Investigate transformations of sinusoidal functions

Ontario Grade 11 Mathematics

Trigonometric Functions

Stretching, Reflecting, and Shifting Sine and Cosine Graphs

You already know that the graphs of

y = \sin x

and

y = \cos x

produce smooth, repeating wave shapes. These base graphs are useful, but real-world periodic patterns such as ocean tides, sound waves, and seasonal temperature changes rarely match the plain base graph exactly. To model those situations, you need to stretch, compress, flip, and slide the base graph. This lesson shows you exactly how to do that using four parameters. By the end, you will be able to look at an equation and immediately picture its graph, or look at a graph and write its equation.

What you will learn

Identify the parameters a, k, d, and c in the equations y = a sin(k(x − d)) + c and y = a cos(k(x − d)) + c and describe what each one does to the graph.
Determine the amplitude, period, phase shift, and vertical shift of a sinusoidal function from its equation.
Sketch the graph of a transformed sinusoidal function by applying transformations in the correct order.
Write the equation of a sinusoidal function from a description of its key features.
Explain how changing each parameter affects the shape and position of the sinusoidal curve.

Prerequisite Bridge: The Base Sinusoidal Graphs

Before transforming anything, make sure you are confident with the two base graphs. The function

y = \sin x

starts at the origin

(0^{\circ}, 0)

, rises to a maximum of

1

at

x = 90^{\circ}

, returns to

0

at

x = 180^{\circ}

, falls to a minimum of

-1

at

x = 270^{\circ}

, and completes one full cycle back at

x = 360^{\circ}

. The function

y = \cos x

follows the same wave shape but starts at its maximum: it equals

1

at

x = 0^{\circ}

, drops to

0

at

x = 90^{\circ}

, reaches

-1

at

x = 180^{\circ}

, and returns to

1

at

x = 360^{\circ}

.

Two key measurements describe these base graphs. The amplitude is the distance from the midline (the horizontal centre of the wave) to either the maximum or the minimum. For both base graphs the amplitude is

1

. The period is the horizontal length of one complete cycle. For both base graphs the period is

360^{\circ}

. Every transformation you apply in this lesson will change one or more of these measurements or shift the graph from its original position.

Base graphs: $y = \sin x$ starts at $(0^{\circ}, 0)$ ; $y = \cos x$ starts at $(0^{\circ}, 1)$ .
Amplitude of the base graphs equals $1$ ; period of the base graphs equals $360^{\circ}$ .
The midline of both base graphs is $y = 0$ .

The Four Parameters: What Each One Does

The general transformed sinusoidal function is written as

y = a\sin(k(x - d)) + c

or

y = a\cos(k(x - d)) + c

. The four letters

a

,

k

,

d

, and

c

are called parameters. Each one controls a different feature of the graph, and you can read those features directly from the equation once you know the rules.

The parameter

a

controls the amplitude. The amplitude equals |a|, which is the absolute value of

a

. If

|a| > 1

, the wave is vertically stretched, meaning it reaches higher and lower than the base graph. If

0 < |a| < 1

, the wave is vertically compressed. When

a

is negative, the graph is also reflected across the midline, meaning the peaks and valleys swap positions. For example,

a = -3

gives an amplitude of

3

and a reflection.

The parameter

k

controls the period. The period of the transformed graph is calculated as

\frac{360^{\circ}}{|k|}

. When

|k| > 1

, the cycle is shorter, giving a horizontal compression; when

0 < |k| < 1

, the cycle is longer, giving a horizontal stretch. For example,

k = 2

gives a period of

180^{\circ}

, and

k = \frac{1}{2}

gives a period of

720^{\circ}

.

The parameter

d

controls the phase shift, which is a horizontal translation. The graph slides

d

degrees to the right when

d > 0

and to the left when

d < 0

. Pay close attention to the sign: the expression inside the brackets is

(x - d)

, so

y = \sin(x - 50^{\circ})

shifts the graph

50^{\circ}

to the right, not to the left.

The parameter

c

controls the vertical shift, also called the vertical translation. It moves the entire midline up by

c

units when

c > 0

and down when

c < 0

. The midline of the transformed graph is the horizontal line

y = c

. The maximum value of the function is

c + |a|

and the minimum value is

c - |a|

.

y = a\sin(k(x - d)) + c

Amplitude equals |a|; if $a < 0$ , the graph is reflected vertically.
Period equals $\frac{360^{\circ}}{|k|}$ ; a larger |k| means a shorter, faster cycle.
Phase shift: the graph moves $d$ degrees to the right for positive $d$ or left for negative $d$ .
Vertical shift: the midline moves to $y = c$ ; maximum equals $c + |a|$ , minimum equals $c - |a|$ .
Read the equation carefully: the bracket form $(x - d)$ means the sign of $d$ gives the direction of the shift.

Reading the Key Features from an Equation

A reliable strategy is to extract each parameter one at a time and record the four key features before you draw anything. Start with

a

to find the amplitude and whether there is a reflection. Then use

k

to calculate the period. Next, read

d

to find the phase shift. Finally, read

c

for the vertical shift and midline.

Consider the function

y = 3\sin(2(x - 40^{\circ})) + 1

. Here

a = 3

, so the amplitude is

3

and there is no reflection. Since

k = 2

, the period is

\frac{360^{\circ}}{2} = 180^{\circ}

. The phase shift is

40^{\circ}

to the right because

d = 40^{\circ}

. The vertical shift is

c = 1

, so the midline is

y = 1

, the maximum is

1 + 3 = 4

, and the minimum is

1 - 3 = -2

.

It is important to factor the bracket correctly before reading

d

. If the equation is written as

y = \sin(2x - 80^{\circ})

, you must factor out the

k

value first:

y = \sin(2(x - 40^{\circ}))

. Now you can clearly see

d = 40^{\circ}

. Failing to factor first is one of the most common errors in this topic.

\text{Period} = \frac{360^{\circ}}{|k|}

Extract $a$ , $k$ , $d$ , $c$ one at a time before sketching.
Always factor the bracket to isolate $(x - d)$ before reading the phase shift.
Midline is $y = c$ ; maximum equals $c + |a|$ ; minimum equals $c - |a|$ .

Sketching the Transformed Graph Step by Step

Once you have the four key features, sketching the graph is a structured process. The goal is to plot one complete cycle accurately and then extend it if needed.

Step 1 is to draw the midline as a dashed horizontal line at

y = c

. This is your new reference line and it replaces

y = 0

from the base graph. Step 2 is to mark the maximum at

c + |a|

and the minimum at

c - |a|

on the vertical axis. Step 3 is to locate where the cycle starts. For a sine function, the cycle starts at the phase shift

x = d

, where the function crosses the midline going upward, assuming

a > 0

. For a cosine function, the cycle starts at

x = d

at the maximum, assuming

a > 0

. Step 4 is to divide the period into four equal quarters. These quarter-period points mark the five key points of one cycle: start at midline, peak at max, middle at midline, valley at min, end at midline for sine; or peak, midline going down, valley, midline going up, peak for cosine. Step 5 is to plot the five key points and draw a smooth wave through them.

For a reflected graph, when

a < 0

, simply swap the peak and valley: the first quarter dips to the minimum instead of rising to the maximum. The shape is otherwise identical.

Draw the midline first because it is the anchor for all other measurements.
Divide the period into four equal quarters to locate the five key points.
For $a > 0$ sine, the cycle begins at the midline going up; for $a > 0$ cosine, the cycle begins at the maximum.
Swap peaks and valleys when $a < 0$ .

Writing an Equation from a Graph or Description

Sometimes you are given a graph or a word description and must produce the equation. Work backwards through the four parameters. First, find the maximum and minimum values. The amplitude is half the total vertical distance:

|a| = \frac{\text{max} - \text{min}}{2}

. The midline, and therefore

c

, is the average:

c = \frac{\text{max} + \text{min}}{2}

.

Next, measure the period from the graph, which is the horizontal length of one complete cycle, and solve for

k

using

k = \frac{360^{\circ}}{\text{period}}

. Then identify the phase shift by locating where the characteristic starting point of your chosen function type appears (midline-crossing for sine, maximum for cosine), and set that equal to

d

. Finally, decide whether

a

is positive or negative by checking whether the graph goes up or down from its starting point.

Choosing between a sine model and a cosine model is often a matter of convenience. If the graph begins at the midline going upward, a sine model with

a > 0

is the most natural choice. If it begins at the maximum, a cosine model with

a > 0

is simpler. Either function can describe any sinusoidal graph, and the phase shift will adjust accordingly.

|a| = \frac{\text{max} - \text{min}}{2}

$|a| = \frac{\text{max} - \text{min}}{2}$ and $c = \frac{\text{max} + \text{min}}{2}$ .
$k = \frac{360^{\circ}}{\text{period}}$ .
Locate the characteristic starting point to determine $d$ .
Choose $a > 0$ or $a < 0$ based on whether the graph initially rises or falls from its starting point.

Summary of the Four Parameters

Parameter	What It Changes	Feature It Controls	Example Value	Effect
$a$	Vertical scale	Amplitude	$a = 3$	Amplitude equals $3$ ; wave is taller
$a$ negative	Vertical scale and flip	Amplitude and reflection	$a = -3$	Amplitude equals $3$ ; graph flipped
$k$	Horizontal scale	Period	$k = 2$	Period equals $180^{\circ}$ ; faster cycle
$d$	Horizontal position	Phase shift	$d = 45^{\circ}$	Graph shifts $45^{\circ}$ to the right
$c$	Vertical position	Midline	$c = -1$	Midline at $y = -1$ ; graph shifts down

Worked example

Example 1: Identifying Features and Sketching from an Equation

For the function

y = -2\cos(3(x - 30^{\circ})) + 5

, state the amplitude, period, phase shift, vertical shift, midline, maximum value, and minimum value. Then describe the five key points of one complete cycle starting at

x = 30^{\circ}

.

Read the parameter a
The value of $a$ is $-2$ . The amplitude is $|a| = |-2| = 2$ . Because $a$ is negative, the graph is reflected, so peaks become valleys and valleys become peaks compared to a standard cosine.
$|a| = 2$
Calculate the period using k
The value of $k$ is $3$ . Substitute into the period formula: period equals $\frac{360^{\circ}}{|k|} = \frac{360^{\circ}}{3} = 120^{\circ}$ . One complete cycle spans $120^{\circ}$ .
$\frac{360^{\circ}}{3} = 120^{\circ}$
Read the phase shift d
The bracket is already factored as $(x - 30^{\circ})$ , so $d = 30^{\circ}$ . The graph is shifted $30^{\circ}$ to the right.
$d = 30^{\circ}$
Read the vertical shift c and find the midline
The value of $c$ is $5$ , so the midline is the horizontal line $y = 5$ . Every output of this function is measured from $y = 5$ , not from $y = 0$ .
$y = 5$
Find the maximum and minimum values
Maximum value equals $c + |a| = 5 + 2 = 7$ . Minimum value equals $c - |a| = 5 - 2 = 3$ . Because of the reflection where $a < 0$ , the graph starts at the minimum rather than the maximum at $x = d$ .
$\text{max} = 7, \ \text{min} = 3$
List the five key points across one cycle
Each quarter of the period is $120^{\circ} \div 4 = 30^{\circ}$ . For a reflected cosine where $a < 0$ , the cycle starts at the minimum, rises through the midline, reaches the maximum, falls through the midline, and returns to the minimum. Starting at $x = 30^{\circ}$ : point 1 is $(30^{\circ}, 3)$ at the minimum; point 2 is $(60^{\circ}, 5)$ at the midline; point 3 is $(90^{\circ}, 7)$ at the maximum; point 4 is $(120^{\circ}, 5)$ at the midline; point 5 is $(150^{\circ}, 3)$ back at the minimum, completing one cycle.

Answer: Amplitude equals

2

; period equals

120^{\circ}

; phase shift is

30^{\circ}

to the right; midline is

y = 5

; maximum is

7

; minimum is

3

. Key points:

(30^{\circ}, 3)

,

(60^{\circ}, 5)

,

(90^{\circ}, 7)

,

(120^{\circ}, 5)

,

(150^{\circ}, 3)

.

Check: Check the period: the cycle runs from

x = 30^{\circ}

to

x = 150^{\circ}

, a span of

150^{\circ} - 30^{\circ} = 120^{\circ}

. This matches

\frac{360^{\circ}}{3} = 120^{\circ}

. Check the maximum:

5 + 2 = 7

. Check the minimum:

5 - 2 = 3

. The reflected cosine correctly dips to the minimum first at

x = 30^{\circ}

. All values are consistent.

Worked example

Example 2: Writing an Equation from a Description

A sinusoidal function has a maximum value of

11

and a minimum value of

3

. It completes one full cycle every

240^{\circ}

. The first maximum after

x = 0^{\circ}

occurs at

x = 20^{\circ}

. Write an equation for this function in the form

y = a\cos(k(x - d)) + c

.

Calculate the amplitude
The amplitude is half the total vertical range. Subtract the minimum from the maximum and divide by two: $|a| = \frac{11 - 3}{2} = \frac{8}{2} = 4$ . The graph rises and falls by $4$ units from the midline.
$|a| = \frac{11 - 3}{2} = 4$
Find the midline and the value of c
The midline sits exactly halfway between the maximum and minimum. Average the two values: $c = \frac{11 + 3}{2} = \frac{14}{2} = 7$ . The midline is $y = 7$ .
$c = \frac{11 + 3}{2} = 7$
Find k from the period
The period is given as $240^{\circ}$ . Rearrange the period formula to solve for $k$ : $k = \frac{360^{\circ}}{\text{period}} = \frac{360^{\circ}}{240^{\circ}} = 1.5$ . So $k = \frac{3}{2}$ .
$k = \frac{360^{\circ}}{240^{\circ}} = \frac{3}{2}$
Determine the phase shift d
A cosine function with $a > 0$ starts each cycle at its maximum. The first maximum occurs at $x = 20^{\circ}$ , so the phase shift is $d = 20^{\circ}$ . The graph is shifted $20^{\circ}$ to the right.
$d = 20^{\circ}$
Determine the sign of a
Because the problem asks for a cosine model and the function begins at its maximum, which is the natural start for $y = \cos x$ with $a > 0$ , use $a = +4$ . No reflection is needed.
$a = 4$
Write the final equation
Substitute $a = 4$ , $k = \frac{3}{2}$ , $d = 20^{\circ}$ , and $c = 7$ into the general form $y = a\cos(k(x - d)) + c$ .
$y = 4\cos\left(\frac{3}{2}(x - 20^{\circ})\right) + 7$

Answer:

y = 4\cos\left(\frac{3}{2}(x - 20^{\circ})\right) + 7

Check: Verify at

x = 20^{\circ}

:

y = 4\cos\left(\frac{3}{2}(20^{\circ} - 20^{\circ})\right) + 7 = 4\cos(0^{\circ}) + 7 = 4(1) + 7 = 11

. This matches the stated maximum of

11

. Check the period: at

x = 20^{\circ} + 240^{\circ} = 260^{\circ}

, the cosine argument is

\frac{3}{2}(260^{\circ} - 20^{\circ}) = \frac{3}{2}(240^{\circ}) = 360^{\circ}

, and

\cos(360^{\circ}) = 1

, giving

y = 11

again, confirming one full cycle. Check minimum:

c - |a| = 7 - 4 = 3

. All checks pass.

Common mistakes and how to avoid them

Reading the phase shift directly from

y = a\sin(kx - b) + c

without factoring, and concluding the shift is

b

degrees.

Correction: Always factor out

k

first to get

y = a\sin(k(x - d)) + c

. The phase shift is

d = \frac{b}{k}

, not

b

.

Confusing the sign of

d

: seeing

(x - 40^{\circ})

and thinking the graph shifts left.

Correction: The form

(x - d)

shifts the graph to the right when

d > 0

. Think of it as the starting

x

-value of the cycle.

Using the amplitude as the maximum value, forgetting to add the vertical shift

c

.

Correction: The maximum value is

c + |a|

and the minimum is

c - |a|

. The amplitude is just the distance from the midline to the peak, not the peak's actual height.

Forgetting to take the absolute value of

a

when a negative

a

is given, and reporting a negative amplitude.

Correction: Amplitude is always a positive number. Amplitude equals |a|. The negative sign tells you about reflection, not about the size of the wave.

Dividing

360^{\circ}

by the period to find

k

but accidentally dividing the period by

360^{\circ}

instead.

Correction: The correct formula is

k = \frac{360^{\circ}}{\text{period}}

. A longer period means a smaller

k

, which makes sense because the wave stretches out horizontally.

Lesson summary

The general forms $y = a\sin(k(x - d)) + c$ and $y = a\cos(k(x - d)) + c$ describe all transformations of the base sinusoidal graphs.
The amplitude equals |a|; a negative $a$ also reflects the graph across the midline.
The period equals $\frac{360^{\circ}}{|k|}$ ; a larger |k| produces a shorter, faster cycle.
The phase shift is $d$ degrees to the right for positive $d$ or left for negative $d$ ; always factor the bracket before reading $d$ .
The midline is $y = c$ ; the maximum value is $c + |a|$ and the minimum value is $c - |a|$ .
To write an equation from a graph: find |a| and $c$ from the max and min, find $k$ from the period, and find $d$ from the position of the characteristic starting point.

Check your understanding

Question 1

What is the amplitude and period of

y = 5\sin(4(x - 15^{\circ})) - 2

?

Amplitude equals $5$ , period equals $90^{\circ}$
Amplitude equals $5$ , period equals $4^{\circ}$
Amplitude equals $-5$ , period equals $90^{\circ}$
Amplitude equals $5$ , period equals $1440^{\circ}$

Show answer and explanation

Amplitude equals

5

, period equals

90^{\circ}

The amplitude is

|a| = |5| = 5

. The period is

\frac{360^{\circ}}{|k|} = \frac{360^{\circ}}{4} = 90^{\circ}

. The value

c = -2

affects the midline, not the amplitude, and

k = 4

goes in the denominator of the period formula.

Question 2

The equation

y = \sin(2x - 60^{\circ})

is written without factoring. What is the correct phase shift after factoring?

$60^{\circ}$ to the right
$30^{\circ}$ to the right
$60^{\circ}$ to the left
$120^{\circ}$ to the right

Show answer and explanation

30^{\circ}

to the right

Factor out

k = 2

:

y = \sin(2(x - 30^{\circ}))

. Now

d = 30^{\circ}

, so the phase shift is

30^{\circ}

to the right. Reading

60^{\circ}

directly without factoring is the classic error this question targets.

Question 3

A sinusoidal function has a maximum of

9

and a minimum of

1

. What are its amplitude and midline?

Amplitude equals $9$ , midline $y = 1$
Amplitude equals $4$ , midline $y = 5$
Amplitude equals $8$ , midline $y = 5$
Amplitude equals $4$ , midline $y = 4$

Show answer and explanation

Amplitude equals

4

, midline

y = 5

Amplitude equals

\frac{9 - 1}{2} = \frac{8}{2} = 4

. Midline equals

\frac{9 + 1}{2} = \frac{10}{2} = 5

. The amplitude is the half-range, not the full range, and the midline is the average of max and min.

Question 4

Which equation represents a cosine graph with amplitude

6

, period

180^{\circ}

, phase shift

0^{\circ}

, and a midline at

y = -3

?

$y = 6\cos(2x) + 3$
$y = 6\cos(2x) - 3$
$y = 6\cos(180x) - 3$
$y = 3\cos(2x) - 6$

Show answer and explanation

y = 6\cos(2x) - 3

Amplitude equals

6

means

a = 6

. Period equals

180^{\circ}

gives

k = \frac{360^{\circ}}{180^{\circ}} = 2

. No phase shift means

d = 0^{\circ}

. Midline at

y = -3

means

c = -3

. Substituting gives

y = 6\cos(2(x - 0^{\circ})) + (-3) = 6\cos(2x) - 3

.

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