Graphs of Trigonometric Functions

Graphs of Trigonometric Functions

• This chapter focuses on using graphs of sinθ, cosθ and tanθ

• We will be seeing how to work out values of these from the graphs

• We are also going to look at transformations of these graphs

Graphs of Trigonometric FunctionsYou need to be able to recognise the graphs of sinθ, cosθ and tanθ

You will have seen all these graphs on your GCSE

The key points to remember are the peaks/troughs of each, and the points of intersection

The Cos graph is the same as the Sin graph, but shifted along (it starts at 1 instead of 0)

The Tan graph has lines called asymptotes. These are points the graph approaches but never reaches (90º, 270º etc…)

8C

1

-190º 180

º270º 360º

y

θ

y

90º 180º

270º 360º-90º-180º-360º

1

0

-1-270º

θ

y = sinθ

y = cosθ

y = tanθ

90º 180º

270º 360º-90º-180º-360º

1

0

-1-270º

θ

0-90º-270º-360º -180º

Period (length of wave) = 360º for Sin and Cos, and 180º for Tan

Graphs of Trigonometric FunctionsYou need to be able to recognise the graphs of sinθ, cosθ and tanθ

These are the same graphs, but with radians instead…

8C

1

-190º 180

º270º 360º

y

θ

y1

0

-1

θ

y = sinθ

y = cosθ

y = tanθ

1

0

-1

θ

0-90º-270º-360º -180º-2π -3π 2

-π -π 2

2π3π 2

ππ 2

90º 180º

270º 360º-90º-270º-360º -180º-2π -3π 2

-π -π 2

2π3π 2

ππ 2

90º 180º

270º 360º-90º-270º-360º -180º-2π -3π 2

-π -π 2

2π3π 2

ππ 2

Graphs of Trigonometric FunctionsYou need to be able to recognise the graphs of

sinθ, cosθ and tanθ

You need to be able to work out larger values of sin, cos

and tan as acute angles (0º - 90º)

Write sin 130º as sine of an acute angle

(sometimes asked as a ‘trigonometric ratio’)

Sin 130º = Sin 50º

8C

y = sinθ

1

90º 180º 270º 360º

y

θ0

-1

13050-40-40

Draw a sketch of the graph Mark on 130º Using the fact that the graph has symmetry, find an acute value of θ which has the same value as sin 130

+30

+30





Write cos (-120)º as cos of an acute angle

Cos(-120)º = -Cos 60º

8C

Draw a sketch of the graph Mark on -120º Using the fact that the graph has symmetry, find an acute value of θ which has the same numerical value as cos (-120)

y = cosθ

-270º

y

1

0

-1

θ90º 180º 270º-180º

-120

-60 60

-90º

+60

+60

The value you find here will have the same digits in it, but will be multiplied by -1





Write tan 4π/3 as tan of an acute angle

Tan 4π/3 = Tan π/3

8C

Draw a sketch of the graph Mark on 4π/3

Using the fact that the graph has symmetry, find an acute value of θ which has the same numerical value as tan 4π/3

y = tanθ

1

0

-1

θ2π3π

2ππ

2

4 3

1 3

1 3+1

3+

Graphs of Trigonometric FunctionsYou need to be able to find the exact values of some Trigonometrical Ratios

Some values of Sin, Cos or Tan can be written using fractions, surds, or combinations of both…

We can use an Equilateral Triangle with sides of length 2 to show this.

Using Pythagoras, the missing side in the right angled triangle is √3 (Square root of 22-12)

8D

60˚ 60˚

60˚

2

2 2

2

1

√3

60˚

30˚Opp Hyp

Opp

Hyp

Sinθ = 1 2Sin30

=√3

2Sin60 =

Opp





8D

60˚ 60˚

60˚

2

2 2

2

1

√3

60˚

30˚Adj Hyp

Adj

Hyp

Cosθ = √3 2Cos30

=1 2Cos60 =

Adj





8D

60˚ 60˚

60˚

2

2 2

2

1

√3

60˚

30˚Opp Adj

AdjOpp

Tanθ = 1 √3Tan30

=√3Tan60 =

OppAdj√3

3=



We can also do a similar demonstration with a right-angled Isosceles triangle, with the equal sides being of length 1 unit.

Using Pythagoras’ Theorem, the hypotenuse will be of length √2 (Square root of 12 + 12)

8D

45˚

1

1

√2

Opp HypSinθ = 1

√2Sin45 =

Opp

Hyp

√2 2=





8D

45˚

1

1

√2

Adj HypCosθ = 1

√2Cos45 =

Adj

Hyp

√2 2=





8D

45˚

1

1

√2

Opp AdjTanθ = 1

1Tan45 =

Opp

Adj

1=

Graphs of Trigonometric FunctionsYou need to be able to recognise transformations of graphs, and sketch them

Transformation type 1

This stretches the graph vertically by a factor ‘a’.

“Multiplying sinθ by a number will affect the y value directly”

8F

siny siny a

3siny

1 sin2

y

1

-190º 180

º270º 360º

y

θ

y = sinθ

0

3

-3

90º 180º

270º 360º

y

θ

y = 3sinθ

0

0.5

-0.590º 180

º270º 360º

y

θ

y = ½sinθ

0

Y values 3 times as big

Y values halved




8F

siny siny a

siny

1

-190º 180

º270º 360º

y

θ

y = sinθ

0

Reflection in the x axis

1

-190º 180

º270º 360º

y

θ

y = -sinθ

0

sin( )y Reflection in the y axis

1

-190º 180

º270º 360º

y

θ

y = sin(-θ)

0

(all the y values will ‘swap sign’)

(You get the same y values for the reversed x value. -90 gives the result 90

would have)




8F

cosy cosy a

cosy

1

-190º 180

º270º 360º

y

θ

y = cosθ

0

Reflection in the x axis

1

-190º 180

º270º 360º

y

θ

y = -cosθ

0

cos( )y 1

-190º 180

º270º 360º

y

θ

y = cos(-θ)

0Reflection in the

y axis



This shifts the graph vertically ‘a’ units. It is important to note that the ‘a’ is added on AFTER doing ‘sinθ’

“Adding an amount onto sinθ is a vertical shift”

8F

siny siny a

1

-190º 180

º270º 360º

y

θ

y = sinθ

0

sin 1y

2 siny

Y values all increase by 1

Y values all decrease by 2

1

-190º 180

º270º 360º

y

θ

y = sinθ + 1

0

-1

-3

y

θ

y = -2 + sinθ

-2



This shifts the graph horizontally ‘-a’ units.NOTE: The ‘a’ is added to θ before we work out the sine value…

“Adding/Subtracting an amount from the bracket is a horizontal shift”

8F

siny sin( )y a

1

-190º 180

º270º 360º

y

θ

y = sinθ

0

sin( 90)y

sin( 30)y

1

-190º 180

º270º 360º

y

θ

y = sin(θ + 90)

0

1

-190º 180

º270º 360º

y

θ

y = sin(θ – 30)

0

Y takes the same set of values, for values of θ that are 90 less than

beforeY takes the same set of values, for values of θ that are 30 more than

before

90

30



This stretches the graph horizontally by a factor ‘1/a’

“Multiplying or dividing θ in the bracket is a horizontal stretch/squash”

8F

siny sin( )y a

1

-190º 180

º270º 360º

y

θ

y = sinθ

0

sin(2 )y

sin3

y

Same set of Y values, for half the θ values

Same set of y values, for triple the θ values

1

-190º 180

º270º

y

θ

y = sin2θ

0 360º

1

-1270º 540º 810º

y

θ

y = sin(θ/3)

0 1080º

Graphs of Trigonometric FunctionsYou need to be able to answer questions with unknowns in

The graph shows the Function:

f(x) = Sinθ + k

a) Write down the value of k 0.5 (Graph 0.5 units higher)b) What is the smallest positive value of θ that gives a minimum point? 270˚c) What is the value of Sinθ at this point? -0.5

8F

1

-190º 180

º270º 360º

y

θ

y = sinθ + k

0

(90, 1.5)

Graphs of Trigonometric FunctionsYou need to be able to answer questions with unknowns in

The graph shows the Function:

f(x) = Cos(θ + k)

a) Write down the value of k 20 (Graph moved 20 units left) f(x) = Cos(θ + 20)b) What is the value of θ at x? x = 250˚c) What are the coordinates of the minimum? (160, -1)d) What is the value of Cosθ at y?

8F

1

-1

70º xº

y

θ

y = cos(θ+k)

0y

250º

f(x) = Cos(θ + k)f(x) = Cos(θ + 20)f(x) = Cos(20)f(x) = 0.94 (2dp)

We know kOn the y axis, θ = 0.Work out the answer!

Summary• We have been reminded of the

graphs for sine, cosine and tan

• We have looked at finding equivalent values on these graphs

• We have also looked at various graph transformations

Graphs of Trigonometric Functions

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Transcript of Graphs of Trigonometric Functions