Trig Graphs (AQA GCSE Maths)

Revision Note

Test Yourself
Jamie W

Author

Jamie W

Expertise

Maths

Drawing Trig Graphs

Why do we need to know what trig graphs look like?

  • Trigonometric graphs (trig graphs) are used in various applications of mathematics
    • for example, the oscillating nature can be used to model how a pendulum swings or tide heights

How do we draw trig graphs?

  • As with other graphs, being familiar with the general style of trigonometric graphs will help you sketch them quickly and you can then use them to find values or angles alongside, or instead of, your calculator
  • All trigonometric graphs follow a pattern – a “starting point” and then “something happens every 90°”

  • y equals sin space x
  • Starts at (0,0)
  • Every 90° it cycles through 1, 0, -1, 0, ...

y-equals-sinx

  • y equals cos space x
  • Starts at (0, 1)
  • Every 90° it cycles through 0, -1, 0, 1, ...

y-equals-cosx

  • y equals tan space x
  • Starts at (0, 0)
  • Every 90° it is either 0 or an asymptote
    • An asymptote is a line that a graph gets ever closer to, without ever crossing or touching it

y-equals-tanx

Worked example

On the axes provided, sketch the graph of y equals sin space x degree for 0 less or equal than x less or equal than 360.

Mark key values on the axes provided; 1, 0 and −1 on the y-axis and 0, 90, 180, 270 and 360 on the x-axis. Try to space them evenly apart

IcJiYEqM_2-14-trig-graphs1

For y equals sin space x degree, the key knowledge is that it starts at (0, 0) then every 90° it cycles though 1, 0 , −1, 0, ... so mark these points on the axes

Finally, join the points with a smooth line. It is best practice to label the curve with its equation

5_-KJTZ8_2-14-trig-graphs2

Solving Trig Equations

You can use the symmetry of trig graphs to find multiple solutions to a trig equation

How are trigonometric equations of the form sin x = k solved?

  • The solutions to the equation sin x = 0.5 in the range 0° < x < 360° are x = 30° and x = 150°
    • If you like, check on a calculator that both sin(30) and sin(150) give 0.5
  • The first solution comes from your calculator (by taking inverse sin of both sides)
    • x = sin-1(0.5) = 30°
  • The second solution comes from the symmetry of the graph y = sin x between 0° and 360°
    • Sketch the graph
    • Draw a vertical line from x = 30° to the curve, then horizontally across to another point on the curve, then vertically back to the x-axis again
    • By the symmetry of the curve, the new value of x is 180° - 30° = 150°
  • In general, if x° is an acute angle that solves sin x = k, then 180° - x° is the obtuse angle that solves the same equation
  • If the calculator gives x as a negative value, continue drawing the curve to the left of the x-axis to help

cie-igcse-3-12-2-solving-trig-equations-1

How are trigonometric equations of the form cos x = k solved?

  • The solutions to the equation cos x = 0.5 in the range 0° < x < 360° are x = 60° and x = 300°
    • If you like, check on a calculator that both cos(60) and cos(300) give 0.5
  • The first solution comes from your calculator (by taking inverse cos of both sides)
    • x = cos-1(0.5) = 60°
  • The second solution comes from the symmetry of the graph y = cos x between 0° and 360°
    • Sketch the graph
    • Draw a vertical line from x = 60° to the curve, then horizontally across to another point on the curve, then vertically back to the x-axis again
    • By the symmetry of the curve, the new value of x is 360° - 60° = 300°
  • In general, if x° is an angle that solves cos x = k, then 360° - x° is another angle that solves the same equation
  • If the calculator gives x as a negative value, continue drawing the curve to the left of the x-axis to help

cie-igcse-3-12-2-solving-trig-equations-2

How are trigonometric equations of the form tan x = k solved?

  • The solutions to the equation tan x = 1 in the range 0° < x < 360° are x = 45° and x = 225°
    • Check on a calculator that both tan(45) and tan(225) give 1
  • The first solution comes from your calculator (by taking inverse tan of both sides)
    • x = tan-1(1) = 45°
  • The second solution comes from the symmetry of the graph y = tan x between 0° and 360°
    • Sketch the graph
    • Draw a vertical line from x = 45° to the curve, then horizontally across to another point on the curve (a different “branch” of tan x), then vertically back to the x-axis again
    • The new value of x is 45° + 180° = 225° as the next “branch” of tan x is shifted 180° to the right
  • In general, if x° is an angle that solves tan x = k, then x° + 180° is another angle that solve the same equation
  • If the calculator gives x as a negative value, continue drawing the curve to the left of the x-axis to help

cie-igcse-3-12-2-solving-trig-equations-3

Exam Tip

  • Use a calculator to check your solutions by substituting them into the original equation
  • For example, 60° and 330° are incorrect solutions of cos x = 0.5, as cos(330) on a calculator is not equal to 0.5

Worked example

Solve sin x = 0.25 in the range 0° < x < 360°, giving your answers correct to 1 decimal place

Use a calculator to find the first solution (by taking inverse sin of both sides)

x = sin-1(0.25) = 14.4775… = 14.48° to 2 dp

Sketch the graph of y = sin x and mark on (roughly) where x = 14.48 and y = 0.25 would be
Draw a vertical line up to the curve, then horizontally across to the next point on the curve, then vertically back down to the x-axis

cie-igcse-3-12-2-solving-trig-equations-4

Find this value using the symmetry of the curve (by taking 14.48 away from 180)

180° – 14.48° = 165.52°

Give both answers correct to 1 decimal place

x = 14.5° or x = 165.5°

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 80,663 Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Jamie W

Author: Jamie W

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.