Year 1 Pure · Chapter 9

Trigonometry

Specification: Edexcel Pure 5 — Trigonometry · AQA section E

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In this chapter

1. The unit circle
2. The graphs
3. Trig identities
4. Solving trig equations
Practice

Key facts — this chapter

On the unit circle, the point at angle $\theta$ is $(\cos\theta, \sin\theta)$ — this defines sin and cos for any angle.
Graphs: $\sin$ and $\cos$ are waves of period $360^\circ$ between $-1$ and $1$ ; $\tan$ has period $180^\circ$ with asymptotes at $90^\circ, 270^\circ, \dots$
Exact values worth memorising: $\sin 30^\circ = \tfrac12$ , $\cos 30^\circ = \tfrac{\sqrt3}{2}$ , $\tan 45^\circ = 1$ , and the rest of the $30/45/60$ family.
Identities: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$ .
Solving on an interval: find the first solution, then use the graph’s symmetry to get the rest within range.
For $\sin(bx)$ or $\cos(bx)$ , widen the interval for $bx$ first, solve there, then divide back.

In a right-angled triangle, $\sin$ , $\cos$ and $\tan$ only make sense for angles between $0^\circ$ and $90^\circ$ . The unit circle frees them — giving every angle a sine and cosine, and turning trig into the wave functions that model anything that repeats.

1. The unit circle

Place a circle of radius $1$ at the origin and mark the point reached by rotating $\theta$ from the positive $x$ -axis.

Key result

The point on the unit circle at angle $\theta$ has coordinates

(\cos\theta,\ \sin\theta).

So $\cos\theta$ is its horizontal position and $\sin\theta$ its vertical position — for any $\theta$ , not just acute ones. This is why sin and cos can be negative: past $90^\circ$ the point moves into other quadrants.

A unit circle with a point P at angle theta, showing cos theta as the horizontal distance and sin theta as the vertical height — cos θ across, sin θ up — the coordinates of the point at angle θ.

Step the angle θ round the unit circle. The point's height is sin θ — watch it trace out the sine wave on the right, while cos θ and tan θ update too.

Exact values

A few angles have exact sin/cos/tan worth knowing by heart — they come from the $30$ – $60$ – $90$ and $45$ – $45$ – $90$ triangles and appear constantly in exam answers.

The values to memorise

\begin{array}{c|ccccc} \theta & 0^\circ & 30^\circ & 45^\circ & 60^\circ & 90^\circ \\ \hline \sin\theta & 0 & \tfrac12 & \tfrac{\sqrt2}{2} & \tfrac{\sqrt3}{2} & 1 \\[2pt] \cos\theta & 1 & \tfrac{\sqrt3}{2} & \tfrac{\sqrt2}{2} & \tfrac12 & 0 \\[2pt] \tan\theta & 0 & \tfrac{1}{\sqrt3} & 1 & \sqrt3 & \text{—} \end{array}

2. The graphs

Let $\theta$ run all the way round the circle (and beyond) and the height $\sin\theta$ and width $\cos\theta$ trace out waves.

Key result

$y = \sin x$ and $y = \cos x$ are waves of period $360^\circ$ , oscillating between $-1$ and $1$ . The cosine graph is the sine graph shifted left by $90^\circ$ .
$y = \tan x$ has period $180^\circ$ and shoots off to $\pm\infty$ at $x = 90^\circ, 270^\circ, \dots$ — vertical asymptotes where $\cos x = 0$ .

Graphs of y = sin x and y = cos x from 0 to 360 degrees, both oscillating between minus 1 and 1 with period 360 degrees — sin and cos: same wave, a 90° shift apart.

Graph of y = tan x with vertical asymptotes at minus 90, 90 and 270 degrees — tan repeats every 180°, with asymptotes wherever cos x = 0.

These shapes are what make solving equations possible: once you know one solution, the graph’s symmetry hands you all the others.

3. Trig identities

Two identities connect the three functions, and both fall straight out of the unit circle.

Key result

\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad\qquad \sin^2\theta + \cos^2\theta = 1

Why it works

On the unit circle the point $(\cos\theta, \sin\theta)$ is exactly $1$ from the origin, so Pythagoras gives $\cos^2\theta + \sin^2\theta = 1$ immediately. And in the right-angled triangle formed by the radius, $\tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{\sin\theta}{\cos\theta}$ .

Worked example 9.1 · Using the identity

Given that $\sin\theta = \tfrac{3}{5}$ and $\theta$ is acute, find the exact values of $\cos\theta$ and $\tan\theta$ .

Use $\sin^2\theta + \cos^2\theta = 1$ :
$\cos^2\theta = 1 - \left(\tfrac{3}{5}\right)^2 = 1 - \tfrac{9}{25} = \tfrac{16}{25}$
$\theta$ is acute, so $\cos\theta$ is positive:
$\cos\theta = \tfrac{4}{5}$
Then
$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \tfrac{3}{4}$

Common mistake

When you square-root to find $\cos\theta$ , the sign depends on the quadrant. For an acute angle cosine is positive, but for an obtuse $\theta$ it would be negative — always check which quadrant $\theta$ lives in before fixing the sign.

4. Solving trig equations

To solve something like $\sin x = c$ on a given interval, find the first solution with your calculator, then read the rest off the graph using its symmetry.

The method

Get the principal value from $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ .
Use the graph’s symmetry for the others in range:
- $\sin$ : a second solution at $180^\circ - x$ ;
- $\cos$ : a second solution at $360^\circ - x$ ;
- $\tan$ : add $180^\circ$ each time.
Add or subtract $360^\circ$ to land any further solutions inside the interval.

The graph of y = sin x with the line y = 0.5 crossing it at 30 and 150 degrees — The line y = 0.5 meets the sine curve twice in 0°–360°: at 30° and 180° − 30°.

Worked example 9.2 · A cosine equation

Solve $\cos x = -\tfrac{1}{2}$ for $0^\circ \leq x \leq 360^\circ$ .

Principal value: $\cos^{-1}\left(\tfrac{1}{2}\right) = 60^\circ$ , so the related acute angle is $60^\circ$ . Since cosine is negative, $x$ is in the second and third quadrants.
Second-quadrant solution: $x = 180^\circ - 60^\circ = 120^\circ$ . Third-quadrant solution: $x = 180^\circ + 60^\circ = 240^\circ$ .

So $x = 120^\circ$ or $x = 240^\circ$ .

Worked example 9.3 · A multiple angle — widen the interval

Solve $\sin 2x = \tfrac{\sqrt3}{2}$ for $0^\circ \leq x \leq 360^\circ$ .

Let $u = 2x$ . As $x$ runs over $0^\circ$ to $360^\circ$ , $u$ runs over $0^\circ$ to $720^\circ$ — solve in this wider interval.
$\sin u = \tfrac{\sqrt3}{2}$ has principal value $u = 60^\circ$ , with sine positive in quadrants 1 and 2:
$u = 60^\circ,\ 120^\circ,\ 60^\circ + 360^\circ,\ 120^\circ + 360^\circ = 60^\circ, 120^\circ, 420^\circ, 480^\circ$
Divide each by $2$ to get $x$ :
$x = 30^\circ,\ 60^\circ,\ 210^\circ,\ 240^\circ$

Common mistake

With $\sin 2x$ , don’t solve for $x$ and then worry about extra solutions — you’ll miss half of them. Widen the interval to cover $2x$ first, collect every solution there, and divide back at the end.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers. Angles are in degrees.

Warm-up

1 [3 marks]

Write down the exact values of:

$\sin 60^\circ$
$\cos 45^\circ$
$\tan 30^\circ$

Show solution

(a) $\sin 60^\circ = \dfrac{\sqrt3}{2}$ .

(b) $\cos 45^\circ = \dfrac{\sqrt2}{2}$ (also written $\dfrac{1}{\sqrt2}$ ).

(c) $\tan 30^\circ = \dfrac{1}{\sqrt3} = \dfrac{\sqrt3}{3}$ .

2 [3 marks]

Solve $\sin x = \tfrac{1}{2}$ for $0^\circ \leq x \leq 360^\circ$ .

Show solution

Principal value $\sin^{-1}\left(\tfrac12\right) = 30^\circ$ . Sine is positive in quadrants 1 and 2, so the second solution is $180^\circ - 30^\circ = 150^\circ$ .

x = 30^\circ \text{ or } 150^\circ

3 [2 marks]

State the period of each graph: (a) $y = \sin x$ , (b) $y = \tan x$ .

Show solution

(a) $360^\circ$ . (b) $180^\circ$ .

4 [3 marks]

Given that $\cos\theta = \tfrac{5}{13}$ and $\theta$ is acute, find $\sin\theta$ .

Show solution

Using $\sin^2\theta + \cos^2\theta = 1$ :

\sin^2\theta = 1 - \left(\tfrac{5}{13}\right)^2 = 1 - \tfrac{25}{169} = \tfrac{144}{169}

$\theta$ is acute, so $\sin\theta = \dfrac{12}{13}$ .

Standard

5 [3 marks]

Solve $\cos x = -\tfrac{\sqrt3}{2}$ for $0^\circ \leq x \leq 360^\circ$ .

Show solution

The related acute angle is $\cos^{-1}\left(\tfrac{\sqrt3}{2}\right) = 30^\circ$ . Cosine is negative in quadrants 2 and 3:

x = 180^\circ - 30^\circ = 150^\circ \quad\text{or}\quad x = 180^\circ + 30^\circ = 210^\circ

6 [4 marks]

Solve $\tan x = \sqrt3$ for $0^\circ \leq x \leq 360^\circ$ .

Show solution

Principal value $\tan^{-1}(\sqrt3) = 60^\circ$ . Tangent has period $180^\circ$ , so add $180^\circ$ for the next solution in range:

x = 60^\circ \quad\text{or}\quad x = 60^\circ + 180^\circ = 240^\circ

7 [4 marks]

Given that $\sin\theta = \tfrac{2}{3}$ and $\theta$ is obtuse, find the exact values of $\cos\theta$ and $\tan\theta$ .

Show solution

$\cos^2\theta = 1 - \left(\tfrac{2}{3}\right)^2 = 1 - \tfrac{4}{9} = \tfrac{5}{9}$ .
$\theta$ is obtuse (second quadrant), so $\cos\theta$ is negative:
$\cos\theta = -\frac{\sqrt5}{3}$
$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{2/3}{-\sqrt5/3} = -\dfrac{2}{\sqrt5} = -\dfrac{2\sqrt5}{5}$ .

8 [5 marks]

Solve $2\sin x - 1 = 0$ for $0^\circ \leq x \leq 360^\circ$ , then state how many solutions $2\sin x - 1 = 0$ has for $0^\circ \leq x \leq 720^\circ$ .

Show solution

Rearrange: $\sin x = \tfrac{1}{2}$ .
On $0^\circ$ – $360^\circ$ : $x = 30^\circ$ or $150^\circ$ (as in Question 2).
Over $0^\circ$ – $720^\circ$ the sine graph repeats once more, adding $30^\circ + 360^\circ = 390^\circ$ and $150^\circ + 360^\circ = 510^\circ$ — so 4 solutions in total.

9 [5 marks]

Solve $\cos 2x = \tfrac{1}{2}$ for $0^\circ \leq x \leq 360^\circ$ .

Show solution

Let $u = 2x$ ; then $0^\circ \leq u \leq 720^\circ$ .
$\cos u = \tfrac12$ has acute angle $60^\circ$ ; cosine is positive in quadrants 1 and 4, so in $0^\circ$ – $360^\circ$ , $u = 60^\circ$ or $300^\circ$ . Adding $360^\circ$ : $u = 420^\circ$ or $660^\circ$ as well.
Divide by $2$ :
$x = 30^\circ,\ 150^\circ,\ 210^\circ,\ 330^\circ$

Exam style

10 [5 marks]

Solve $\sin x = \cos x$ for $0^\circ \leq x \leq 360^\circ$ .

Show solution

Divide both sides by $\cos x$ (valid where $\cos x \neq 0$ ):

\frac{\sin x}{\cos x} = 1 \;\;\Longrightarrow\;\; \tan x = 1

Principal value $45^\circ$ ; tangent period $180^\circ$ :

x = 45^\circ \quad\text{or}\quad x = 225^\circ

11 [6 marks]

Solve $2\sin^2 x + \sin x - 1 = 0$ for $0^\circ \leq x \leq 360^\circ$ .

Show solution

This is a quadratic in $\sin x$ . Let $s = \sin x$ : $2s^2 + s - 1 = 0$ .
Factorise: $(2s - 1)(s + 1) = 0$ , so $s = \tfrac12$ or $s = -1$ .
Solve each on $0^\circ$ – $360^\circ$ :
- $\sin x = \tfrac12$ : $x = 30^\circ$ or $150^\circ$ .
- $\sin x = -1$ : $x = 270^\circ$ .

So $x = 30^\circ,\ 150^\circ,\ 270^\circ$ .

12 [6 marks]

Solve $\sin(x + 30^\circ) = \tfrac{\sqrt3}{2}$ for $0^\circ \leq x \leq 360^\circ$ .

Show solution

Let $u = x + 30^\circ$ . As $x$ runs $0^\circ$ – $360^\circ$ , $u$ runs $30^\circ$ – $390^\circ$ .
$\sin u = \tfrac{\sqrt3}{2}$ has acute angle $60^\circ$ ; sine positive in quadrants 1 and 2. In range $30^\circ$ – $390^\circ$ :
$u = 60^\circ,\ 120^\circ,\ 60^\circ + 360^\circ = 420^\circ\ (\text{too big})$
so $u = 60^\circ$ or $120^\circ$ (note $u = 420^\circ$ exceeds $390^\circ$ ).
Subtract $30^\circ$ :
$x = 30^\circ \quad\text{or}\quad x = 90^\circ$