact values#

StandardDfE GCE Mathematics E2AQA 7357 E

The sine and cosine graphs

Fig. 2

y = \sin x

and

y = \cos x

have period

2\pi

; cosine leads sine by

\tfrac{\pi}{2}

The graphs of

y = \sin x

and

y = \cos x

are periodic with period

2\pi

and oscillate between

-1

and

1

;

\cos x

is the

\sin

graph translated

\tfrac{\pi}{2}

to the left. The graph of

y = \tan x

has period

\pi

and vertical asymptotes at

x = \tfrac{\pi}{2} + n\pi

, where

\cos x = 0

. Knowing these shapes lets you predict how many solutions a trigonometric equation has in a given interval.

The symmetries of the graphs encode useful relationships:

\sin

is an odd function,

\sin(-x) = -\sin x

, while

\cos

is even,

\cos(-x) = \cos x

. The graphs also give the standard 'related angle' results such as

\sin(\pi - x) = \sin x

and

\cos(\pi - x) = -\cos x

, which are indispensable when finding all solutions of an equation.

Exact values for the special angles must be memorised. In particular

\sin 30^\circ = \tfrac{1}{2}

\cos 30^\circ = \tfrac{\sqrt{3}}{2}

\tan 30^\circ = \tfrac{1}{\sqrt{3}}

\sin 45^\circ = \cos 45^\circ = \tfrac{1}{\sqrt{2}}

, and

\sin 60^\circ = \tfrac{\sqrt{3}}{2}

. These come from the 1-1-

\sqrt{2}

and 1-2-

\sqrt{3}

right triangles and are expected in exact-value questions with no calculator credit for decimals.

Transformations apply to trigonometric graphs as to any others:

y = a\sin(bx) + c

has amplitude

|a|

, period

\dfrac{2\pi}{b}

and is shifted vertically by

c

. Recognising amplitude, period and shift lets you both sketch and model oscillations such as tides or alternating current.

\sin 30^\circ = \tfrac{1}{2},\ \cos 30^\circ = \tfrac{\sqrt3}{2},\ \tan 45^\circ = 1,\ \sin 60^\circ = \tfrac{\sqrt3}{2}

Exact values

Read from the two standard right-angled triangles; required in exact form.

Worked example

Reading amplitude and period

State the amplitude and period of $y = 3\cos(2x)$ and its maximum value.

Amplitude
The amplitude is $|a| = 3$ .
Period
The period is $\dfrac{2\pi}{b} = \dfrac{2\pi}{2} = \pi$ .
Maximum
The greatest value of $\cos$ is 1, so the maximum of $y$ is $3$ .

Result: Amplitude 3, period $\pi$ , maximum value 3.

Typical mistakes

Confusing the period of $\tan x$ ( $\pi$ ) with that of $\sin x$ and $\cos x$ ( $2\pi$ ).
Mixing up which of $\sin$ and $\cos$ is odd and which is even.

Active revision

Sketch $y = 2\sin\left(x + \tfrac{\pi}{3}\right)$ for $0 \leq x \leq 2\pi$ , stating the amplitude, period and coordinates of the maximum.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Identities and reciprocal and inverse functions#

StandardDfE GCE Mathematics E5-E7AQA 7357 E

The tangent function and its asymptotes

Fig. 3

y = \tan x

has period

\pi

and vertical asymptotes where

\cos x = 0

Two identities underlie almost all trigonometric manipulation:

\sin^2\theta + \cos^2\theta = 1

(from Pythagoras on the unit circle) and

\tan\theta = \dfrac{\sin\theta}{\cos\theta}

. Dividing the first by

\cos^2\theta

or by

\sin^2\theta

gives the two derived identities

1 + \tan^2\theta = \sec^2\theta

and

1 + \cot^2\theta = \csc^2\theta

, which are the standard tools for problems involving the reciprocal functions.

The reciprocal functions are defined as

\sec\theta = \dfrac{1}{\cos\theta}

\csc\theta = \dfrac{1}{\sin\theta}

and

\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}

. Each is undefined wherever its denominator is zero, giving vertical asymptotes, and each inherits the periodicity of its parent function. The memory aid is that the third letter of

\sec

and

\csc

points to

\cos

and

\sin

respectively.

Proving an identity means transforming one side into the other by legitimate steps, never treating it as an equation to solve. The reliable strategy is to convert everything to sines and cosines, combine over a common denominator, and use

\sin^2 + \cos^2 = 1

to simplify. You work on the more complicated side until it matches the simpler one.

The inverse functions

\arcsin

\arccos

and

\arctan

reverse the trigonometric functions, but only on restricted domains where the parent function is one-to-one:

\arcsin

and

\arctan

take values in

\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]

and

\arccos

[0, \pi]

. These restrictions are why a calculator returns only a principal value, and why you must generate the other solutions of an equation yourself.

\sin^2\theta + \cos^2\theta = 1

The Pythagorean identity

Holds for every $\theta$ ; the foundation of trigonometric proof.

1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta

Derived identities

Obtained by dividing the Pythagorean identity by $\cos^2\theta$ and by $\sin^2\theta$ .

Worked example

Proving an identity

Prove that $\sec^2\theta + \csc^2\theta \equiv \sec^2\theta\,\csc^2\theta$ .

Convert to sines and cosines
$\sec^2\theta + \csc^2\theta = \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta}$ .
Common denominator
$= \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta}$ using $\sin^2 + \cos^2 = 1$ .
$\frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}$
Rewrite as reciprocals
$\dfrac{1}{\sin^2\theta\cos^2\theta} = \dfrac{1}{\cos^2\theta}\cdot\dfrac{1}{\sin^2\theta} = \sec^2\theta\,\csc^2\theta$ .

Result: Both sides equal $\dfrac{1}{\sin^2\theta\cos^2\theta}$ , so the identity holds.

Typical mistakes

Writing $\sec\theta = \dfrac{1}{\sin\theta}$ (it is $\dfrac{1}{\cos\theta}$ ).
Assuming the calculator's inverse value is the only solution of an equation.

Active revision

Prove the identity $\dfrac{1}{1 - \sin\theta} + \dfrac{1}{1 + \sin\theta} \equiv 2\sec^2\theta$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Compound- and double-angle formulae#

AdvancedDfE GCE Mathematics E8-E9AQA 7357 E

Combining two sinusoids into one

Fig. 4

3\sin x + 4\cos x = 5\sin(x + \alpha)

has amplitude

R = \sqrt{3^2 + 4^2} = 5

The compound-angle (addition) formulae express the trigonometric functions of a sum or difference of angles:

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

and

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

, with the important sign switch in the cosine formula. From these follows

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

. They let you find exact values of angles such as

75^\circ = 45^\circ + 30^\circ

and are the source of every other identity in this section.

Setting

B = A

gives the double-angle formulae:

\sin 2A = 2\sin A\cos A

and

\cos 2A = \cos^2 A - \sin^2 A

, the latter also written as

2\cos^2 A - 1

1 - 2\sin^2 A

using the Pythagorean identity. These three forms of

\cos 2A

are chosen to match whatever else is in a problem, and they are the standard route to integrating

\cos^2 x

\sin^2 x

later.

A power of the method is the harmonic form: any expression

a\sin\theta + b\cos\theta

can be written as a single sinusoid

R\sin(\theta + \alpha)

, where

R = \sqrt{a^2 + b^2}

and

\tan\alpha = \dfrac{b}{a}

. This immediately gives the maximum (

R

) and minimum (

-R

) of the expression and turns an awkward two-term equation into a single-term one that is easy to solve.

Finally, for very small angles measured in radians the approximations

\sin\theta \approx \theta

\tan\theta \approx \theta

and

\cos\theta \approx 1 - \tfrac{1}{2}\theta^2

hold, coming from the first terms of the series expansions. They are used to simplify limits and to model situations such as a pendulum swinging through a small angle.

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

Compound-angle formula for sine

The corresponding cosine formula switches the sign: $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$ .

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

The harmonic (R) form

Combines two terms into one sinusoid, revealing amplitude $R$ and phase $\alpha$ .

Worked example

Exact value from a compound angle

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

Write 15 as a difference
$15^\circ = 45^\circ - 30^\circ$ , so $\cos 15^\circ = \cos(45^\circ - 30^\circ)$ .
Apply the formula
$\cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ = \dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2} + \dfrac{1}{\sqrt2}\cdot\dfrac{1}{2}$ .
$\cos 15^\circ = \frac{\sqrt3}{2\sqrt2} + \frac{1}{2\sqrt2}$
Simplify
$= \dfrac{\sqrt3 + 1}{2\sqrt2} = \dfrac{\sqrt6 + \sqrt2}{4}$ .

Result: $\cos 15^\circ = \dfrac{\sqrt6 + \sqrt2}{4}$ .

Typical mistakes

Getting the sign wrong in $\cos(A + B) = \cos A\cos B - \sin A\sin B$ .
Using degrees inside the small-angle approximations, which require radians.

Active revision

Express $7\sin\theta - 24\cos\theta$ in the form $R\sin(\theta - \alpha)$ , and hence find its maximum value and the smallest positive $\theta$ at which it occurs.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Solving trigonometric equations#

AdvancedDfE GCE Mathematics E7AQA 7357 E

Multiple solutions in an interval

Fig. 5

\sin x = \tfrac{1}{2}

has two solutions in

[0, 2\pi]

x = \tfrac{\pi}{6}

and

x = \tfrac{5\pi}{6}

Solving a trigonometric equation over a given interval means finding every value that works, not just the one the calculator returns. The principal value from

\arcsin

\arccos

\arctan

is one solution; the others are generated using the symmetry of the graph — for sine, a second solution at

\pi -

(principal value), and then all solutions differ by the period (

2\pi

for

\sin

and

\cos

\pi

for

\tan

When the argument is a multiple or shift of the variable, such as

\sin(2x + \tfrac{\pi}{6})

, widen the working interval to match the argument before solving. If

0 \leq x \leq 2\pi

then

2x + \tfrac{\pi}{6}

ranges over

\tfrac{\pi}{6} \leq \cdot \leq 4\pi + \tfrac{\pi}{6}

, so you must find all solutions of the inner equation across that wider range and only then solve back for

x

. Forgetting to widen the interval is the single biggest cause of lost solutions.

Many equations first need an identity to reduce them to a single function. An equation mixing

\sin

and

\cos

, or containing

\cos 2x

, is rearranged with a double-angle formula into a quadratic in one ratio; for example

2\cos 2x + 1 = 3\sin x

becomes, via

\cos 2x = 1 - 2\sin^2 x

, a quadratic in

\sin x

. Solve the quadratic, then solve each resulting simple equation.

Always finish by checking that each solution lies in the required interval and by discarding any that arose spuriously (for instance from squaring). Presenting the full solution set, in the units the question uses, is where the final method and accuracy marks are earned.

\sin x = k \ \Rightarrow\ x = \arcsin k \ \text{or}\ x = \pi - \arcsin k \ (+\,2\pi n)

All solutions for sine

The two base solutions per period come from the symmetry of the sine graph.

Worked example

A quadratic in cosine

Solve $2\cos^2 x - \cos x - 1 = 0$ for $0 \leq x \leq 2\pi$ .

Factorise as a quadratic
Let $c = \cos x$ : $2c^2 - c - 1 = (2c + 1)(c - 1) = 0$ , so $\cos x = -\tfrac{1}{2}$ or $\cos x = 1$ .
Solve cos x = 1
$x = 0$ (and $x = 2\pi$ is the endpoint).
Solve cos x = -1/2
Principal value $\arccos(-\tfrac{1}{2}) = \tfrac{2\pi}{3}$ ; the second is $2\pi - \tfrac{2\pi}{3} = \tfrac{4\pi}{3}$ .

Result: $x = 0,\ \tfrac{2\pi}{3},\ \tfrac{4\pi}{3},\ 2\pi$ .

Typical mistakes

Giving only the principal value and missing the other solutions in the interval.
Failing to extend the interval for the argument, so multiple-angle solutions are lost.

Active revision

Solve $3\tan^2 x - \tan x - 2 = 0$ for $0 \leq x \leq 2\pi$ , giving answers to 3 significant figures.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Trigonometry

Key points

Arc, area and segment

Key points

Reading amplitude and period

Key points

Proving an identity

Key points

Exact value from a compound angle

Key points

A quadratic in cosine

Sources