DoAssignment.ca

D2.7 · Sketch transformed sine and cosine functions

Learn to sketch transformed sine and cosine functions through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Trigonometric Functions

Understanding How Parameters Shape the Graph

You already know what the basic graphs of y=sin⁡xy = \sin x and y=cos⁡xy = \cos x look like — smooth, repeating waves that cycle between −1-1 and 11. In the real world, however, almost nothing follows that exact basic shape. Ocean tides, sound waves, seasonal temperatures, and the swing of a pendulum all repeat in wave-like patterns, but with different heights, different widths, and different starting positions. In this lesson you will learn to read a transformed equation and use it to produce an accurate sketch — without a calculator or graphing technology. All angles in this course are measured in degrees.

What you will learn

Prerequisite Bridge: The Parent Graphs

Before adding transformations, recall the two parent graphs you studied earlier. The function y=sin⁡xy = \sin x starts at the origin (0∘,0)(0^\circ, 0), rises to a maximum of 11 at x=90∘x = 90^\circ, returns to zero at x=180∘x = 180^\circ, dips to a minimum of −1-1 at x=270∘x = 270^\circ, and completes one full cycle back at x=360∘x = 360^\circ.
The function y=cos⁡xy = \cos x starts at a maximum of 11 when x=0∘x = 0^\circ, crosses zero at x=90∘x = 90^\circ, reaches a minimum of −1-1 at x=180∘x = 180^\circ, crosses zero again at x=270∘x = 270^\circ, and completes one full cycle back at x=360∘x = 360^\circ.
Both graphs repeat forever. The length of one complete cycle is called the period, which equals 360∘360^\circ for both parent functions. The distance from the midline (the horizontal middle of the wave) to a peak or trough is 11 for both parents. These two facts — period and midline distance — are the foundation for everything that follows.

The Transformation Equation and Its Four Parameters

A transformed sine or cosine function is written in the form y=asin⁡(k(x−d))+cy = a\sin(k(x - d)) + c or y=acos⁡(k(x−d))+cy = a\cos(k(x - d)) + c. Each of the four letters aa, kk, dd, and cc controls one specific feature of the graph. Understanding what each parameter does — independently — is the key to sketching quickly and accurately.
The parameter aa is the amplitude factor. It tells you how tall the wave is, measured from the midline to a peak or trough. If a=3a = 3, the wave reaches 33 units above and 33 units below the midline. If aa is negative, the wave is also reflected vertically — a peak becomes a trough and vice versa. Always use |a| when stating amplitude because amplitude is never negative.
The parameter kk controls the period. A larger kk squeezes the cycle horizontally, making it shorter. The period of the transformed function is 360∘∣k∣\frac{360^\circ}{|k|}. For example, k=2k = 2 gives a period of 180∘180^\circ, meaning the wave completes a full cycle in half the usual horizontal distance.
The parameter dd is the phase shift, also called the horizontal shift. It moves the entire graph left or right. When the equation is written as k(x−d)k(x - d), a positive dd shifts the graph to the right by dd degrees and a negative dd shifts it to the left. Many errors come from misreading the sign here — always rewrite the bracket in the form (x−d)(x - d) first.
The parameter cc is the vertical shift. It moves the entire graph up (if c>0c > 0) or down (if c<0c < 0) from the xx-axis. The midline of the graph — the horizontal line running through the middle of the wave — is the line y=cy = c. The maximum value of the function is c+∣a∣c + |a| and the minimum value is c−∣a∣c - |a|.
y=asin⁡(k(x−d))+cy = a\sin(k(x - d)) + c

A Strategy for Sketching: Five Key Points Per Cycle

Rather than plotting dozens of points, experienced sketchers locate exactly five key points that define one full cycle: the start of the cycle, the first quarter-point, the halfway point, the three-quarter point, and the end of the cycle. These positions are found by dividing the period into four equal parts. Each quarter-period step moves you from one key feature to the next.
For a sine-based function, the five key points within one cycle follow this pattern along the midline and peaks: midline → max → midline → min → midline (assuming a>0a > 0). For a cosine-based function with a>0a > 0, the pattern is: max → midline → min → midline → max. If a<0a < 0, every maximum becomes a minimum and vice versa.
Here is the step-by-step sketching strategy. First, identify aa, kk, dd, and cc from the equation. Second, calculate the amplitude |a|, the period P=360∘∣k∣P = \frac{360^\circ}{|k|}, and the quarter-period P4\frac{P}{4}. Third, draw the midline y=cy = c as a dashed horizontal reference line. Fourth, mark the maximum y=c+∣a∣y = c + |a| and minimum y=c−∣a∣y = c - |a| as dashed boundary lines. Fifth, place the five key points starting at x=dx = d (the phase shift) and adding P4\frac{P}{4} each time. Sixth, draw a smooth wave through the five points. Finally, extend the wave left and right as needed to show at least one full cycle.
P=360∘∣k∣P = \frac{360^\circ}{|k|}

Reading the Equation Carefully: Common Traps

One of the most common difficulties is extracting kk and dd correctly when the equation is not already fully factored. For example, y=sin⁡(2x−90∘)y = \sin(2x - 90^\circ) looks like the phase shift is 90∘90^\circ to the right, but that is wrong. You must factor out kk first: y=sin⁡(2(x−45∘))y = \sin(2(x - 45^\circ)). Now you can read k=2k = 2 and d=45∘d = 45^\circ correctly. The phase shift is 45∘45^\circ to the right, not 90∘90^\circ.
Another trap involves the sign of aa. The equation y=−4cos⁡(x)+1y = -4\cos(x) + 1 has amplitude ∣−4∣=4|-4| = 4, a midline at y=1y = 1, and a reflection. Because aa is negative, the cosine graph is flipped: instead of starting at a maximum of 1+4=51 + 4 = 5, it starts at a minimum of 1−4=−31 - 4 = -3.
Also watch the sign inside the bracket for the phase shift. The equation y=sin⁡(k(x−d))+cy = \sin(k(x - d)) + c has a phase shift of dd to the right. If the bracket reads (x+30∘)(x + 30^\circ), rewrite it as (x−(−30∘))(x - (-30^\circ)), giving a phase shift of −30∘-30^\circ, which is a shift to the left by 30∘30^\circ.

Connecting the Sketch to Real Meaning

Once you can sketch these graphs reliably, you can interpret what each feature means in context. The amplitude tells you the size of the variation from average — for example, how many degrees above or below average a temperature swings. The period tells you how long one complete cycle takes — hours for a tide, months for a seasonal pattern. The vertical shift tells you the average or baseline value. The phase shift tells you when the cycle starts — for instance, which hour of the day a tide first reaches its maximum.
Being able to sketch from an equation — without technology — is a foundational skill. It trains you to reason about the behaviour of a function directly from its algebraic form, which becomes important throughout the rest of this course and beyond.

Summary of the Four Transformation Parameters

ParameterNameEffect on GraphHow to Calculate
aaAmplitude factorStretches or compresses vertically; negative aa reflects the graphAmplitude =∣a∣= |a|
kkPeriod factorCompresses (∣k∣>1|k|>1) or stretches (∣k∣<1|k|<1) the cycle horizontallyP=360∘∣k∣P = \frac{360^\circ}{|k|}
ddPhase shiftShifts the graph right (positive dd) or left (negative dd)Factor out kk first, then read dd
ccVertical shiftMoves the entire graph up or down; sets the midline at y=cy = cMax =c+∣a∣= c+|a|; Min =c−∣a∣= c-|a|

Worked example

Sketching a Transformed Sine Function

Sketch one full cycle of y=3sin⁡(2(x−30∘))+1y = 3\sin(2(x - 30^\circ)) + 1. Identify the amplitude, period, phase shift, midline, maximum, and minimum before sketching.
  1. Identify the four parameters
    Read the values directly from the equation y=3sin⁡(2(x−30∘))+1y = 3\sin(2(x - 30^\circ)) + 1. Here a=3a = 3, k=2k = 2, d=30∘d = 30^\circ, and c=1c = 1. The equation is already factored correctly, so no extra algebra is needed.
    a=3,k=2,d=30∘,c=1a = 3, \quad k = 2, \quad d = 30^\circ, \quad c = 1
  2. Calculate amplitude, period, and quarter-period
    The amplitude is ∣a∣=∣3∣=3|a| = |3| = 3. The period is 360∘∣k∣=360∘2=180∘\frac{360^\circ}{|k|} = \frac{360^\circ}{2} = 180^\circ. Dividing the period into four equal parts gives a quarter-period of 180∘4=45∘\frac{180^\circ}{4} = 45^\circ.
    P=360∘2=180∘,P4=45∘P = \frac{360^\circ}{2} = 180^\circ, \qquad \frac{P}{4} = 45^\circ
  3. State the midline, maximum, and minimum
    The midline is the horizontal line y=c=1y = c = 1. The maximum value is c+∣a∣=1+3=4c + |a| = 1 + 3 = 4. The minimum value is c−∣a∣=1−3=−2c - |a| = 1 - 3 = -2. Draw dashed horizontal lines at y=4y = 4, y=1y = 1, and y=−2y = -2 on your sketch.
    ymax=4,ymid=1,ymin=−2y_{\text{max}} = 4, \quad y_{\text{mid}} = 1, \quad y_{\text{min}} = -2
  4. List the five key x-values
    The cycle starts at the phase shift x=d=30∘x = d = 30^\circ. Add the quarter-period of 45∘45^\circ repeatedly to find the remaining four x-values.
    30°,75°,120°,165°,210°30°, 75°, 120°, 165°, 210°
  5. Assign y-values to each key x-value
    Because a>0a > 0 and the base function is sine, the pattern of y-values for one cycle is: midline, max, midline, min, midline. Pair each x-value with its y-value to get the five key points.
    (30°, 1), (75°, 4), (120°, 1), (165°, −2), (210°, 1)(30°,\ 1),\ (75°,\ 4),\ (120°,\ 1),\ (165°,\ -2),\ (210°,\ 1)
  6. Sketch the curve
    Plot the five key points on a coordinate grid, with the x-axis labelled in degrees and the y-axis showing values from −2-2 to 44. Draw a smooth, continuous wave through the points — rising from the midline, arching to the maximum, falling back to the midline, dipping to the minimum, then returning to the midline. This completes one full cycle.
Answer: Amplitude =3= 3, Period =180∘= 180^\circ, Phase shift =30∘= 30^\circ right, Midline y=1y = 1. Key points: (30∘,1)(30^\circ, 1), (75∘,4)(75^\circ, 4), (120∘,1)(120^\circ, 1), (165∘,−2)(165^\circ, -2), (210∘,1)(210^\circ, 1).
Check: Substitute the maximum point into the original equation to verify. At x=75∘x = 75^\circ: y=3sin⁡(2(75∘−30∘))+1=3sin⁡(90∘)+1=3(1)+1=4y = 3\sin(2(75^\circ - 30^\circ)) + 1 = 3\sin(90^\circ) + 1 = 3(1) + 1 = 4. This matches the maximum point (75∘,4)(75^\circ, 4). ✓

Worked example

Sketching a Reflected, Shifted Cosine Function

Sketch one full cycle of y=−2cos⁡(3x+90∘)y = -2\cos(3x + 90^\circ). Identify all four parameters. Note: the equation is not yet in standard factored form.
  1. Rewrite in standard factored form
    The bracket 3x+90∘3x + 90^\circ is not factored, so you cannot read the phase shift directly. Factor out k=3k = 3: write 3x+90∘=3(x+30∘)=3(x−(−30∘))3x + 90^\circ = 3(x + 30^\circ) = 3(x - (-30^\circ)). The equation becomes y=−2cos⁡(3(x−(−30∘)))y = -2\cos(3(x - (-30^\circ))).
    y=−2cos⁡(3(x+30∘))y = -2\cos(3(x + 30^\circ))
  2. Identify the four parameters
    From the factored equation, read: a=−2a = -2, k=3k = 3, d=−30∘d = -30^\circ (shift left 30∘30^\circ), and c=0c = 0 (no vertical shift, so the midline is the x-axis).
    a=−2,k=3,d=−30∘,c=0a = -2, \quad k = 3, \quad d = -30^\circ, \quad c = 0
  3. Calculate amplitude, period, and quarter-period
    The amplitude is ∣a∣=∣−2∣=2|a| = |-2| = 2. The period is 360∘∣k∣=360∘3=120∘\frac{360^\circ}{|k|} = \frac{360^\circ}{3} = 120^\circ. The quarter-period is 120∘4=30∘\frac{120^\circ}{4} = 30^\circ.
    P=360∘3=120∘,P4=30∘P = \frac{360^\circ}{3} = 120^\circ, \qquad \frac{P}{4} = 30^\circ
  4. State the midline, maximum, and minimum
    The midline is y=c=0y = c = 0 (the x-axis). The maximum value is 0+2=20 + 2 = 2 and the minimum value is 0−2=−20 - 2 = -2. These are the boundary lines for the wave.
    ymax=2,ymin=−2y_{\text{max}} = 2, \quad y_{\text{min}} = -2
  5. List the five key x-values
    Start at the phase shift x=d=−30∘x = d = -30^\circ. Add the quarter-period of 30∘30^\circ repeatedly to get the x-values for one complete cycle.
    −30°,0°,30°,60°,90°-30°, 0°, 30°, 60°, 90°
  6. Assign y-values, accounting for the reflection
    The base cosine pattern for a>0a > 0 would be: max, midline, min, midline, max. Because a=−2a = -2 is negative, the entire wave is reflected across the midline, flipping the pattern to: min, midline, max, midline, min. Apply this reflected pattern to the five x-values.
    (−30°, −2), (0°, 0), (30°, 2), (60°, 0), (90°, −2)(-30°,\ -2),\ (0°,\ 0),\ (30°,\ 2),\ (60°,\ 0),\ (90°,\ -2)
  7. Sketch the curve
    Plot the five key points. The wave starts at a minimum, rises through zero to a maximum, then falls back through zero to a minimum — the mirror image of a standard cosine. Draw a smooth curve through the points from x=−30∘x = -30^\circ to x=90∘x = 90^\circ.
Answer: Amplitude =2= 2, Period =120∘= 120^\circ, Phase shift =30∘= 30^\circ left, Midline y=0y = 0. Key points (reflected): (−30∘,−2)(-30^\circ, -2), (0∘,0)(0^\circ, 0), (30∘,2)(30^\circ, 2), (60∘,0)(60^\circ, 0), (90∘,−2)(90^\circ, -2).
Check: Substitute x=30∘x = 30^\circ into the original equation: y=−2cos⁡(3(30∘)+90∘)=−2cos⁡(90∘+90∘)=−2cos⁡(180∘)=−2(−1)=2y = -2\cos(3(30^\circ) + 90^\circ) = -2\cos(90^\circ + 90^\circ) = -2\cos(180^\circ) = -2(-1) = 2. This matches the maximum point (30∘,2)(30^\circ, 2). ✓

Common mistakes and how to avoid them

Reading the phase shift directly from an unfactored bracket — for example, reading y=sin⁡(2x−60∘)y = \sin(2x - 60^\circ) as a shift of 60∘60^\circ instead of factoring to get y=sin⁡(2(x−30∘))y = \sin(2(x - 30^\circ)) and reading the shift as 30∘30^\circ.
Correction: Always factor kk out of the bracket first. Divide the constant term inside by kk to find the true phase shift dd.
Stating a negative amplitude — for example, writing amplitude =−3= -3 when a=−3a = -3.
Correction: Amplitude is always a positive number. Write amplitude =∣a∣=3= |a| = 3 and note separately that the negative sign causes a reflection.
Forgetting to reflect the key-point pattern when aa is negative — for instance, starting a cosine sketch at a maximum even though a<0a < 0.
Correction: When a<0a < 0, flip the entire pattern. A cosine with negative aa starts at a minimum, not a maximum.
Calculating the period as 360∘×∣k∣360^\circ \times |k| instead of 360∘∣k∣\frac{360^\circ}{|k|}, which gives a period that is far too large.
Correction: Divide 360∘360^\circ by |k|. A larger |k| makes the period shorter — more cycles fit in the same horizontal space.
Placing the first key point at x=0x = 0 regardless of the phase shift.
Correction: The cycle begins at x=dx = d, the phase shift value. Add P4\frac{P}{4} to dd successively to find the remaining four key x-values.

Lesson summary

Check your understanding

Question 1

What is the period of the function y=5sin⁡(4x)−3y = 5\sin(4x) - 3?
  1. 1440∘1440^\circ
  2. 90∘90^\circ
  3. 360∘360^\circ
  4. 45∘45^\circ
Show answer and explanation
90∘90^\circ
The period is 360∘∣k∣=360∘4=90∘\frac{360^\circ}{|k|} = \frac{360^\circ}{4} = 90^\circ. Multiplying by kk instead of dividing gives the incorrect 1440∘1440^\circ. The amplitude and vertical shift do not affect the period.

Question 2

What is the maximum value of the function y=−3cos⁡(x)+2y = -3\cos(x) + 2?
  1. 55
  2. −1-1
  3. 33
  4. −3-3
Show answer and explanation
55
Here a=−3a = -3 and c=2c = 2, so the amplitude is ∣a∣=3|a| = 3. The maximum value is always c+∣a∣=2+3=5c + |a| = 2 + 3 = 5, regardless of the sign of aa. The negative sign causes a reflection in the shape of the graph, but the range still extends from c−∣a∣=−1c - |a| = -1 to c+∣a∣=5c + |a| = 5.

Question 3

The equation y=sin⁡(3x+45∘)y = \sin(3x + 45^\circ) is rewritten in standard factored form. What is the phase shift?
  1. 45∘45^\circ to the right
  2. 15∘15^\circ to the right
  3. 45∘45^\circ to the left
  4. 15∘15^\circ to the left
Show answer and explanation
15∘15^\circ to the left
Factor out k=3k = 3: 3x+45∘=3(x+15∘)=3(x−(−15∘))3x + 45^\circ = 3(x + 15^\circ) = 3(x - (-15^\circ)). So d=−15∘d = -15^\circ, which means a shift of 15∘15^\circ to the left. Reading the phase shift before factoring gives the incorrect value of 45∘45^\circ.

Question 4

For the function y=4sin⁡(2(x−20∘))−1y = 4\sin(2(x - 20^\circ)) - 1, which set of values is correct?
  1. Amplitude =4= 4, Period =180∘= 180^\circ, Midline y=−1y = -1
  2. Amplitude =4= 4, Period =720∘= 720^\circ, Midline y=−1y = -1
  3. Amplitude =4= 4, Period =180∘= 180^\circ, Midline y=1y = 1
  4. Amplitude =2= 2, Period =180∘= 180^\circ, Midline y=−1y = -1
Show answer and explanation
Amplitude =4= 4, Period =180∘= 180^\circ, Midline y=−1y = -1
From the equation: a=4a = 4 so amplitude =4= 4; k=2k = 2 so period =360∘2=180∘= \frac{360^\circ}{2} = 180^\circ; c=−1c = -1 so the midline is y=−1y = -1. The other options each contain one incorrect value — period 720∘720^\circ comes from multiplying instead of dividing, midline y=1y = 1 confuses the sign of cc, and amplitude 22 mistakes kk for aa.

Key terms

Amplitude
The distance from the midline of a sinusoidal graph to its maximum or minimum value. Always a positive number; equal to |a| in the standard form equation.
Period
The horizontal length of one complete cycle of a repeating function. For y=asin⁡(k(x−d))+cy = a\sin(k(x-d))+c, the period equals 360∘∣k∣\frac{360^\circ}{|k|}.
Phase shift
A horizontal translation of the graph. In y=asin⁡(k(x−d))+cy = a\sin(k(x-d))+c, the phase shift is dd degrees. Positive dd shifts right; negative dd shifts left.
Vertical shift
A translation of the entire graph up or down. Determined by cc in the standard form; it sets the position of the midline at y=cy = c.
Midline
The horizontal line y=cy = c that runs through the middle of the wave, halfway between the maximum and minimum values.
Reflection
A flip of the graph across the midline, caused by a negative value of aa. It swaps the positions of peaks and troughs without changing the amplitude or period.
Key points
The five specific points per cycle — located at each quarter-period interval — that are sufficient to sketch one full cycle of a sinusoidal function.
Parent graph
The basic, untransformed version of a function. For trigonometry, the parent graphs are y=sin⁡xy = \sin x and y=cos⁡xy = \cos x, each with amplitude 11, period 360∘360^\circ, and no shifts.

Continue through MCR3U

View the complete Ontario Grade 11 Mathematics learning path

About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation D2.7. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question