What are the basic trigonometric graphs and their characteristics?

The basic trigonometric graphs include the sine, cosine, and tangent functions. The sine and cosine graphs are periodic with a period of 360 degrees or 2π radians, and they have an amplitude of 1. The sine graph starts at the origin (0,0), while the cosine graph starts at its maximum value (1,0). The tangent graph has a period of 180 degrees or π radians and features vertical asymptotes where the function is undefined. Understanding these characteristics is crucial for Cambridge IGCSE Mathematics students.

How do you determine the amplitude and period of a trigonometric graph?

The amplitude of a trigonometric graph, such as sine or cosine, is the maximum distance from the midline to the peak of the graph. It is determined by the coefficient in front of the sine or cosine function. The period is the length of one complete cycle of the graph and is calculated by dividing 360 degrees (or 2π radians) by the absolute value of the coefficient of the angle variable. For Cambridge IGCSE Mathematics, understanding these concepts helps in graph transformations.

How do phase shifts affect trigonometric graphs?

Phase shifts in trigonometric graphs occur when the graph is horizontally shifted along the x-axis. This shift is determined by the horizontal translation in the function's equation. For example, in y = sin(x - c), the graph is shifted c units to the right. Understanding phase shifts is essential for Cambridge IGCSE students to accurately graph and interpret trigonometric functions.

What is the significance of vertical shifts in trigonometric graphs?

Vertical shifts occur when a trigonometric graph is moved up or down along the y-axis. This is represented by a constant added or subtracted from the function, such as y = sin(x) + d. The graph shifts up by d units if d is positive and down if d is negative. Recognizing vertical shifts is important for Cambridge IGCSE Mathematics students to understand how trigonometric functions can be transformed.

How can trigonometric graphs be used to model real-world phenomena?

Trigonometric graphs are used to model periodic phenomena such as sound waves, tides, and seasonal patterns. By adjusting the amplitude, period, and phase shift, these graphs can represent various cycles and oscillations. Cambridge IGCSE Mathematics students learn to apply these concepts to solve real-world problems, enhancing their understanding of how mathematics describes natural patterns.

Why is "Giving only the principal value when more solutions exist" a common mistake in Trigonometric Graphs?

Why it happens: Calculator only returns one angle. How to avoid it: Always check the range. For = k in [0, 360], there are usually TWO solutions. Source: 0580/42 — recurring

Why is "Using period $360\degree$ for $\tan$" a common mistake in Trigonometric Graphs?

Why it happens: Generalising from and . How to avoid it: has period 180. Add 180 for the next solution, not 360.

Why is "Confusing $\sin(180 - \theta)$ with $\sin(180 + \theta)$" a common mistake in Trigonometric Graphs?

Why it happens: Memorising the wrong identity. How to avoid it: is positive in 1st & 2nd quadrants. So second solution = 180 - principal.

Why is "Plotting $\tan 90\degree$ as a finite value" a common mistake in Trigonometric Graphs?

Why it happens: Calculator shows ERROR, students guess. How to avoid it: 90 is UNDEFINED. The graph has a vertical asymptote there.

Trigonometric Graphs – Study Notes & Past Paper Style Questions | Cambridge IGCSE Maths 0580 Extended

Short Notes - Trigonometric Graphs

Detailed Study Notes

Detailed notes on Trigonometry for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Trigonometric Graphs — Cambridge IGCSE 0580 Maths Extended (2026)

Sine, cosine and tangent graphs from $0\degree$ to $360\degree$ (and beyond). Period, amplitude, key values and where the graphs cross zero — all examinable.

At a glance

Sine $y = \sin x$ : wave between $-1$ and $1$ , period $360\degree$ , starts at $(0, 0)$ .
Cosine $y = \cos x$ : wave between $-1$ and $1$ , period $360\degree$ , starts at $(0, 1)$ .
Tangent $y = \tan x$ : period $180\degree$ , asymptotes at $90\degree, 270\degree, \ldots$
Sine and cosine are 'shifted by $90\degree$ ' — $\cos x = \sin(x + 90\degree)$ .
Sine: positive in $(0, 180)$ , negative in $(180, 360)$ .
Cosine: positive in $(0, 90) \cup (270, 360)$ , negative in $(90, 270)$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E8.6 — Recognise, sketch and interpret graphs of $y = \sin x$ , $y = \cos x$ , $y = \tan x$ for $0\degree \le x \le 360\degree$ .

$y = \sin x$

Wave from $-1$ to $1$ , period $360\degree$ . Passes through $(0, 0), (180, 0), (360, 0)$ .

Key features.

Range: $-1 \le y \le 1$ .
Period: $360\degree$ (one full wave from $0\degree$ to $360\degree$ ).
Maximum $y = 1$ at $x = 90\degree$ .
Minimum $y = -1$ at $x = 270\degree$ .
Zeros: $x = 0\degree, 180\degree, 360\degree$ (every $180\degree$ ).

Symmetry.

$\sin(180\degree - x) = \sin x$ (mirror across $90\degree$ ).
$\sin(360\degree - x) = -\sin x$ (mirror across $180\degree$ , then below axis).

y = sin x: starts at (0, 0), max +1 at 90°, min −1 at 270°, zeros every 180°.

Worked. Solve $\sin x = 0.5$ for $0\degree \le x \le 360\degree$ .

Principal value: $x = \sin^{-1}(0.5) = 30\degree$ .
Second solution: $180\degree - 30\degree = 150\degree$ .
Both in range: $x = 30\degree, 150\degree$ .

Range $[-1, 1]$ .
Period $360\degree$ .
Max at $90\degree$ , min at $270\degree$ .
Zeros every $180\degree$ .

$y = \cos x$

Wave from $-1$ to $1$ , period $360\degree$ . Starts at $(0, 1)$ .

Key features.

Range: $-1 \le y \le 1$ .
Period: $360\degree$ .
Maximum $y = 1$ at $x = 0\degree, 360\degree$ .
Minimum $y = -1$ at $x = 180\degree$ .
Zeros: $x = 90\degree, 270\degree$ .

Relationship to sine. $\cos x$ is $\sin x$ shifted left by $90\degree$ : $\cos x = \sin(x + 90\degree).$

y = cos x: starts at (0, 1) — it is the sine curve shifted left by 90°.

Symmetry.

$\cos(360\degree - x) = \cos x$ (even-like, mirror across $0\degree$ and $360\degree$ ).
$\cos(180\degree - x) = -\cos x$ .

Worked. Solve $\cos x = -0.5$ for $0\degree \le x \le 360\degree$ .

$\cos^{-1}(-0.5) = 120\degree$ .
Second solution: $360\degree - 120\degree = 240\degree$ .
$x = 120\degree, 240\degree$ .

Range $[-1, 1]$ .
Period $360\degree$ .
Max at $0\degree$ and $360\degree$ , min at $180\degree$ .
Zeros at $90\degree$ and $270\degree$ .

$y = \tan x$

Period $180\degree$ . Vertical asymptotes at $90\degree$ , $270\degree$ , …

Key features.

Range: all real numbers.
Period: $180\degree$ (NOT $360\degree$ ).
Asymptotes: $x = 90\degree, 270\degree, \ldots$ ( $\cos x = 0$ , so tan undefined).
Zeros: $x = 0\degree, 180\degree, 360\degree$ .
The graph rises from $-\infty$ at just after $90\degree$ (e.g. at $91\degree$ ) to $+\infty$ at just before $90\degree$ (e.g. at $89\degree$ ). Then it repeats.

y = tan x repeats every 180°, with vertical asymptotes where cos x = 0.

Worked. Solve $\tan x = 1$ for $0\degree \le x \le 360\degree$ .

$\tan^{-1}(1) = 45\degree$ .
Period $180\degree$ , so next: $45\degree + 180\degree = 225\degree$ .
$x = 45\degree, 225\degree$ .

Range: all reals.
Period $180\degree$ .
Asymptotes at $90\degree, 270\degree$ .
Zeros at $0, 180, 360\degree$ .

Key values to remember

Memorise sin/cos/tan of $0, 30, 45, 60, 90, 180, 270, 360\degree$ .

$x$	$\sin x$	$\cos x$	$\tan x$
$0\degree$	$0$	$1$	$0$
$30\degree$	$\dfrac{1}{2}$	$\dfrac{\sqrt 3}{2}$	$\dfrac{1}{\sqrt 3}$
$45\degree$	$\dfrac{\sqrt 2}{2}$	$\dfrac{\sqrt 2}{2}$	$1$
$60\degree$	$\dfrac{\sqrt 3}{2}$	$\dfrac{1}{2}$	$\sqrt 3$
$90\degree$	$1$	$0$	undef
$180\degree$	$0$	$-1$	$0$
$270\degree$	$-1$	$0$	undef
$360\degree$	$0$	$1$	$0$

Sketch checklist — when asked to sketch a trig graph for $0\degree \le x \le 360\degree$ :

Mark the axes with $90\degree, 180\degree, 270\degree, 360\degree$ on the $x$ -axis.
Mark $-1, 0, 1$ on the $y$ -axis (for sine/cosine).
Plot the maximum, minimum and zero points clearly.
Draw a smooth curve through them.
Label the graph $y = \sin x$ (or whatever).

As P travels round the unit circle, its height traces y = sin x and its horizontal position traces y = cos x.

Memorise $0, 30, 45, 60, 90\degree$ values.
Sketch: mark axes, plot key points, draw smooth curve.
Label the curve.

Quick recap

Sine: range $[-1, 1]$ , period $360\degree$ , zero at $0, 180, 360\degree$ .
Cosine: range $[-1, 1]$ , period $360\degree$ , zero at $90, 270\degree$ .
Tangent: all reals, period $180\degree$ , asymptotes at $90, 270\degree$ .
$\cos x = \sin(x + 90\degree)$ .
Memorise key values for $0, 30, 45, 60, 90\degree$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Period — the horizontal length of one full repetition of a graph.
Amplitude — half the vertical distance between max and min (for sine/cosine: $1$ ).
Asymptote — a line the graph approaches but never touches (vertical for tan).
Symmetry — sine: rotational about $(180\degree, 0)$ ; cosine: line symmetry about $x = 0$ .

How it’s examined

Trig graphs appear most years on Paper 2 (2-3 marks: sketch or read a value) and Paper 4 (4-6 marks combined with solving trig equations). Examiner reports flag forgetting the period of tan ( $180\degree$ , not $360\degree$ ) and missing the second solution when solving sine/cosine equations in range.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E8.6); 0580/22 May/Jun 2024 — Q12 (trig graph sketch); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Trigonometric Graphs

Step-by-step solutions to past-paper-style questions on trigonometric graphs, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Trigonometric Graphs

Almost every trigonometric graphs exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Graph or diagram

Recognise it by

A sketch, complete the table, state the max/min/zeros or read off the graph instruction for $\sin\theta$ , $\cos\theta$ or $\tan\theta$ — including transformed graphs such as $y=\sin\theta+c$ .

How to approach it

Anchor the curve with its key points — $(0\degree,0)$ , $(90\degree,1)$ , $(180\degree,0)$ , $(270\degree,-1)$ , $(360\degree,0)$ for sine — apply any shift, then draw a smooth wave. State range, period and asymptotes explicitly.

Common trap

Plotting $\tan 90\degree$ as a finite value instead of an asymptote, or forgetting a vertical shift moves the max/min floor. Examiner reports flag jagged sketches that lose the shape mark.

Direct calculation

Recognise it by

A solve $\sin\theta=k$ , use symmetry to write the second solution, or state the amplitude and period instruction — a short calculation using the graph's properties.

How to approach it

Find the principal value with the inverse function, then use the symmetry rule for the function ( $180\degree-\theta$ for $\sin$ , $360\degree-\theta$ for $\cos$ ) to get all solutions in range. For $y=a\sin(b\theta)$ , amplitude is $|a|$ and period is $\dfrac{360\degree}{|b|}$ .

Common trap

Quoting only the principal value, or mixing the $\sin$ and $\cos$ symmetry rules. Examiner reports note there are usually two solutions in $[0\degree,360\degree]$ and the period is stated with degree units.

1Identify key features of $y = \sin\theta$
ExtendedGraph or diagram• sin graph
Question
State the maximum value, minimum value, and the values of $\theta$ between $0\degree$ and $360\degree$ for which $\sin\theta = 0$ .
Step-by-step solution
1. Step 1
  $\sin\theta$ oscillates between $-1$ and $1$ .
  $\max = 1,\ \min = -1$
2. Step 2
  $\sin\theta = 0$ at $\theta = 0\degree, 180\degree, 360\degree$ .
Answer
Max $1$ , min $-1$ ; zeros at $0\degree, 180\degree, 360\degree$ .
2Use symmetry of $\cos\theta$
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q17• cos graph
Question
$\cos 70\degree = 0.342$ . Use symmetry to write $\cos 290\degree$ .
Step-by-step solution
1. Step 1
  $\cos$ is symmetric about $\theta = 0$ , so $\cos(360\degree - \theta) = \cos\theta$ .
  $\cos 290\degree = \cos 70\degree = 0.342$
Answer
$0.342$
3Solve a basic trig equation using the graph
ExtendedDirect calculation• solve
Question
Solve $\sin\theta = 0.5$ for $0\degree \le \theta \le 360\degree$ .
Step-by-step solution
1. Step 1
  Principal value.
  $\theta_{1} = \sin^{-1}(0.5) = 30\degree$
2. Step 2
  Second value (sin positive in 1st & 2nd quadrants): $180\degree - 30\degree$ .
  $\theta_{2} = 150\degree$
Answer
$\theta = 30\degree, 150\degree$
4Identify the asymptotes of $\tan\theta$
ExtendedGraph or diagram• tan graph
Question
State the values of $\theta$ between $0\degree$ and $360\degree$ where $\tan\theta$ is undefined.
Step-by-step solution
1. Step 1
  $\tan = \sin/\cos$ , undefined where $\cos = 0$ .
  $\theta = 90\degree,\ 270\degree$
Answer
$90\degree$ and $270\degree$
5Key features of $y = \cos\theta$
CoreGraph or diagram• cos graph
Question
State the maximum value, minimum value, and the values of $\theta$ between $0\degree$ and $360\degree$ at which $\cos\theta = 0$ .
Step-by-step solution
1. Step 1
  $\cos\theta$ oscillates between $-1$ and $1$ .
  $\max = 1\ (\text{at}\ \theta = 0\degree, 360\degree),\ \min = -1\ (\text{at}\ \theta = 180\degree)$
2. Step 2
  $\cos\theta = 0$ when the graph crosses the horizontal axis.
  $\theta = 90\degree,\ 270\degree$
Answer
Max $1$ at $\theta = 0\degree, 360\degree$ ; min $-1$ at $\theta = 180\degree$ ; zeros at $\theta = 90\degree, 270\degree$ .
Examiner tip
Sketch with labelled key points $(0\degree, 1)$ , $(90\degree, 0)$ , $(180\degree, -1)$ , $(270\degree, 0)$ , $(360\degree, 1)$ . Marking these earns the structure marks even before the curve is drawn.
6Find the period and amplitude of $y = a\sin(b\theta)$
ExtendedDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q20• period, amplitude
Question
State the amplitude and the period of $y = 3\sin(2\theta)$ for $\theta$ measured in degrees.
Step-by-step solution
1. Step 1
  Amplitude is the coefficient $|a|$ in front of the trig function.
  $\text{amplitude} = |3| = 3$
2. Step 2
  Period of $\sin(b\theta)$ in degrees is $\dfrac{360\degree}{|b|}$ .
  $\text{period} = \dfrac{360\degree}{2} = 180\degree$
Answer
Amplitude $3$ ; period $180\degree$ .
Examiner tip
0580 mark schemes accept amplitude in either form (positive number or 'distance from axis to max') but require the period stated with degree units.
7Vertical transformation $y = \sin\theta + c$
ExtendedGraph or diagram• transformation
Question
The graph of $y = \sin\theta + 2$ is sketched for $0\degree \le \theta \le 360\degree$ . State the maximum and minimum values of $y$ and the values of $\theta$ at which they occur.
Step-by-step solution
1. Step 1
  Adding $2$ to $\sin\theta$ shifts the whole curve up by $2$ units; the maximum and minimum locations are unchanged.
2. Step 2
  Maximum of $\sin\theta$ is $1$ at $\theta = 90\degree$ , so max of $y = 1 + 2 = 3$ .
3. Step 3
  Minimum of $\sin\theta$ is $-1$ at $\theta = 270\degree$ , so min of $y = -1 + 2 = 1$ .
Answer
Max $y = 3$ at $\theta = 90\degree$ ; min $y = 1$ at $\theta = 270\degree$ .
Examiner tip
The shifted graph never falls below $1$ — many candidates forget the floor is no longer zero and report the wrong minimum.
8Use symmetry to find a second solution
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2023 Q18• symmetry, solve
Question
$\cos\theta = 0.6$ and $\theta_{1} = 53.1\degree$ is one solution in $[0\degree, 360\degree]$ . Use the symmetry of the cosine graph to write the second solution.
Step-by-step solution
1. Step 1
  $\cos\theta$ is symmetric about $\theta = 0$ (and about $\theta = 360\degree$ ). So $\cos(360\degree - \theta) = \cos\theta$ .
2. Step 2
  Apply the symmetry.
  $\theta_{2} = 360\degree - 53.1\degree = 306.9\degree$
Answer
$\theta_{2} = 306.9\degree$
Examiner tip
Confusing $360\degree - \theta$ with $180\degree - \theta$ (the sine rule) is the most common slip. Match the rule to the function: $\cos$ uses $360\degree - \theta$ .
9A* solve $\sin x = -0.5$ using the graph
ChallengeGraph or diagram• Adapted from 0580/42 May/Jun 2023 Q23• solve, challenge
Question
Solve $\sin x = -0.5$ for $0\degree \le x \le 360\degree$ by reading from the graph of $y = \sin x$ .
Step-by-step solution
1. Step 1
  Sketch $y = \sin x$ and the horizontal line $y = -0.5$ . They intersect in the 3rd and 4th quadrants (where $\sin$ is negative).
2. Step 2
  Reference acute angle from $\sin^{-1}(0.5) = 30\degree$ .
3. Step 3
  3rd quadrant solution: $x = 180\degree + 30\degree = 210\degree$ .
4. Step 4
  4th quadrant solution: $x = 360\degree - 30\degree = 330\degree$ .
Answer
$x = 210\degree,\ 330\degree$
Examiner tip
The calculator returns $\sin^{-1}(-0.5) = -30\degree$ , which is OUTSIDE the required range. Use the symmetry of the sine graph (or the CAST diagram) to find the two valid solutions in $[0\degree, 360\degree]$ .
10Sketch $y = \sin\theta$ from a table of values
CoreGraph or diagram• sin graph, sketch
Question
Complete the table of values for $y = \sin\theta$ and use them to describe the shape of the graph for $0\degree \le \theta \le 360\degree$ .\n\n| $\theta$ | $0\degree$ | $90\degree$ | $180\degree$ | $270\degree$ | $360\degree$ |\n|---|---|---|---|---|---|\n| $\sin\theta$ | ? | ? | ? | ? | ? |
Step-by-step solution
1. Step 1
  Read each value directly: $\sin 0\degree = 0$ , $\sin 90\degree = 1$ , $\sin 180\degree = 0$ , $\sin 270\degree = -1$ , $\sin 360\degree = 0$ .
2. Step 2
  Plot the points. The curve rises from $(0, 0)$ to a maximum of $1$ at $90\degree$ , falls back to zero at $180\degree$ , drops to a minimum of $-1$ at $270\degree$ and returns to zero at $360\degree$ .
3. Step 3
  Key features: range $-1 \le y \le 1$ , period $360\degree$ , passes through the origin.
Answer
Values: $0, 1, 0, -1, 0$ . Graph is a smooth wave with amplitude $1$ and period $360\degree$ .
Examiner tip
Mark schemes award the structure marks for showing the maximum, minimum and three zeros clearly. A jagged or angular sketch loses the shape mark.

Key Formulae — Trigonometric Graphs

The formulae you need to memorise for trigonometric graphs on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

$y = \sin\theta$
$\text{Range: } -1 \le y \le 1; \text{ Period: } 360\degree$
When to use
Sketching or solving sine equations.
$y = \cos\theta$
$\text{Range: } -1 \le y \le 1; \text{ Period: } 360\degree$
When to use
Sketching or solving cosine equations.
$y = \tan\theta$
$\text{Range: all reals; Period: } 180\degree;\text{ asymptotes at } 90\degree, 270\degree$
When to use
Sketching or solving tangent equations.
Useful symmetries
$\sin(180\degree - \theta) = \sin\theta;\ \cos(360\degree - \theta) = \cos\theta$
When to use
Find a second solution from a known principal value.

Key Definitions and Keywords — Trigonometric Graphs

Definitions to memorise and the exact keywords mark schemes credit for trigonometric graphs answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Period
Examiner keyword
The smallest positive interval over which the graph repeats.
Amplitude
Examiner keyword
Half the distance between the maximum and minimum values (1 for $\sin$ and $\cos$ ).
Asymptote
Examiner keyword
A line the graph approaches but never touches. $\tan\theta$ has vertical asymptotes where $\cos\theta = 0$ .
Principal value
Examiner keyword
The single value returned by an inverse trig function on a calculator (e.g. $\sin^{-1}$ returns a value between $-90\degree$ and $90\degree$ ).

Common Mistakes and Misconceptions — Trigonometric Graphs

The traps other students keep falling into on trigonometric graphs questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Giving only the principal value when more solutions exist
0580/42 — recurring
Why it happens
Calculator only returns one angle.
How to avoid it
Always check the range. For $\sin\theta = k$ in $[0\degree, 360\degree]$ , there are usually TWO solutions.
✕Using period $360\degree$ for $\tan$
Why it happens
Generalising from $\sin$ and $\cos$ .
How to avoid it
$\tan$ has period $180\degree$ . Add $180\degree$ for the next solution, not $360\degree$ .
✕Confusing $\sin(180 - \theta)$ with $\sin(180 + \theta)$
Why it happens
Memorising the wrong identity.
How to avoid it
$\sin$ is positive in 1st & 2nd quadrants. So second solution = $180\degree -$ principal.
✕Plotting $\tan 90\degree$ as a finite value
Why it happens
Calculator shows ERROR, students guess.
How to avoid it
$\tan 90\degree$ is UNDEFINED. The graph has a vertical asymptote there.

Trigonometric Graphs — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Trigonometric Graphs

Short Notes - Trigonometric Graphs

Detailed Study Notes

At a glance

What you’ll learn

y=sin⁡xy = \sin xy=sinx

y=cos⁡xy = \cos xy=cosx

y=tan⁡xy = \tan xy=tanx

Key values to remember

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Trigonometric Graphs

1Identify key features of y=sin⁡θy = \sin\thetay=sinθ

2Use symmetry of cos⁡θ\cos\thetacosθ

3Solve a basic trig equation using the graph

4Identify the asymptotes of tan⁡θ\tan\thetatanθ

5Key features of y=cos⁡θy = \cos\thetay=cosθ

6Find the period and amplitude of y=asin⁡(bθ)y = a\sin(b\theta)y=asin(bθ)

7Vertical transformation y=sin⁡θ+cy = \sin\theta + cy=sinθ+c

8Use symmetry to find a second solution

9A* solve sin⁡x=−0.5\sin x = -0.5sinx=−0.5 using the graph

10Sketch y=sin⁡θy = \sin\thetay=sinθ from a table of values

y=sin⁡θy = \sin\thetay=sinθ

y=cos⁡θy = \cos\thetay=cosθ

y=tan⁡θy = \tan\thetay=tanθ

Useful symmetries

Period

Amplitude

Asymptote

Principal value

✕Giving only the principal value when more solutions exist

✕Using period 360°360\degree360° for tan⁡\tantan

✕Confusing sin⁡(180−θ)\sin(180 - \theta)sin(180−θ) with sin⁡(180+θ)\sin(180 + \theta)sin(180+θ)

✕Plotting tan⁡90°\tan 90\degreetan90° as a finite value

Trigonometric Graphs — frequently asked questions

Trigonometric Graphs

Short Notes - Trigonometric Graphs

Detailed Study Notes

At a glance

What you’ll learn

y=sin⁡xy = \sin xy=sinx

y=cos⁡xy = \cos xy=cosx

y=tan⁡xy = \tan xy=tanx

Key values to remember

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Trigonometric Graphs

1Identify key features of y=sin⁡θy = \sin\thetay=sinθ

2Use symmetry of cos⁡θ\cos\thetacosθ

3Solve a basic trig equation using the graph

4Identify the asymptotes of tan⁡θ\tan\thetatanθ

5Key features of y=cos⁡θy = \cos\thetay=cosθ

6Find the period and amplitude of y=asin⁡(bθ)y = a\sin(b\theta)y=asin(bθ)

7Vertical transformation y=sin⁡θ+cy = \sin\theta + cy=sinθ+c

8Use symmetry to find a second solution

9A* solve sin⁡x=−0.5\sin x = -0.5sinx=−0.5 using the graph

10Sketch y=sin⁡θy = \sin\thetay=sinθ from a table of values

y=sin⁡θy = \sin\thetay=sinθ

y=cos⁡θy = \cos\thetay=cosθ

y=tan⁡θy = \tan\thetay=tanθ

Useful symmetries

Period

Amplitude

Asymptote

Principal value

✕Giving only the principal value when more solutions exist

✕Using period 360°360\degree360° for tan⁡\tantan

✕Confusing sin⁡(180−θ)\sin(180 - \theta)sin(180−θ) with sin⁡(180+θ)\sin(180 + \theta)sin(180+θ)

✕Plotting tan⁡90°\tan 90\degreetan90° as a finite value

Trigonometric Graphs — frequently asked questions

$y = \sin x$

$y = \cos x$

$y = \tan x$

1Identify key features of $y = \sin\theta$

2Use symmetry of $\cos\theta$

4Identify the asymptotes of $\tan\theta$

5Key features of $y = \cos\theta$

6Find the period and amplitude of $y = a\sin(b\theta)$

7Vertical transformation $y = \sin\theta + c$

9A* solve $\sin x = -0.5$ using the graph

10Sketch $y = \sin\theta$ from a table of values

$y = \sin\theta$

$y = \cos\theta$

$y = \tan\theta$

✕Using period $360\degree$ for $\tan$

✕Confusing $\sin(180 - \theta)$ with $\sin(180 + \theta)$

✕Plotting $\tan 90\degree$ as a finite value

$y = \sin x$

$y = \cos x$

$y = \tan x$

1Identify key features of $y = \sin\theta$

2Use symmetry of $\cos\theta$

4Identify the asymptotes of $\tan\theta$

5Key features of $y = \cos\theta$

6Find the period and amplitude of $y = a\sin(b\theta)$

7Vertical transformation $y = \sin\theta + c$

9A* solve $\sin x = -0.5$ using the graph

10Sketch $y = \sin\theta$ from a table of values

$y = \sin\theta$

$y = \cos\theta$

$y = \tan\theta$

✕Using period $360\degree$ for $\tan$

✕Confusing $\sin(180 - \theta)$ with $\sin(180 + \theta)$

✕Plotting $\tan 90\degree$ as a finite value