Add Maths · 10 Trigonometry

10 Trigonometry

Before you start

Sine, cosine and tangent in a right-angled triangle (SOHCAHTOA), and Pythagoras' theorem.
The exact values for $30 °$ , $45 °$ and $60 °$ , from the half-square and half-equilateral triangles: $\sin 30 ° = \frac{1}{2}$ , $\cos 30 ° = \frac{\sqrt{3}}{2}$ , $\tan 30 ° = \frac{1}{\sqrt{3}}$ ; $\sin 45 ° = \cos 45 ° = \frac{1}{\sqrt{2}}$ , $\tan 45 ° = 1$ ; $\sin 60 ° = \frac{\sqrt{3}}{2}$ , $\cos 60 ° = \frac{1}{2}$ , $\tan 60 ° = \sqrt{3}$ . They are not on the List of formulas.
Radians from topic 9: $π$ radians $= 180 °$ , so $\frac{π}{6} = 30 °$ , $\frac{π}{4} = 45 °$ , $\frac{π}{3} = 60 °$ and $\frac{π}{2} = 90 °$ .
Solving a quadratic equation by factorising, or by the quadratic formula (which is on the List of formulas).
Adding algebraic fractions over a common denominator, and the difference of two squares $(1 - s) (1 + s) = 1 - s^{2}$ .

Syllabus objectives covered (6)

Know and use the six trigonometric functions of angles of any magnitude.
Understand and use the amplitude and period of a trigonometric function, including the relationship between graphs of related trigonometric functions.
Draw and use the graphs of y = a sin bx + c y = a cos bx + c y = a tan bx + c where a is a positive integer, b is a simple fraction or integer, and c is an integer.
Use the relationships: • sin² A + cos² A = 1 • sec² A = 1 + tan² A • cosec² A = 1 + cot² A
Solve, for a given domain, trigonometric equations involving the six trigonometric functions.
Prove trigonometric relationships involving the six trigonometric functions.

Part 1 of 8

Angles of any size: the unit circle

SOHCAHTOA works only inside a right-angled triangle, where every angle is less than $90 °$ . To give $\sin 150 °$ or $\cos (- 60 °)$ a meaning, you need a definition that works for any angle.
Draw a circle of radius $1$ with its centre at the origin. Start on the positive $x$ -axis and turn through the angle $θ$ : anticlockwise for a positive angle, clockwise for a negative one. Call the point you reach $P$ . Its coordinates define cosine and sine, and tangent is their quotient.
$\cos θ = x -coordinate of P, \sin θ = y -coordinate of P, \tan θ = \frac{\sin θ}{\cos θ}$
graph diagram — not drawn yet
Square-scaled axes for −1.5 ≤ x ≤ 1.5 and −1.5 ≤ y ≤ 1.5 with the unit circle (centre the origin O, radius 1). Draw the radius OP at 150° from the positive x-axis, mark θ = 150° with an anticlockwise arc from the positive x-axis, drop a dashed perpendicular from P to the x-axis, label P with (cos θ, sin θ) ≈ (−0.87, 0.5), and mark the 30° basic angle between OP and the negative x-axis. In each quadrant write the letter of the functions that are positive there: A (top right), S (top left), T (bottom left), C (bottom right).
$P$ is at $150 °$ . Its $y$ -coordinate, $\sin 150 °$ , is positive; its $x$ -coordinate, $\cos 150 °$ , is negative. The basic angle is $30 °$ .
For an acute angle this is the SOHCAHTOA you already know. The radius is the hypotenuse and has length $1$ , so the opposite side is $\sin θ$ and the adjacent side is $\cos θ$ . The new definition simply keeps going past $90 °$ .
The sign comes from the quadrant $P$ is in. In the first quadrant ( $0 °$ to $90 °$ ) $x$ and $y$ are both positive, so all three functions are positive. In the second ( $90 °$ to $180 °$ ) $x$ is negative and $y$ is positive, so only sine is positive.
In the third quadrant ( $180 °$ to $270 °$ ) $x$ and $y$ are both negative, so sine and cosine are negative and tangent, a negative divided by a negative, is positive. In the fourth ( $270 °$ to $360 °$ ) only cosine is positive. Going anticlockwise from the first quadrant that spells All, Sine, Tangent, Cosine: the CAST diagram.
The size comes from the basic angle: the acute angle between $O P$ and the $x$ -axis. Any trig function of $θ$ equals its value at the basic angle, with the sign of the quadrant attached. Between $0 °$ and $360 °$ the basic angle is $θ$ , $180 ° - θ$ , $θ - 180 °$ or $360 ° - θ$ , quadrant by quadrant.
Three examples. $\sin 150 °$ : basic angle $30 °$ , second quadrant, sine positive, so $\sin 150 ° = \frac{1}{2}$ . $\cos 240 °$ : basic angle $60 °$ , third quadrant, cosine negative, so $\cos 240 ° = - \frac{1}{2}$ . $\tan (- 45 °)$ : turn $45 °$ clockwise into the fourth quadrant, where tangent is negative, so $\tan (- 45 °) = - 1$ .
A whole turn brings $P$ back to where it started, so adding or subtracting $360 °$ never changes sine, cosine or tangent: $\cos 420 ° = \cos 60 ° = \frac{1}{2}$ . Radians work the same way, with $2 π$ for a whole turn and $π$ for a half turn. $\frac{5 π}{6}$ is $150 °$ , so its basic angle is $π - \frac{5 π}{6} = \frac{π}{6}$ and $\sin \frac{5 π}{6} = \frac{1}{2}$ .

Part 2 of 8

The other three: sec, cosec and cot

The other three of the six functions are reciprocals of the first three.
$\sec θ = \frac{1}{\cos θ}, cosec θ = \frac{1}{\sin θ}, \cot θ = \frac{1}{\tan θ} = \frac{\cos θ}{\sin θ}$
To pair them up, look at the third letter of each name: sec goes with cos, cosec with sin, cot with tan. The pairing that looks natural, sec with sin, is the wrong one.
A reciprocal has the same sign as the function underneath it, so CAST still gives the sign. $\sec 120 ° = \frac{1}{\cos 120 °} = \frac{1}{- \frac{1}{2}} = - 2$ , and $\cot 150 ° = \frac{1}{\tan 150 °} = \frac{1}{- \frac{1}{\sqrt{3}}} = - \sqrt{3}$ .
A reciprocal does not exist where the function underneath is $0$ . $\cos 90 ° = 0$ , so $\sec 90 °$ is undefined; $\sin 180 ° = 0$ , so $cosec 180 °$ is undefined. Their graphs have vertical asymptotes at those angles, and no solution of an equation in $\sec$ or $cosec$ can sit there.
Your calculator has no sec, cosec or cot key. Work out the partner and take its reciprocal: $cosec 50 ° = 1 \div \sin 50 ° = 1.305 \dots$ . To solve, turn the equation over first: $\sec θ = 3$ means $\cos θ = \frac{1}{3}$ .
The $\sin^{- 1}$ key is the inverse function: it gives the angle whose sine is a number, so $\sin^{- 1} 0.5 = 30 °$ . It is not the reciprocal. The reciprocal of $\sin 30 °$ is $cosec 30 ° = 2$ .
Given one function and the quadrant, you can find all the others without finding the angle. Draw a right-angled triangle for the basic angle using sizes only, find the third side by Pythagoras, then attach each sign from CAST.
Say $\sin θ = \frac{3}{5}$ and $θ$ is obtuse. The triangle has opposite $3$ and hypotenuse $5$ , so the adjacent side is $\sqrt{25 - 9} = 4$ . An obtuse angle is in the second quadrant, where only sine, and so cosec, is positive.
$\cos θ = - \frac{4}{5}, \tan θ = - \frac{3}{4}, \sec θ = - \frac{5}{4}, cosec θ = \frac{5}{3}, \cot θ = - \frac{4}{3}$

Part 3 of 8

Period, amplitude, and how related graphs compare

Turn $P$ steadily round the circle and plot its height against the angle: that is $y = \sin x$ . It starts at $0$ , rises to $1$ at $90 °$ , is back to $0$ at $180 °$ , reaches $- 1$ at $270 °$ and returns to $0$ at $360 °$ . Plotting the $x$ -coordinate of $P$ instead gives $y = \cos x$ , which starts at $1$ .
graph diagram — not drawn yet
Axes for 0° ≤ x ≤ 360° with ticks every 90° labelled in both degrees and radians (0, π/2, π, 3π/2, 2π), and −1.5 ≤ y ≤ 1.5 with ticks at −1, 0 and 1. Solid curve y = sin x through (0, 0), (90°, 1), (180°, 0), (270°, −1), (360°, 0); dashed curve y = cos x through (0, 1), (90°, 0), (180°, −1), (270°, 0), (360°, 1). Label each curve with its equation.
Both waves repeat every $360 °$ ( $2 π$ ) and stay between $- 1$ and $1$ . The cosine wave is the sine wave moved $90 °$ to the left.
The period is the horizontal length after which a graph repeats exactly. For $\sin x$ and $\cos x$ it is $360 °$ , or $2 π$ radians, because that is one full turn of $P$ .
The amplitude is the greatest distance of the curve from its centre line, which is half of (maximum $-$ minimum). For $\sin x$ and $\cos x$ the centre line is $y = 0$ and the amplitude is $1$ .
$y = \tan x$ behaves differently. It repeats every $180 °$ ( $π$ ), because $P$ and the point directly opposite it give the same quotient $\frac{y}{x}$ . Where $\cos x = 0$ , at $x = 90 °$ and $x = 270 °$ , $\tan x$ is undefined and the graph has vertical asymptotes. It has no maximum or minimum, so it has no amplitude.
graph diagram — not drawn yet
Axes for 0° ≤ x ≤ 360° with ticks every 45° (labels at 90°, 180°, 270°, 360°) and −4 ≤ y ≤ 4. Curve y = tan x in three branches: from (0, 0) through (45°, 1) rising towards the dashed vertical asymptote x = 90°; a middle branch rising from far below next to x = 90° through (135°, −1), (180°, 0), (225°, 1) towards the dashed asymptote x = 270°; a last branch rising from far below next to x = 270° through (315°, −1) to (360°, 0). Label the asymptotes x = 90° and x = 270°.
Tangent repeats every $180 °$ , is $0$ at $0 °$ , $180 °$ and $360 °$ , and has an asymptote wherever $\cos x = 0$ .
Now the three numbers in $y = a \sin b x + c$ . Multiplying by $a$ stretches the graph vertically: every height is multiplied by $a$ , so the amplitude becomes $a$ .
Replacing $x$ by $b x$ squeezes the graph horizontally. $\sin b x$ completes a cycle when $b x$ has gone through $360 °$ , which happens when $x$ has gone through only $\frac{360 °}{b}$ .
$period of \sin b x and \cos b x = \frac{360 °}{b} = \frac{2 π}{b}, period of \tan b x = \frac{180 °}{b} = \frac{π}{b}$
Adding $c$ moves the whole graph up by $c$ (down if $c$ is negative). The centre line becomes $y = c$ , so a sine or cosine curve has maximum $c + a$ and minimum $c - a$ . The amplitude stays $a$ .
So $y = \sin x$ and $y = 3 \sin 2 x$ are related like this: $y = 3 \sin 2 x$ has amplitude $3$ instead of $1$ and period $180 °$ instead of $360 °$ . It fits two waves, each three times as tall, into the space where $\sin x$ fits one. A fraction for $b$ works the other way: $y = \cos \frac{x}{2}$ has period $\frac{360 °}{1 / 2} = 720 °$ , so it is stretched out to half as many waves.
The same thinking links the six functions to each other. $y = \cos x$ is $y = \sin x$ moved $90 °$ to the left, so $\cos x = \sin (x + 90 °)$ . $y = \sec x$ is large in size where $\cos x$ is close to $0$ : it equals $1$ where $\cos x = 1$ , equals $- 1$ where $\cos x = - 1$ , and has an asymptote wherever $\cos x = 0$ . It also has period $360 °$ , the same as $\cos x$ .

Part 4 of 8

Sketching the graphs, and reading from them

A sketch needs the right shape and the key points labelled, not graph paper. Before you draw, work out four things: the centre line $y = c$ , the maximum $c + a$ and minimum $c - a$ , the period, and how many periods fit into the domain you are given.
Split each period into quarters. A sine curve goes centre line, maximum, centre line, minimum, centre line at the five quarter points. A cosine curve goes maximum, centre line, minimum, centre line, maximum.
Take $y = 2 \sin 3 x - 1$ for $0 ° \leq x \leq 120 °$ . Centre line $y = - 1$ , maximum $1$ , minimum $- 3$ , period $\frac{360 °}{3} = 120 °$ . The domain is exactly one cycle and the quarter points are $30 °$ apart.
$(0 °, - 1), (30 °, 1), (60 °, - 1), (90 °, - 3), (120 °, - 1)$
Join those points with a smooth wave, rounded at the top and bottom, never pointed. Label the maximum and minimum with their coordinates. If the curve crosses an axis, find and label the crossings: here $2 \sin 3 x - 1 = 0$ gives $\sin 3 x = \frac{1}{2}$ , so $3 x = 30 °$ or $150 °$ , and the curve cuts the $x$ -axis at $x = 10 °$ and $x = 50 °$ .
For $y = a \tan b x + c$ , find the asymptotes first. Tangent is undefined when its angle is $90 °$ , $270 °$ and so on, so solve $b x = 90 °$ , $b x = 270 °$ , … for $x$ . Label the $x$ -coordinate of every asymptote in the domain: the syllabus requires the asymptote's $x$ -coordinate to be clearly labelled.
Take $y = \tan 2 x + 1$ for $0 ° \leq x \leq 180 °$ . The period is $\frac{180 °}{2} = 90 °$ . The asymptotes are where $2 x = 90 °$ and $2 x = 270 °$ , so at $x = 45 °$ and $x = 135 °$ . Where $\tan 2 x = 0$ , at $x = 0 °$ , $90 °$ and $180 °$ , the curve passes through $y = 1$ , and every branch rises from left to right.
graph diagram — not drawn yet
Axes for 0° ≤ x ≤ 180° with ticks every 22.5° (labels at 45°, 90°, 135°, 180°) and −4 ≤ y ≤ 6. Dashed vertical asymptotes at x = 45° and x = 135°, each labelled with its equation. Curve y = tan 2x + 1 in three branches: from (0, 1) through (22.5°, 2) rising towards x = 45°; from far below next to x = 45° through (67.5°, 0), (90°, 1), (112.5°, 2) rising towards x = 135°; from far below next to x = 135° through (157.5°, 0) to (180°, 1). Dots at (0, 1), (90°, 1) and (180°, 1), and the x-intercepts (67.5°, 0) and (157.5°, 0) labelled.
Period $90 °$ : the asymptotes are $90 °$ apart at $x = 45 °$ and $x = 135 °$ , and each branch is the tangent shape lifted by $1$ .
Using a graph: the number of times the curve meets the horizontal line $y = k$ is the number of solutions of $f (x) = k$ in the domain. On $y = 2 \sin 3 x - 1$ for $0 ° \leq x \leq 120 °$ , the line $y = 0$ crosses twice, so $2 \sin 3 x - 1 = 0$ has two solutions. The line $y = 1$ only touches the top, so $2 \sin 3 x - 1 = 1$ has one. The line $y = 2$ is above the maximum, so $2 \sin 3 x - 1 = 2$ has none.
You can also read $a$ , $b$ and $c$ off a graph by running the rules backwards: $c$ is halfway between the maximum and minimum, $a$ is half their difference, and $b = \frac{360 °}{period}$ . A sine curve with maximum $5$ , minimum $- 1$ and period $120 °$ has $c = 2$ , $a = 3$ and $b = 3$ : it is $y = 3 \sin 3 x + 2$ .

Part 5 of 8

The three identities and where they come from

$P$ lies on a circle of radius $1$ centred at the origin, so by Pythagoras its coordinates satisfy $x^{2} + y^{2} = 1$ . Its coordinates are $(\cos A, \sin A)$ , which gives the first identity, true for every angle $A$ .
$\sin^{2} A + \cos^{2} A = 1$
$\sin^{2} A$ means ${(\sin A)}^{2}$ : find the sine, then square it. It is not $\sin (A^{2})$ .
Divide every term by $\cos^{2} A$ . Because $\frac{\sin A}{\cos A} = \tan A$ and $\frac{1}{\cos A} = \sec A$ , this gives the second identity.
$\frac{\sin^{2} A}{\cos^{2} A} + \frac{\cos^{2} A}{\cos^{2} A} = \frac{1}{\cos^{2} A} \Rightarrow \tan^{2} A + 1 = \sec^{2} A$
Divide the first identity by $\sin^{2} A$ instead, and $\frac{\cos A}{\sin A} = \cot A$ gives the third.
$1 + \cot^{2} A = {cosec}^{2} A$
All three are on the List of formulas, written as $\sin^{2} A + \cos^{2} A = 1$ , $\sec^{2} A = 1 + \tan^{2} A$ and ${cosec}^{2} A = 1 + \cot^{2} A$ . The List does not give $\tan A = \frac{\sin A}{\cos A}$ or the definitions of sec, cosec and cot, so those you must know.
What you actually use are the rearranged forms: $\cos^{2} A = 1 - \sin^{2} A$ , $\sin^{2} A = 1 - \cos^{2} A$ , $\tan^{2} A = \sec^{2} A - 1$ and $\cot^{2} A = {cosec}^{2} A - 1$ . Find the squared term you want to get rid of, and swap it for the function that matches the rest of the expression.
Example: write $3 \cos^{2} A + 2 \sin^{2} A$ in terms of $\sin A$ only. The unwanted term is $\cos^{2} A$ , so replace it by $1 - \sin^{2} A$ .
$3 (1 - \sin^{2} A) + 2 \sin^{2} A = 3 - 3 \sin^{2} A + 2 \sin^{2} A = 3 - \sin^{2} A$
The identities also find one function from another without the angle. If $\sec A = - 3$ and $A$ is obtuse, then $\tan^{2} A = \sec^{2} A - 1 = 9 - 1 = 8$ , so $\tan A = \pm 2 \sqrt{2}$ . Tangent is negative in the second quadrant, so $\tan A = - 2 \sqrt{2}$ . An equation such as $\tan^{2} A = 8$ always has two roots, one of each sign; the quadrant picks one.

Part 6 of 8

Solving an equation: every solution in the domain

An equation such as $\sin x = 0.4$ has infinitely many solutions, because the sine wave keeps crossing the line $y = 0.4$ . The question gives a domain, such as $0 ° \leq x \leq 360 °$ , and you must find every solution inside it and none outside it.
First get a single trig function on its own. Set your calculator to degrees or radians to match the domain. Then find the principal value, the one angle your calculator gives: $\sin^{- 1} 0.4 = 23.578 \dots °$ . For $\sin^{- 1}$ and $\tan^{- 1}$ it lies between $- 90 °$ and $90 °$ ; for $\cos^{- 1}$ it lies between $0 °$ and $180 °$ .
Find the second angle from the symmetry of the graph, where $α$ is the principal value.
$\sin x = k : x = α or 180 ° - α, \cos x = k : x = α or 360 ° - α, \tan x = k : x = α or α + 180 °$
graph diagram — not drawn yet
Axes for 0° ≤ x ≤ 360° (ticks every 90°) and −1.5 ≤ y ≤ 1.5, with the curve y = sin x and a dashed horizontal line y = 0.4 across the domain. Dots at the two intersections, with dashed verticals down to the x-axis labelled α ≈ 23.6° and 180° − α ≈ 156.4°; a short double-headed arrow from 0° to α and another of the same length from 156.4° to 180°. Label the curve y = sin x and the line y = 0.4.
The sine curve is symmetrical about $x = 90 °$ , so the second solution is as far below $180 °$ as the first is above $0 °$ .
Then add or subtract the period ( $360 °$ for sine and cosine, $180 °$ for tangent) as often as the domain needs, and discard anything outside it. For $\sin x = 0.4$ in $0 ° \leq x \leq 360 °$ the answers are $23.6 °$ and $180 ° - 23.6 ° = 156.4 °$ ; adding $360 °$ to either goes past the end of the domain.
A negative principal value works the same way, but it is itself outside $0 °$ to $360 °$ . For $\tan x = - 2$ , $\tan^{- 1} (- 2) = - 63.4 °$ . Add $180 °$ to get $116.6 °$ , and again to get $296.6 °$ . Those two are the solutions in $0 ° \leq x \leq 360 °$ ; $- 63.4 °$ is not one of them.
Reciprocal functions: turn them over first. $cosec x = 4$ becomes $\sin x = \frac{1}{4}$ , and $\cot x = 3$ becomes $\tan x = \frac{1}{3}$ . Then solve as before.
Multiple angles. For $\cos 2 x = 0.5$ with $0 ° \leq x \leq 360 °$ , call the whole angle $u = 2 x$ . If $x$ runs from $0 °$ to $360 °$ , then $u$ runs from $0 °$ to $720 °$ , so find every $u$ in that wider range before you halve.
$u = 60 °, 300 °, 420 °, 660 ° \Rightarrow x = 30 °, 150 °, 210 °, 330 °$
Changing the domain first is what keeps every solution: a period of $180 °$ fits into $360 °$ twice, so $\cos 2 x = k$ has twice as many solutions as $\cos x = k$ . A shift works the same way. For $\sin (x + 30 °)$ with $0 ° \leq x \leq 360 °$ , the angle $x + 30 °$ runs from $30 °$ to $390 °$ ; find the angles in that range, then subtract $30 °$ .
With a radian domain, work in radians throughout and give radian answers. Without a calculator you can finish only when the value is a known exact value, such as $\frac{1}{2}$ , $\frac{\sqrt{3}}{2}$ or $\sqrt{3}$ , so give the angles exactly, for example $\frac{π}{4}$ . On the calculator paper, give angles in degrees to $1$ decimal place and other non-exact answers, including radians, to $3$ significant figures, unless the question says otherwise.

Part 7 of 8

Equations that need an identity or a factorisation first

When an equation mixes two functions, use an identity or a definition to turn it into one function. Choose by what is squared: $\cos^{2} x$ beside $\sin x$ becomes $1 - \sin^{2} x$ ; $\sec^{2} x$ beside $\tan x$ becomes $1 + \tan^{2} x$ ; ${cosec}^{2} x$ beside $\cot x$ becomes $1 + \cot^{2} x$ .
Example: solve $2 \cos^{2} x + 3 \sin x = 3$ for $0 ° \leq x \leq 360 °$ . Replace $\cos^{2} x$ by $1 - \sin^{2} x$ and collect everything on one side: $2 - 2 \sin^{2} x + 3 \sin x - 3 = 0$ . Multiply by $- 1$ so the squared term is positive.
$2 \sin^{2} x - 3 \sin x + 1 = 0$
This is a quadratic in $\sin x$ . Factorise it exactly as you would $2 s^{2} - 3 s + 1$ : $(2 \sin x - 1) (\sin x - 1) = 0$ , so $\sin x = \frac{1}{2}$ or $\sin x = 1$ . Solve each: $\sin x = \frac{1}{2}$ gives $x = 30 °, 150 °$ , and $\sin x = 1$ gives $x = 90 °$ .
Check each root is possible before solving it. $\sin x$ and $\cos x$ always lie between $- 1$ and $1$ , so a root such as $\cos x = 3$ gives no solutions: say so and reject it. $\sec x$ and $cosec x$ never lie strictly between $- 1$ and $1$ . $\tan x$ and $\cot x$ can take any value.
Two functions with no squares, such as $5 \sin θ + 2 \cos θ = 0$ , become one by dividing by $\cos θ$ : $5 \tan θ + 2 = 0$ , so $\tan θ = - \frac{2}{5}$ . That division is safe because $\cos θ = 0$ would force $\sin θ = 0$ as well, and $\sin^{2} θ + \cos^{2} θ = 1$ says both cannot be $0$ at once.
Reciprocals clear by multiplying. $3 \cot θ = \tan θ$ is $\frac{3}{\tan θ} = \tan θ$ ; multiply by $\tan θ$ to get $\tan^{2} θ = 3$ , so $\tan θ = \sqrt{3}$ or $\tan θ = - \sqrt{3}$ . Keep both signs: in $0 °$ to $360 °$ they give $60 °, 240 °$ and $120 °, 300 °$ .
Never divide by a trig function unless you have shown it cannot be zero at a solution; factorise instead. In $\sin x \cos x = \frac{1}{2} \sin x$ , dividing by $\sin x$ throws away the solutions of $\sin x = 0$ . Write $\sin x (\cos x - \frac{1}{2}) = 0$ instead, so $\sin x = 0$ or $\cos x = \frac{1}{2}$ . In $0 ° \leq x \leq 360 °$ that gives $0 °, 180 °, 360 °$ and $60 °, 300 °$ .

Part 8 of 8

Proving an identity

An identity is true for every angle at which both sides are defined. The symbol $\equiv$ means exactly that. Proving one means starting from one side and changing it, one valid step at a time, until it is exactly the other side.
Do not treat it as an equation and do the same thing to both sides, or cross-multiply. That starts from the statement you are trying to prove, so it proves nothing.
Start from the more complicated side, the one with fractions, reciprocals or more terms, because simplifying is easier than complicating. Write $LHS = \dots$ and finish with $\dots = RHS$ .
Three moves do almost all the work: write sec, cosec, cot and tan in terms of sin and cos; put fractions over a common denominator; and use $\sin^{2} A + \cos^{2} A = 1$ , in either direction, whenever a square appears.
Example: prove that $\frac{\sec θ - \cos θ}{\tan θ} \equiv \sin θ$ . Start on the left, the complicated side, and write everything in sin and cos.
$LHS = \frac{\frac{1}{\cos θ} - \cos θ}{\frac{\sin θ}{\cos θ}}$
Multiply the top and the bottom by $\cos θ$ to clear the small fractions. The top becomes $1 - \cos^{2} θ$ , which the first identity turns into $\sin^{2} θ$ .
$= \frac{1 - \cos^{2} θ}{\sin θ} = \frac{\sin^{2} θ}{\sin θ} = \sin θ = RHS$
Each line makes one visible change. Because the result is printed for you, the marks are for the steps that reach it: show the identity you used, and do not jump from the start to the answer.
In this lesson's exam questions a proof is set as Show (that), which means provide structured evidence that leads to a given result. If you get stuck halfway, rewrite the other side in sin and cos too and aim for the same middle expression; then write the two chains as one.

Worked examples

Example 1. For $0 ° \leq x \leq 180 °$ , the curve $C$ has equation $y = 3 \cos 2 x + 1$ . (a) Write down the amplitude and the period. (b) Sketch $C$ , labelling its maximum and minimum points and where it meets the axes. (c) Determine the number of solutions of $3 \cos 2 x + 1 = 2$ for $0 ° \leq x \leq 180 °$ .

Show the working

(a) Read off $a = 3$ , $b = 2$ and $c = 1$ . The amplitude is $a = 3$ . The period is $\frac{360 °}{2} = 180 °$ , so the domain is exactly one full cycle.
(b) The centre line is $y = 1$ , the maximum is $1 + 3 = 4$ and the minimum is $1 - 3 = - 2$ . A cosine curve starts at its maximum. Split the period into quarters of $45 °$ .
$(0 °, 4), (45 °, 1), (90 °, - 2), (135 °, 1), (180 °, 4)$
The minimum $- 2$ is below the $x$ -axis, so the curve crosses it. Solve $3 \cos 2 x + 1 = 0$ : $\cos 2 x = - \frac{1}{3}$ . With $u = 2 x$ and $0 ° \leq u \leq 360 °$ , $u = 109.47 °$ or $360 ° - 109.47 ° = 250.53 °$ , so $x = 54.7 °$ or $x = 125.3 °$ .
It meets the $y$ -axis at $(0 °, 4)$ . Draw one smooth wave from $(0 °, 4)$ down through $(54.7 °, 0)$ to the minimum $(90 °, - 2)$ , and back up through $(125.3 °, 0)$ to $(180 °, 4)$ .
graph diagram — not drawn yet
Axes for 0° ≤ x ≤ 180° with ticks every 45° and −3 ≤ y ≤ 5. Solid smooth curve y = 3cos 2x + 1 through (0, 4), (45°, 1), (90°, −2), (135°, 1), (180°, 4). Dashed horizontal line y = 2 crossing the curve twice (near x = 35.3° and x = 144.7°). Dots with coordinate labels at (0, 4), (90°, −2) and (180°, 4); the x-intercepts labelled 54.7° and 125.3°.
One full cycle of $y = 3 \cos 2 x + 1$ . The line $y = 2$ crosses it twice, so $3 \cos 2 x + 1 = 2$ has two solutions.
(c) The solutions of $3 \cos 2 x + 1 = 2$ are where $C$ meets the line $y = 2$ . That line lies between the minimum $- 2$ and the maximum $4$ , and in one cycle it crosses the curve once on the way down and once on the way up.
To be certain, solve it: $\cos 2 x = \frac{1}{3}$ , so $2 x = 70.5 °$ or $289.5 °$ , and $x = 35.3 °$ or $x = 144.7 °$ . Both lie in the domain, which confirms the count.

Answer: (a) Amplitude $3$ , period $180 °$ . (b) A cosine wave with maximum points $(0 °, 4)$ and $(180 °, 4)$ , minimum point $(90 °, - 2)$ , crossing the $x$ -axis at $x = 54.7 °$ and $x = 125.3 °$ . (c) Two solutions.

Example 2. Solve $3 \sec^{2} θ - 5 \tan θ - 1 = 0$ for $0 ° \leq θ \leq 360 °$ .

Show the working

There are two functions, and the squared one is $\sec^{2} θ$ . The identity $\sec^{2} θ = 1 + \tan^{2} θ$ swaps it for tangent, which matches the other term.
$3 (1 + \tan^{2} θ) - 5 \tan θ - 1 = 0$
Expand and collect: $3 + 3 \tan^{2} θ - 5 \tan θ - 1 = 0$ .
$3 \tan^{2} θ - 5 \tan θ + 2 = 0$
Factorise exactly as you would $3 t^{2} - 5 t + 2$ .
$(3 \tan θ - 2) (\tan θ - 1) = 0 \Rightarrow \tan θ = \frac{2}{3} or \tan θ = 1$
$\tan θ = 1$ : the principal value is $45 °$ and tangent repeats every $180 °$ , so $θ = 45 °$ or $225 °$ .
$\tan θ = \frac{2}{3}$ : $\tan^{- 1} \frac{2}{3} = 33.69 °$ , so $θ = 33.7 °$ or $33.69 ° + 180 ° = 213.69 °$ , which is $213.7 °$ . Add the $180 °$ to the unrounded value, then round.
$\sec θ$ is undefined only at $90 °$ and $270 °$ , and none of the four answers is there, so all four stand. List them in order.

Answer: $θ = 33.7 °, 45 °, 213.7 °, 225 °$

Example 3. Solve $cosec (2 x + \frac{π}{3}) = 2$ for $0 \leq x \leq π$ , giving your answers in terms of $π$ .

Show the working

Turn the reciprocal over: $\sin (2 x + \frac{π}{3}) = \frac{1}{2}$ .
Call the whole angle $u = 2 x + \frac{π}{3}$ and change the domain to match. When $x = 0$ , $u = \frac{π}{3}$ ; when $x = π$ , $u = 2 π + \frac{π}{3} = \frac{7 π}{3}$ .
$\frac{π}{3} \leq u \leq \frac{7 π}{3}$
Solve $\sin u = \frac{1}{2}$ . The basic angle is $\frac{π}{6}$ , and sine is positive in the first and second quadrants, so $u = \frac{π}{6}$ or $π - \frac{π}{6} = \frac{5 π}{6}$ . Adding $2 π$ gives $\frac{13 π}{6}$ and $\frac{17 π}{6}$ .
Keep only the values between $\frac{π}{3} = \frac{2 π}{6}$ and $\frac{7 π}{3} = \frac{14 π}{6}$ . That leaves $u = \frac{5 π}{6}$ and $u = \frac{13 π}{6}$ : $\frac{π}{6}$ is too small and $\frac{17 π}{6}$ is too large.
Undo the shift, then the multiple: subtract $\frac{π}{3}$ , then halve.
$2 x = \frac{5 π}{6} - \frac{2 π}{6} = \frac{π}{2} \Rightarrow x = \frac{π}{4}, 2 x = \frac{13 π}{6} - \frac{2 π}{6} = \frac{11 π}{6} \Rightarrow x = \frac{11 π}{12}$
Check one: $x = \frac{11 π}{12}$ gives $u = \frac{11 π}{6} + \frac{2 π}{6} = \frac{13 π}{6}$ , and $\sin \frac{13 π}{6} = \sin \frac{π}{6} = \frac{1}{2}$ , so $cosec u = 2$ .

Answer: $x = \frac{π}{4}$ or $x = \frac{11 π}{12}$

Example 4. Show that $\frac{\cos θ}{1 - \sin θ} - \frac{\cos θ}{1 + \sin θ} \equiv 2 \tan θ$ .

Show the working

The left-hand side has the fractions, so start there. Combine them over the common denominator $(1 - \sin θ) (1 + \sin θ)$ .
$LHS = \frac{\cos θ (1 + \sin θ) - \cos θ (1 - \sin θ)}{(1 - \sin θ) (1 + \sin θ)}$
Expand the top. The minus sign applies to both terms of the second bracket: $\cos θ + \sin θ \cos θ - \cos θ + \sin θ \cos θ = 2 \sin θ \cos θ$ .
The bottom is a difference of two squares, $1 - \sin^{2} θ$ , and the first identity turns it into $\cos^{2} θ$ .
$LHS = \frac{2 \sin θ \cos θ}{\cos^{2} θ}$
Cancel one $\cos θ$ and recognise $\frac{\sin θ}{\cos θ}$ .
$= \frac{2 \sin θ}{\cos θ} = 2 \tan θ = RHS$

Answer: The left-hand side simplifies to $2 \tan θ$ , which is the right-hand side, so the identity is proved.

Where people go wrong

Mistake: Taking $\sec θ$ to mean $\frac{1}{\sin θ}$ , pairing the functions by their first letters.

Instead: $\sec θ = \frac{1}{\cos θ}$ and $cosec θ = \frac{1}{\sin θ}$ . The third letter names the partner: sec–cos, cosec–sin, cot–tan.

Every later step then uses the wrong function, so you end up solving a different equation.

Mistake: Using the $\sin^{- 1}$ key to work out $cosec x$ .

Instead: $cosec x = 1 \div \sin x$ . The $\sin^{- 1}$ key gives the angle that has a given sine.

The $^{- 1}$ on the key means the inverse function, not the reciprocal: $\sin^{- 1} 0.5 = 30 °$ , but $\frac{1}{\sin 30 °} = 2$ .

Mistake: Solving $\cos 2 x = 0.5$ for $0 ° \leq x \leq 360 °$ by finding $2 x = 60 °, 300 °$ and stopping, which gives only $x = 30 °, 150 °$ .

Instead: Change the domain first: $0 ° \leq 2 x \leq 720 °$ , so $2 x = 60 °, 300 °, 420 °, 660 °$ and $x = 30 °, 150 °, 210 °, 330 °$ .

Halving the period doubles the number of cycles inside the domain, so there are twice as many solutions.

Mistake: Dividing both sides of $\sin x \cos x = \frac{1}{2} \sin x$ by $\sin x$ .

Instead: Factorise: $\sin x (\cos x - \frac{1}{2}) = 0$ , and solve $\sin x = 0$ as well as $\cos x = \frac{1}{2}$ .

Dividing by an expression that can be zero throws away the solutions where it is zero, here $0 °$ , $180 °$ and $360 °$ .

Mistake: Proving an identity by treating it as an equation: doing the same thing to both sides, or cross-multiplying.

Instead: Start from one side, the more complicated one, and change it step by step until it is exactly the other side.

Working on both sides at once begins by assuming the identity is true, which is the very thing you have to show.

Check yourself

Without a calculator, find the exact values of $\cos 300 °$ , $\tan 225 °$ and $cosec \frac{7 π}{6}$ .
Answer
$\cos 300 ° = \frac{1}{2}$ (basic angle $60 °$ , fourth quadrant, cosine positive). $\tan 225 ° = 1$ (basic angle $45 °$ , third quadrant, tangent positive). $\frac{7 π}{6}$ is $210 °$ , where $\sin 210 ° = - \frac{1}{2}$ , so $cosec \frac{7 π}{6} = - 2$ .
Given that $\cos A = \frac{5}{13}$ and $270 ° < A < 360 °$ , find the exact values of $\sin A$ , $\tan A$ and $\cot A$ .
Answer
$\sin^{2} A = 1 - \frac{25}{169} = \frac{144}{169}$ . Sine is negative in the fourth quadrant, so $\sin A = - \frac{12}{13}$ . Then $\tan A = \frac{\sin A}{\cos A} = - \frac{12}{5}$ and $\cot A = - \frac{5}{12}$ .
State the period of each graph, and its amplitude if it has one: (a) $y = 5 \sin \frac{x}{3}$ , in degrees; (b) $y = 2 \tan 4 x + 3$ , in radians.
Answer
(a) Amplitude $5$ ; period $\frac{360 °}{1 / 3} = 1080 °$ . (b) A tangent graph has no maximum or minimum, so no amplitude; period $\frac{π}{4}$ .
The curve $y = a \cos b x + c$ , where $a$ and $b$ are positive, has a maximum value of $7$ , a minimum value of $- 1$ and period $\frac{π}{2}$ . Find $a$ , $b$ and $c$ .
Answer
$c = \frac{7 + (- 1)}{2} = 3$ , $a = \frac{7 - (- 1)}{2} = 4$ , and $\frac{2 π}{b} = \frac{π}{2}$ gives $b = 4$ . The curve is $y = 4 \cos 4 x + 3$ .
Write down the equations of the asymptotes of $y = 3 \tan \frac{x}{2} - 1$ for $- 2 π \leq x \leq 2 π$ .
Answer
Tangent is undefined when $\frac{x}{2} = \pm \frac{π}{2}$ ; the next ones, $\frac{x}{2} = \pm \frac{3 π}{2}$ , are outside the domain. So the asymptotes are $x = - π$ and $x = π$ .
Solve $2 \sin (x - \frac{π}{6}) + 1 = 0$ for $0 \leq x \leq 2 π$ .
Answer
$\sin u = - \frac{1}{2}$ with $u = x - \frac{π}{6}$ , and the domain becomes $- \frac{π}{6} \leq u \leq \frac{11 π}{6}$ . In that range $u = - \frac{π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$ , so $x = 0, \frac{4 π}{3}, 2 π$ . Both end points of the domain are solutions.
Solve $9 \cot θ = \tan θ$ for $0 ° < θ < 360 °$ .
Answer
Multiply $\frac{9}{\tan θ} = \tan θ$ by $\tan θ$ : $\tan^{2} θ = 9$ , so $\tan θ = 3$ or $\tan θ = - 3$ . $\tan^{- 1} 3 = 71.57 °$ , so $θ = 71.6 °, 108.4 °, 251.6 °, 288.4 °$ .
Show that $\frac{1 + \tan^{2} θ}{1 + \cot^{2} θ} \equiv \tan^{2} θ$ .
Answer
$LHS = \frac{\sec^{2} θ}{{cosec}^{2} θ} = \frac{1 / \cos^{2} θ}{1 / \sin^{2} θ} = \frac{\sin^{2} θ}{\cos^{2} θ} = \tan^{2} θ = RHS$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The curve $C$ has equation $y = 3 \sin 2 x - 1$ for $0 \leq x \leq π$ .

(a) Write down the amplitude and the period of $y = 3 \sin 2 x - 1$ . 2 marks No calculator

Mark scheme

B1 amplitude 3
B1 period π (accept 180°)

(b) Sketch $C$ , stating the coordinates of its maximum point and its minimum point. 3 marks No calculator

Mark scheme

B1 one complete sine-shaped cycle, starting at (0, −1), rising first, and ending at (π, −1)
B1 maximum point (π/4, 2)
B1 minimum point (3π/4, −4)

(c) State the number of solutions of $3 \sin 2 x - 1 = - 2$ for $0 \leq x \leq π$ . 1 mark No calculator

Mark scheme

B1 2

Question 2.

Solve $6 \sin^{2} x + \cos x - 4 = 0$ for $0 ° \leq x \leq 360 °$ . 5 marks

Mark scheme

M1 uses sin²x = 1 − cos²x to obtain an equation in cos x only
A1 6cos²x − cos x − 2 = 0 oe, three terms equal to 0
M1 solves their three-term quadratic in cos x, e.g. (3cos x − 2)(2cos x + 1) = 0
A1 awrt 48.2° and 311.8°
A1 120° and 240°, and no others in the range

Question 3.

(a) Show that $\frac{\sec θ}{\cot θ + \tan θ} \equiv \sin θ$ . 4 marks No calculator

Mark scheme

M1 cot θ and tan θ written as cos θ/sin θ and sin θ/cos θ (or another valid first step, e.g. numerator and denominator multiplied by tan θ to give sec θ tan θ/(1 + tan²θ))
M1 combined over the common denominator sin θ cos θ: (cos²θ + sin²θ)/(sin θ cos θ), or equivalent progress on an alternative route
A1 sin²θ + cos²θ = 1 used to give 1/(sin θ cos θ) (on the alternative route, 1 + tan²θ = sec²θ used to give tan θ/sec θ)
A1 (1/cos θ) × sin θ cos θ = sin θ, given result reached with all steps shown; nfww

(b) Hence solve $\frac{\sec θ}{\cot θ + \tan θ} = 3 \cos θ$ for $0 ° < θ < 360 °$ . 3 marks

Mark scheme

M1 uses (a) to obtain sin θ = 3cos θ
M1 divides by cos θ to obtain tan θ = 3 oe
A1 awrt 71.6° and 251.6°, and no others in the range

How this is examined

Solving an equation that needs an identity first, then finding every angle in the domain: a Solve over a stated domain where two functions are mixed, such as $\sec^{2}$ with $\tan$ , ${cosec}^{2}$ with $\cot$ , or $\sin^{2}$ with $\cos$ ; 5–6 marks The correct identity substituted (M), a three-term quadratic in one function, equal to $0$ (A), a valid method of solving it (M), then for each possible root every angle in the domain: the calculator value and the second angle from symmetry. Impossible roots such as $\cos x = 3$ rejected; angles in degrees to $1$ decimal place on the calculator paper.
Proving an identity one visible step at a time: a Show (that) involving sec, cosec, cot or two algebraic fractions; 3–5 marks One side only transformed; reciprocals rewritten in sin and cos; fractions combined over a common denominator; $\sin^{2} θ + \cos^{2} θ = 1$ (or a rearrangement) visibly used; the last line exactly the other side. A step that jumps to the printed answer cannot earn the final mark.
Multiple or shifted angles: changing the domain before solving: a Solve such as $cosec (2 x + \frac{π}{3}) = 2$ or $\tan 3 x = k$ over a domain in degrees or radians; 3–5 marks The reciprocal turned over; the domain rewritten for the whole angle; every value of the whole angle in that range; then the shift undone before dividing by the multiple. Radian answers stay in radians, given exactly in terms of $π$ when no calculator is allowed.
Sketching a trig graph with its key features labelled: a Sketch of $y = a \sin b x + c$ , $y = a \cos b x + c$ or $y = a \tan b x + c$ over a given domain in degrees or radians, which may come with a Write down for the amplitude and period; 2–4 marks The right number of cycles for the domain, the correct value at the start, maximum and minimum at $c + a$ and $c - a$ in the right places, and for tangent every asymptote in the domain drawn with its $x$ -coordinate labelled.
Exact values from one given ratio and a quadrant: a non-calculator Find the exact value of one function, given another and a range such as $90 ° < θ < 180 °$ ; 2–3 marks An identity or a right-angled triangle giving the size; the sign chosen from the quadrant, not left as $\pm$ ; the answer as an exact fraction or surd such as $\frac{2 \sqrt{5}}{5}$ , not a decimal.
Using your graph to count solutions: a short State or Determine: how many solutions $f (x) = k$ has in the domain; 1 mark The number of times your curve meets the horizontal line $y = k$ , which must agree with your own sketch.

Words to know

Unit circle: The circle of radius $1$ centred at the origin. The point at angle $θ$ on it has coordinates $(\cos θ, \sin θ)$ .
Basic angle: The acute angle between the radius and the $x$ -axis. A trig function of any angle equals plus or minus its value at the basic angle.
CAST diagram: The four quadrants labelled with the positive functions, anticlockwise from the first: All, Sine, Tangent, Cosine.
Reciprocal functions: $\sec θ = \frac{1}{\cos θ}$ , $cosec θ = \frac{1}{\sin θ}$ and $\cot θ = \frac{1}{\tan θ}$ .
Period: The horizontal length after which a graph repeats exactly: $\frac{360 °}{b}$ for $\sin b x$ and $\cos b x$ , $\frac{180 °}{b}$ for $\tan b x$ .
Amplitude: The greatest distance of a sine or cosine curve from its centre line, half of (maximum $-$ minimum); $a$ in $y = a \sin b x + c$ . Tangent has none.
Asymptote: A line that a graph gets closer and closer to as it heads off towards infinity. For $y = \tan x$ these are the vertical lines where $\cos x = 0$ , which the graph never reaches.
Principal value: The single angle your calculator gives for $\sin^{- 1}$ , $\cos^{- 1}$ or $\tan^{- 1}$ , from which the other solutions are found.
Identity: A statement, written with $\equiv$ , that is true for every value of the variable at which both sides are defined.

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