Transformations of Trig Graphs

Graph any sine or cosine function of the form y = A sin(B(x - C)) + D: what each parameter does, the right order of transformations and the five key points method.

Advanced30 min18 problems
Practice Problems

Definition

A transformed sine or cosine graph has the general form
y=Asin⁡(B(x−C))+Dory=Acos⁡(B(x−C))+Dy = A\sin(B(x - C)) + D \quad \text{or} \quad y = A\cos(B(x - C)) + D
Each of the four parameters changes the parent graph y=sin⁡(x)y = \sin(x) or y=cos⁡(x)y = \cos(x) in exactly one way (in this lesson B>0B > 0):
ParameterWhat it doesFeature
AAvertical stretch or compression; A<0A < 0 also reflects the wave across its midlineamplitude =∣A∣= \lvert A \rvert
BBhorizontal compression (B>1B > 1) or stretch (B<1B < 1)period =2πB= \frac{2\pi}{B}
CChorizontal shift: right if C>0C > 0, left if C<0C < 0phase shift =C= C
DDvertical shiftmidline y=Dy = D

Order: stretch first, then shift

Inside the bracket, first compress or stretch by BB, then shift by CC. Outside, first stretch by AA (and reflect if A<0A < 0), then shift by DD. Always **factor out BB** before you read the shift: sin⁡(2x−π)=sin⁡(2(x−π2))\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2})), so the shift is π2\frac{\pi}{2}, not π\pi.

The five key points method

  1. 1.One cycle starts at x=Cx = C and ends at x=C+periodx = C + \text{period}.
  2. 2.Split it into four equal steps: step=period4\text{step} = \frac{\text{period}}{4}.
  3. 3.Fill in the y-values. Sine: midline, max, midline, min, midline. Cosine: max, midline, min, midline, max. If A<0A < 0, swap max and min.
  4. 4.Max =D+∣A∣= D + |A|, min =D−∣A∣= D - |A|, midline value =D= D.
Domain: all real numbers. Range: [D−∣A∣, D+∣A∣][D - |A|,\ D + |A|].

Try it now

What is the amplitude of y=−4sin⁡(2x)+1y = -4\sin(2x) + 1?

Worked Examples

For y=3sin⁡(2(x−π6))−1y = 3\sin(2(x - \frac{\pi}{6})) - 1, find the amplitude, period, phase shift, midline and range.
1

Match the general form

Compare with y=Asin⁡(B(x−C))+Dy = A\sin(B(x - C)) + D → A=3, B=2, C=π6, D=−1A = 3,\ B = 2,\ C = \frac{\pi}{6},\ D = -1
2

Amplitude and period

Amplitude =∣3∣=3= |3| = 3, period =2π2=π= \frac{2\pi}{2} = \pi → Amplitude 33, period π\pi
3

Phase shift and midline

C=π6>0C = \frac{\pi}{6} > 0, so the shift is to the right; D=−1D = -1 → Shift π6\frac{\pi}{6} right, midline y=−1y = -1
4

Range

[D−∣A∣, D+∣A∣]=[−1−3, −1+3][D - |A|,\ D + |A|] = [-1 - 3,\ -1 + 3] → [−4,2][-4, 2]

Common Mistakes

Reading the phase shift of y=sin⁡(2x−π)y = \sin(2x - \pi) as π\pi

Why it's wrong: The shift is the number subtracted from xx itself, not from 2x2x. BB multiplies the whole bracket, so it also squeezes the shift.

Correct: Factor out BB first: sin⁡(2(x−π2))\sin(2(x - \frac{\pi}{2})), so the shift is π2\frac{\pi}{2} to the right.

Shifting y=cos⁡(x+π3)y = \cos(x + \frac{\pi}{3}) to the right

Why it's wrong: The general form has x−Cx - C. Here x+π3=x−(−π3)x + \frac{\pi}{3} = x - (-\frac{\pi}{3}), so C=−π3C = -\frac{\pi}{3} is negative.

Correct: A plus sign inside the bracket means a shift to the left: π3\frac{\pi}{3} to the left.

Starting the key points at x=0x = 0 or stepping by a whole period

Why it's wrong: After a phase shift the cycle no longer starts at 00, and five points must split ONE period into four equal parts.

Correct: Start at x=Cx = C and step by period4\frac{\text{period}}{4}. For period π\pi the step is π4\frac{\pi}{4}.

Keeping the usual max/min order when A<0A < 0

Why it's wrong: A negative AA reflects the wave across its midline, so every maximum becomes a minimum and the other way round.

Correct: For y=−2sin⁡(x)y = -2\sin(x) the pattern is midline, min, midline, max, midline. The amplitude is still 22.

Interactive Visual

Function graph

y = sin(x)

Challenge 1 of 6

Stretch and lift

Move the sliders until your curve lies exactly on the dashed target curve.

How close

0 of 6 solved

  • A: bigger
  • B: right
  • C: right
  • D: bigger
-2π-π0π2π-4-2024xy

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y = A·sin(B(x − C)) + DStretch and lift

Function graph

0π/2π3π/22π-2-101234xy12345

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Table of values

x0.7853981.57082.356193.141593.92699
sin(x)0.70710710.707107-1.02068 × 10^-11-0.707107
2·sin(2(x − π/4)) + 1131-11
Midline y = 111111

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Interactive Grapher

-3π-2π-π0π2π3π-4-2024xy

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2sin(2(x - pi/4)) + 1

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Practice Problems

18 problems
Problem 1 of 18
Easy
What is the amplitude of y=−4sin⁡(2x)+1y = -4\sin(2x) + 1?

Why It Matters

Almost every repeating process in nature and technology is modeled by a transformed sine or cosine, and the four parameters always have a physical meaning:
  • Ferris wheels: the radius is the amplitude, the height of the center is the midline, one turn is the period
  • Tides: the water depth rises and falls around an average depth about every 12 hours
  • Daylight: the length of the day swings around 12 hours with a period of 365 days
  • Sound and electricity: amplitude is loudness or voltage, the period sets the pitch or the frequency
When you can turn an equation into a sketch in a minute, you can read the highest and lowest values and the times when they happen straight from the formula. That is exactly what engineers, oceanographers and climate scientists do with these models.

Real World Applications

Ferris Wheel Height

The height of a seat on a turning Ferris wheel is a transformed cosine: the center of the wheel gives the midline and the radius gives the amplitude.

Example:

A wheel with radius 20 m whose center is 22 m above the ground turns once every 30 minutes. A seat that starts at the bottom has height h(t)=−20cos⁡(π15t)+22h(t) = -20\cos(\frac{\pi}{15}t) + 22 meters after tt minutes, and the period is 2ππ/15=30\frac{2\pi}{\pi/15} = 30.

1Try It Yourself

The height of the seat is h(t)=−20cos⁡(π15t)+22h(t) = -20\cos(\frac{\pi}{15}t) + 22 meters, with tt in minutes.

How high above the ground is the seat at its highest point (in meters)?

Step 1: Write the mathematical expression

Highest point = midline + amplitude:

Tides in a Harbor

Harbor masters predict the water depth with a sine model whose period is about 12 hours, so ships know when they can enter.

Example:

The depth d(t)=1.5sin⁡(π6(t−3))+4d(t) = 1.5\sin(\frac{\pi}{6}(t - 3)) + 4 meters, tt hours after midnight, has midline 4 m, amplitude 1.5 m and period 2ππ/6=12\frac{2\pi}{\pi/6} = 12 hours. The depth swings between 2.5 m and 5.5 m.

2Try It Yourself

A large ship needs high tide. The depth is d(t)=1.5sin⁡(π6(t−3))+4d(t) = 1.5\sin(\frac{\pi}{6}(t - 3)) + 4 meters, tt hours after midnight.

At what time tt (in hours after midnight) is the first high tide after t=3t = 3?

Step 1: Write the mathematical expression

Start of the cycle plus a quarter of the period:

Hours of Daylight

The number of daylight hours over the year follows a sine wave with a period of 365 days. The midline is 12 hours, the day length at the equinoxes.

Example:

In a city in northern Europe, L(t)=4sin⁡(2π365(t−80))+12L(t) = 4\sin(\frac{2\pi}{365}(t - 80)) + 12 hours, where tt is the day of the year. Day 80 (around March 21) is the spring equinox, and the day length swings between 8 and 16 hours.

3Try It Yourself

The day length is L(t)=4sin⁡(2π365(t−80))+12L(t) = 4\sin(\frac{2\pi}{365}(t - 80)) + 12 hours, where tt is the day of the year.

On which day of the year is daylight the longest? Round to a whole day.

Step 1: Write the mathematical expression

Start of the cycle plus a quarter of the period, rounded:

Key Takeaways

  • 1In y=Asin⁡(B(x−C))+Dy = A\sin(B(x - C)) + D: ∣A∣|A| is the amplitude, 2πB\frac{2\pi}{B} the period, CC the phase shift and y=Dy = D the midline
  • 2Stretch first, then shift: inside the bracket BB before CC, outside AA before DD
  • 3Always factor out BB before reading the phase shift: sin⁡(2x−π)=sin⁡(2(x−π2))\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2}))
  • 4x−Cx - C shifts right by CC, x+Cx + C shifts left by CC
  • 5Five key points: start at x=Cx = C, step =period4= \frac{\text{period}}{4}. Sine: midline, max, midline, min, midline; cosine: max, midline, min, midline, max; A<0A < 0 swaps max and min
  • 6The domain is all real numbers and the range is [D−∣A∣, D+∣A∣][D - |A|,\ D + |A|]

Frequently Asked Questions

Because B=2B = 2 multiplies the whole bracket. Factor it out: sin⁡(2x−π)=sin⁡(2(x−π2))\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2})). The cycle starts where the bracket is 00, at x=π2x = \frac{\pi}{2}, so the shift is π2\frac{\pi}{2}.
Because B=2B = 2 multiplies the whole bracket. Factor it out: sin⁡(2x−π)=sin⁡(2(x−π2))\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2})). The cycle starts where the bracket is 00, at x=π2x = \frac{\pi}{2}, so the shift is π2\frac{\pi}{2}.
Yes, for a shift and a stretch in the same direction. Shifting sin⁡(x)\sin(x) right by π2\frac{\pi}{2} and THEN compressing by 22 gives sin⁡(2x−π2)=sin⁡(2(x−π4))\sin(2x - \frac{\pi}{2}) = \sin(2(x - \frac{\pi}{4})), a shift of only π4\frac{\pi}{4}. Horizontal and vertical changes do not affect each other, so you may do the two groups in either order.
Yes. Start at x=Cx = C and step by period4\frac{\text{period}}{4} as for sine. Only the pattern of y-values changes: cosine goes max, midline, min, midline, max (or the reverse if A<0A < 0).
Rewrite it first. Sine is odd, sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta), so sin⁡(−2x)=−sin⁡(2x)\sin(-2x) = -\sin(2x). Cosine is even, cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta), so cos⁡(−2x)=cos⁡(2x)\cos(-2x) = \cos(2x). Then B>0B > 0 and the method works as usual.

Glossary

Amplitude
Distance from the midline to a maximum or minimum: ∣A∣|A|
Period
Horizontal length of one full cycle: 2πB\frac{2\pi}{B}
Phase shift
Horizontal shift of the graph: CC in x−Cx - C (right if positive, left if negative)
Midline
The horizontal line y=Dy = D halfway between the maximum and the minimum
Key points
The five points that split one cycle into four equal steps: start, quarter, half, three quarters, end
Reflection
Flipping the wave across its midline, caused by a negative AA

Formula Card

General form

y=Asin⁡(B(x−C))+Dy = A\sin(B(x - C)) + D

Same letters for cosine: $y = A\cos(B(x - C)) + D$

Amplitude

∣A∣|A|

Vertical stretch; $A < 0$ also reflects the wave

Period

2πB\frac{2\pi}{B}

Length of one cycle

Phase shift and midline

CC and y=Dy = D

Right if $C > 0$, left if $C < 0$; up if $D > 0$, down if $D < 0$

Key point step

period4\frac{\text{period}}{4}

Start at $x = C$ and add the step four times

Range

[D−∣A∣, D+∣A∣][D - |A|,\ D + |A|]

From the minimum to the maximum

Today's challengeOne problem, three tries

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