Maths · 7 Trigonometry

7.4 Trigonometric functions

Before you start

The exact values of sin, cos and tan at $0 °$ , $30 °$ , $45 °$ , $60 °$ and $90 °$ (7.3).
Using the inverse trigonometric functions on a calculator (7.2).
That an asymptote is a line the curve approaches but never meets (3.5).

Syllabus objectives covered (2)

Recognise, sketch and interpret the following graphs for 0° ⩽ x ⩽ 360°: • y = sin x • y = cos x • y = tan x.
Solve trigonometric equations involving sin x, cos x or tan x, for 0° ⩽ x ⩽ 360°.

Part 1 of 3

The three graphs

Beyond right-angled triangles, sine, cosine and tangent are functions defined for every angle — and their graphs are what let you solve equations properly.
** $y = \sin x$ .** Starts at $(0 °, 0)$ , rises to a maximum of 1 at $90 °$ , back to 0 at $180 °$ , down to a minimum of $- 1$ at $270 °$ , and back to 0 at $360 °$ .
** $y = \cos x$ .** The same wave shifted: starts at $(0 °, 1)$ , falls to 0 at $90 °$ , to a minimum of $- 1$ at $180 °$ , back to 0 at $270 °$ , and to 1 at $360 °$ .
Both have amplitude 1 and period $360 °$ , and both stay between $- 1$ and $1$ . That last fact is doing real work later on.
Sine starts at zero, cosine starts at one, tangent breaks at 90° and 270°.
** $y = \tan x$ .** Completely different in character. It rises from 0, shoots up towards infinity approaching $90 °$ , reappears from below at $90 °$ , passes through 0 at $180 °$ , and repeats.
It has vertical asymptotes at $90 °$ and $270 °$ , is unbounded — no maximum or minimum — and its period is $180 °$ , not $360 °$ .
The symmetries are what you use to solve equations, so notice them now. The sine curve is symmetrical about $x = 90 °$ ; the cosine curve is symmetrical about $x = 180 °$ .

Part 2 of 3

Why one answer is never enough

Ask your calculator for $\sin^{- 1} (0.5)$ and it says $30 °$ . That is correct — but it is not the whole answer.
Draw the line $y = 0.5$ across the sine graph from $0 °$ to $360 °$ and it cuts the curve twice. There is a second solution, and the calculator will never tell you about it.
This is the single most important idea in the subtopic: the calculator gives one solution; the graph tells you how many there are and where.
So the method is always: get the first solution from the calculator, then use the graph or the symmetry rules to find the rest.
**For $\sin x = k$ **, the second solution is $180 ° - (first)$ . Since $\sin 30 ° = 0.5$ , the other is $180 ° - 30 ° = 150 °$ .
**For $\cos x = k$ **, the second solution is $360 ° - (first)$ . Since $\cos 60 ° = 0.5$ , the other is $360 ° - 60 ° = 300 °$ .
**For $\tan x = k$ **, the second solution is $(first) + 180 °$ , because the tangent graph repeats every $180 °$ .
Then check every answer lies within the range asked for — here $0 °$ to $360 °$ — and discard any that do not.

Part 3 of 3

Working through an equation

Step 1: rearrange so the trigonometric function is alone. $2 \cos x + 1 = 0$ becomes $\cos x = - \frac{1}{2}$ .
This step is easy to rush. Everything afterwards depends on having the correct value of $k$ .
Step 2: find the first solution with the inverse function, or from the exact values in 7.3 if you recognise the number.
Step 3: find the others using the symmetry rule for that function.
Step 4: check the range and discard anything outside $0 °$ to $360 °$ .
A caution with negative values. For $\cos x = - \frac{1}{2}$ the calculator gives $120 °$ , which is already in range; applying the rule gives $360 ° - 120 ° = 240 °$ . Both are valid.
For $\sin x = - \frac{1}{2}$ the calculator gives $- 30 °$ , which is outside the range. Use the graph: the solutions are $180 ° + 30 ° = 210 °$ and $360 ° - 30 ° = 330 °$ .
That is why sketching the graph is worth the thirty seconds. A negative calculator answer is a signal to reason from the picture rather than apply a rule blindly.
One more check that costs nothing: if $| k | > 1$ , then $\sin x = k$ and $\cos x = k$ have no solutions at all, because those curves never leave the band from $- 1$ to $1$ . Tangent has no such limit.

Worked examples

Example 1. Solve $\sin x = \frac{\sqrt{3}}{2}$ for $0 ° \leq x \leq 360 °$ .

Show the working

Recognise the value from 7.3: $\sin 60 ° = \frac{\sqrt{3}}{2}$ , so the first solution is $x = 60 °$ .
Sketch $y = \sin x$ and the horizontal line $y = \frac{\sqrt{3}}{2}$ . The line cuts the curve twice in the range, so expect two answers.
Apply the sine rule for the second solution.
$x = 180 ° - 60 ° = 120 °$
Check both lie in $0 °$ to $360 °$ . They do.
Verify one: $\sin 120 °$ does equal $\frac{\sqrt{3}}{2}$ , since $120 °$ is the mirror image of $60 °$ about the line $x = 90 °$ .

Answer: $x = 60 °$ or $x = 120 °$ .

Example 2. Solve $2 \cos x + 1 = 0$ for $0 ° \leq x \leq 360 °$ .

Show the working

Rearrange first so that cosine stands alone.
$2 \cos x = - 1 \Rightarrow \cos x = - \frac{1}{2}$
Find the first solution. From the exact values, $\cos 60 ° = \frac{1}{2}$ , and the negative sign puts the angle in the second quadrant: the calculator gives $120 °$ .
Sketch $y = \cos x$ with the line $y = - \frac{1}{2}$ . It crosses twice.
Apply the cosine rule for the second solution.
$x = 360 ° - 120 ° = 240 °$
Both are in range. Check the symmetry: the cosine curve is symmetrical about $x = 180 °$ , and $120 °$ and $240 °$ are equally spaced either side of it.

Answer: $x = 120 °$ or $x = 240 °$ .

Example 3. Solve $\tan x = - 1$ for $0 ° \leq x \leq 360 °$ .

Show the working

The calculator gives $\tan^{- 1} (- 1) = - 45 °$ , which is outside the required range — so do not stop there.
The tangent graph repeats every $180 °$ , so add $180 °$ to bring the solution into range.
$- 45 ° + 180 ° = 135 °$
Add another $180 °$ for the next solution.
$135 ° + 180 ° = 315 °$
Adding $180 °$ again would give $495 °$ , which is beyond $360 °$ , so stop.
Two solutions in range. Note that tangent gives solutions $180 °$ apart, not $360 °$ — a wider spread than sine or cosine would give.

Answer: $x = 135 °$ or $x = 315 °$ .

Where people go wrong

Mistake: Giving only the calculator's answer.

Instead: Sketch the graph and find every solution in the range.

The inverse function returns one value. A sine or cosine equation almost always has two solutions in $0 °$ to $360 °$ , and the missing one costs a mark.

Mistake: Using $180 ° - x$ for a cosine equation.

Instead: Sine uses $180 ° - x$ ; cosine uses $360 ° - x$ ; tangent uses $x + 180 °$ .

Each rule comes from that curve's own symmetry — sine about $x = 90 °$ , cosine about $x = 180 °$ .

Mistake: Accepting a negative calculator answer as a solution.

Instead: Bring it into the range using the graph or by adding $180 °$ or $360 °$ .

The question specifies $0 ° \leq x \leq 360 °$ , so anything outside must be converted or discarded.

Mistake: Solving without rearranging first.

Instead: Get $\sin x$ , $\cos x$ or $\tan x$ alone before touching the inverse function.

Applying $\cos^{- 1}$ to the whole of $2 \cos x + 1$ is meaningless and produces nonsense.

Mistake: Looking for solutions to $\sin x = 1.5$ .

Instead: There are none — sine and cosine never exceed 1 or fall below $- 1$ .

Recognising this immediately saves time and is itself the answer the question wants.

Mistake: Assuming the tangent graph has period $360 °$ like the others.

Instead: Its period is $180 °$ .

That is why tangent solutions are $180 °$ apart, and why there are asymptotes at $90 °$ and $270 °$ .

Check yourself

State the maximum and minimum values of $y = \sin x$ , and the values of $x$ at which they occur in $0 ° \leq x \leq 360 °$ .
Answer
Maximum $1$ at $x = 90 °$ ; minimum $- 1$ at $x = 270 °$ .
State the period of $y = \cos x$ and of $y = \tan x$ .
Answer
$360 °$ for cosine, $180 °$ for tangent.
Where are the asymptotes of $y = \tan x$ in the range $0 °$ to $360 °$ ?
Answer
At $x = 90 °$ and $x = 270 °$ .
Solve $\sin x = 0.5$ for $0 ° \leq x \leq 360 °$ .
Answer
$x = 30 °$ and $x = 180 ° - 30 ° = 150 °$ .
Solve $\cos x = 0.5$ for $0 ° \leq x \leq 360 °$ .
Answer
$x = 60 °$ and $x = 360 ° - 60 ° = 300 °$ .
Solve $\tan x = 1$ for $0 ° \leq x \leq 360 °$ .
Answer
$x = 45 °$ and $x = 45 ° + 180 ° = 225 °$ .
How many solutions does $\cos x = 2$ have, and why?
Answer
None. The cosine curve never rises above 1.

Exam questions

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

Question 1. The graph of $y = \sin x$ is drawn for $0 ° \leq x \leq 360 °$ .

(a) Sketch the graph, marking the values where it crosses the $x$ -axis. 2 marks No calculator

Mark scheme

B1 correct sine shape, one full cycle, max 1 and min -1
B1 crossings marked at 0°, 180° and 360°

(b) Solve $\sin x = 0.8$ for $0 ° \leq x \leq 360 °$ , giving your answers correct to one decimal place. 3 marks

Mark scheme

M1 sin⁻¹(0.8) attempted
A1 53.1° awrt
A1 126.9° awrt

(c) Explain how the graph shows that $\sin x = 0.8$ has exactly two solutions in this range. 1 mark No calculator

Mark scheme

B1 the line y = 0.8 crosses the curve twice between 0° and 360° oe

Question 2. Solve the following equations for $0 ° \leq x \leq 360 °$ .

(a) $3 \tan x = 3$ 2 marks No calculator

Mark scheme

B1 45°
B1 225°

(b) $4 \cos x + 3 = 1$ 3 marks

Mark scheme

M1 cos x = -0.5 seen
A1 120°
A1 240°

(c) Explain why $2 \sin x = 3$ has no solutions. 2 marks No calculator

Mark scheme

B1 rearranges to sin x = 1.5
B1 the sine of an angle never exceeds 1 oe

Question 3. A student solves $\sin x = - 0.4$ for $0 ° \leq x \leq 360 °$ and writes down the calculator answer $- 23.6 °$ as the solution.

(a) Comment on the student's answer. 1 mark No calculator

Mark scheme

B1 -23.6° is outside the given range, so it is not a valid solution oe

(b) Solve the equation correctly for $0 ° \leq x \leq 360 °$ . 3 marks

Mark scheme

M1 uses the graph or symmetry to bring the solution into range
A1 203.6° awrt
A1 336.4° awrt

Words to know

Period: The horizontal distance after which a repeating graph starts to repeat: $360 °$ for sine and cosine, $180 °$ for tangent.
Amplitude: Half the vertical distance between maximum and minimum — 1 for $y = \sin x$ and $y = \cos x$ .
Inverse trigonometric function: $\sin^{- 1}$ , $\cos^{- 1}$ , $\tan^{- 1}$ , which return one angle for a given ratio.
Asymptote: A line the curve approaches but never meets, as at $90 °$ and $270 °$ on $y = \tan x$ .

All Maths lessons Next: 7.5 Non-right-angled triangles