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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 Functions
y
1
You need to be able to recognise
the graphs of sinθ, cosθ and tanθ
-360º
-270º
-180º
-90º
You will have seen all these graphs on
your GCSE
The Tan graph has lines called
asymptotes. These are points the
graph approaches but never reaches
(90º, 270º etc…)
Period (length of wave) = 360º for
Sin and Cos, and 180º for Tan
0
90º
180º
270º
θ
360º
-1
y
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)
y = sinθ
y = cosθ
1
-360º
-270º
-180º
-90º
0
90º
180º
270º
θ
360º
-1
y = tanθ
1
-360º
-270º
-180º
-90º
0
90º
180º
270º
θ
360º
-1
8C
Graphs of Trigonometric Functions
y
You need to be able to
recognise the graphs of sinθ,
cosθ and tanθ
y = sinθ
1
-2π
-360º
-3π
-270º
2
-π
-180º
-π
-90º
2
0
-1
These are the same graphs, but
with radians instead…
π
90º
2
π
180º
y
-2π
-360º
-3π
-270º
2
-π
-90º
2
0
-1
θ
360º
2π
y = cosθ
1
-π
-180º
3π
270º
2
π
90º
2
π
180º
3π
270º
2
θ
360º
2π
y = tanθ
1
-2π
-360º
-3π
-270º
2
-π
-180º
-π
-90º
2
0
-1
π
90º
2
π
180º
3π
270º
2
θ
360º
2π
8C
Graphs of Trigonometric Functions
You 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)
y = sinθ
y
1
-40
-40
50
0
130
θ
90º
180º
270º
360º
-1
 Write sin 130º as sine of an
acute angle
(sometimes asked as a
‘trigonometric ratio’)
Sin 130º = Sin 50º
 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
8C
Graphs of Trigonometric Functions
y = cosθ
y
You 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)
+60
-60
+30
-270º
-180º +30 -90º
-120
1
+60
60
0
θ
90º
180º
270º
-1
 Write cos (-120)º as cos of an
acute angle
Cos(-120)º = -Cos 60º
 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)
 The value you find here will have the
same digits in it, but will be multiplied by -1
8C
Graphs of Trigonometric Functions
You need to be able to
recognise the graphs of sinθ,
cosθ and tanθ
y = tanθ
1
0
You need to be able to work out
larger values of sin, cos and tan
as acute angles (0º - 90º)
-1
1
3
+1
3
4
3
π
2
π
+1
3
θ
3π
2
2π
 Write tan 4π/3 as tan of an
acute angle
Tan
4π/
3
= Tan π/3
 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
8C
Graphs of Trigonometric Functions
You need to be able to find the exact
values of some Trigonometrical Ratios
60˚
2
Some values of Sin, Cos or Tan can be
written using fractions, surds, or
combinations of both…
60˚
Opp
Sinθ = Hyp
1
Sin30 = 2
√3
Sin60 = 2
60˚
2
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)
2
Hyp
2 30˚
√3 Opp
60˚
Opp
1
8D
Graphs of Trigonometric Functions
You need to be able to find the exact
values of some Trigonometrical Ratios
60˚
2
Some values of Sin, Cos or Tan can be
written using fractions, surds, or
combinations of both…
60˚
Adj
Cosθ = Hyp
√3
Cos30 = 2
Cos60 =
1
2
60˚
2
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)
2
Hyp
2 30˚
√3 Adj
60˚
Adj
1
8D
Graphs of Trigonometric Functions
You 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…
60˚
2
60˚
Opp
Tanθ = Adj
1
√3
Tan30 = √3 = 3
Tan60 = √3
60˚
2
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)
2
2 30˚
Adj
√3 Opp
60˚
Opp
Adj
1
8D
Graphs of Trigonometric Functions
You 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 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)
Hyp
Opp
1
√2
45˚
1
Opp
Sinθ = Hyp
1
Sin45 = √2
= √2
2
8D
Graphs of Trigonometric Functions
You 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 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)
Hyp
1
√2
45˚
1
Adj
Adj
Cosθ = Hyp
1
Cos45 = √2
= √2
2
8D
Graphs of Trigonometric Functions
You 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 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)
Opp 1
√2
45˚
1
Adj
Opp
Tanθ = Adj
1
Tan45 = 1
= 1
8D
Graphs of Trigonometric Functions
y
You need to be able to recognise
transformations of graphs, and
sketch them
Transformation type 1
y  sin 
y  a sin 
This stretches the graph vertically by
a factor ‘a’.
y  3sin 
1
y  sin 
2
Y values 3 times
as big
Y values halved
“Multiplying sinθ by a number will
affect the y value directly”
y = sinθ
1
0
90º
180º
270º
θ
360º
-1
y
3
y = 3sinθ
0
90º
180º
270º
θ
360º
-3
y = ½sinθ
y
0.5
0
-0.5
90º
180º
270º
θ
360º
8F
Graphs of Trigonometric Functions
y
You need to be able to recognise
transformations of graphs, and sketch
them
Transformation type 1
y  sin 
y   sin 
1
0
Reflection in the
x axis
y  sin( )
(You get the same y values for the reversed x
value. -90 gives the result 90 would have)
180º
y
270º
θ
360º
y = -sinθ
1
0
90º
180º
270º
θ
360º
-1
(all the y values will ‘swap sign’)
Reflection in the
y axis
90º
-1
y  a sin 
This stretches the graph vertically by a
factor ‘a’.
y = sinθ
y
y = sin(-θ)
1
0
-1
90º
180º
270º
θ
360º
8F
Graphs of Trigonometric Functions
y
You need to be able to recognise
transformations of graphs, and sketch
them
Transformation type 1
y  cos 
y   cos 
1
0
Reflection in the
x axis
90º
180º
y
y  cos( )
θ
360º
y = -cosθ
1
0
90º
180º
270º
θ
360º
-1
y
Reflection in the
y axis
270º
-1
y  a cos 
This stretches the graph vertically by a
factor ‘a’.
y = cosθ
y = cos(-θ)
1
0
-1
90º
180º
270º
θ
360º
8F
Graphs of Trigonometric Functions
y
You need to be able to recognise
transformations of graphs, and sketch
them
Transformation type 2
y  sin 
y  sin   a
This shifts the graph vertically ‘a’ units.
 It is important to note that the ‘a’ is
added on AFTER doing ‘sinθ’
y  sin   1
Y values all increase
by 1
y  2  sin 
Y values all decrease
by 2
“Adding an amount onto sinθ is a vertical
shift”
y = sinθ
1
0
90º
180º
270º
θ
360º
-1
y
y = sinθ + 1
1
0
90º
180º
270º
θ
360º
-1
y
y = -2 + sinθ
-1
-2
θ
-3
8F
Graphs of Trigonometric Functions
y
You need to be able to recognise
transformations of graphs, and sketch
them
Transformation type 3
y  sin 
y  sin(  a)
This shifts the graph horizontally ‘-a’
units.
NOTE: The ‘a’ is added to θ before we
work out the sine value…
Y takes the same set of
values, for values of θ
y  sin(  90)
that are 90 less than
before
Y takes the
same set of
y  sin(  30)
values, for values of θ
that are 30 more than
before
“Adding/Subtracting an amount from
the bracket is a horizontal shift”
y = sinθ
1
0
90º
180º
270º
θ
360º
-1
y
90
y = sin(θ + 90)
1
0
90º
180º
270º
θ
360º
-1
y
30
y = sin(θ – 30)
1
0
90º
180º
270º
θ
360º
-1
8F
Graphs of Trigonometric Functions
y
You need to be able to recognise
transformations of graphs, and sketch
them
1
0
90º
180º
270º
θ
360º
-1
Transformation type 4
y  sin 
y = sinθ
y  sin(a )
This stretches the graph horizontally
by a factor ‘1/a’
y  sin(2 )
Same set of Y values, for
half the θ values
 
y  sin  
3
Same set of y values, for
triple the θ values
“Multiplying or dividing θ in the bracket
is a horizontal stretch/squash”
y
y = sin2θ
1
0
90º
180º
270º
θ
360º
-1
y
y = sin(θ/3)
1
0
270º
540º
810º
θ
1080º
-1
8F
Graphs of Trigonometric Functions
y
You need to be able to answer
questions with unknowns in
The graph shows the Function:
(90, 1.5)
y = sinθ + k
1
0
90º
180º
270º
θ
360º
-1
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
Graphs of Trigonometric Functions
y
You 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?
y = cos(θ+k)
1 y
0
70º
250º
xº
θ
-1
f(x) = Cos(θ + k)
f(x) = Cos(θ + 20)
f(x) = Cos(20)
f(x) = 0.94 (2dp)
We know k
On the y
axis, θ = 0.
Work out
the answer!
8F
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