Does the order of transformations really matter?

Yes, for a shift and a stretch in the same direction. Shifting $\sin(x)$ right by $\frac{\pi}{2}$ and THEN compressing by $2$ gives $\sin(2x - \frac{\pi}{2}) = \sin(2(x - \frac{\pi}{4}))$, a shift of only $\frac{\pi}{4}$. Horizontal and vertical changes do not affect each other, so you may do the two groups in either order.

Do the key points work the same way for cosine?

Yes. Start at $x = C$ and step by $\frac{\text{period}}{4}$ as for sine. Only the pattern of y-values changes: cosine goes max, midline, min, midline, max (or the reverse if $A < 0$).

What if $B$ is negative?

Rewrite it first. Sine is odd, $\sin(-\theta) = -\sin(\theta)$, so $\sin(-2x) = -\sin(2x)$. Cosine is even, $\cos(-\theta) = \cos(\theta)$, so $\cos(-2x) = \cos(2x)$. Then $B > 0$ and the method works as usual.

Teacher Resources

Transformations of Trig Graphs

Q: Why is the phase shift of $\sin(2x - \pi)$ not $\pi$?

Because $B = 2$ multiplies the whole bracket. Factor it out: $\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2}))$. The cycle starts where the bracket is $0$, at $x = \frac{\pi}{2}$, so the shift is $\frac{\pi}{2}$.

Q: What if $B$ is negative?

Rewrite it first. Sine is odd, $\sin(-\theta) = -\sin(\theta)$, so $\sin(-2x) = -\sin(2x)$. Cosine is even, $\cos(-\theta) = \cos(\theta)$, so $\cos(-2x) = \cos(2x)$. Then $B > 0$ and the method works as usual.

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.

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Definition

A transformed sine or cosine graph has the general form

y = 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 = \cos(x)

in exactly one way (in this lesson

B > 0

Parameter	What it does	Feature
$A$	vertical stretch or compression; $A < 0$ also reflects the wave across its midline	amplitude $= \lvert A \rvert$
$B$	horizontal compression ( $B > 1$ ) or stretch ( $B < 1$ )	period $= \frac{2\pi}{B}$
$C$	horizontal shift: right if $C > 0$ , left if $C < 0$	phase shift $= C$
$D$	vertical shift	midline $y = D$

Order: stretch first, then shift

Inside the bracket, first compress or stretch by

B

, then shift by

C

. Outside, first stretch by

A

(and reflect if

A < 0

), then shift by

D

. Always **factor out

B

** before you read the shift:

\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2}))

, so the shift is

\frac{\pi}{2}

, not

\pi

The five key points method

1.One cycle starts at $x = C$ and ends at $x = C + \text{period}$ .
2.Split it into four equal steps: $\text{step} = \frac{\text{period}}{4}$ .
3.Fill in the y-values. Sine: midline, max, midline, min, midline. Cosine: max, midline, min, midline, max. If $A < 0$ , swap max and min.
4.Max $= D + |A|$ , min $= D - |A|$ , midline value $= D$ .

Domain: all real numbers. Range:

[D - |A|,\ D + |A|]

Try it now

What is the amplitude of

y = -4\sin(2x) + 1

Worked Examples

For

y = 3\sin(2(x - \frac{\pi}{6})) - 1

, find the amplitude, period, phase shift, midline and range.

Match the general form

Compare with

y = A\sin(B(x - C)) + D

→

A = 3,\ B = 2,\ C = \frac{\pi}{6},\ D = -1

Amplitude and period

Amplitude

= |3| = 3

, period

= \frac{2\pi}{2} = \pi

→ Amplitude

3

, period

\pi

Phase shift and midline

C = \frac{\pi}{6} > 0

, so the shift is to the right;

D = -1

→ Shift

\frac{\pi}{6}

right, midline

y = -1

Range

[D - |A|,\ D + |A|] = [-1 - 3,\ -1 + 3]

→

[-4, 2]

Answer: Amplitude $3$ , period $\pi$ , phase shift $\frac{\pi}{6}$ to the right, midline $y = -1$ , range $[-4, 2]$ .

Common Mistakes

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

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

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

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

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

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

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

Why it's wrong: After a phase shift the cycle no longer starts at $0$ , and five points must split ONE period into four equal parts.

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

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

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

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

Interactive Visual

Function graph

A = 1B = 1C = 0D = 0

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

Tap or hover over the graph to read off values.

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

Function graph

Tap or hover over the graph to read off values.

Tap a name to show or hide its graph.

Table of values

x	0.785398	1.5708	2.35619	3.14159	3.92699
sin(x)	0.707107	1	0.707107	-1.02068 × 10^-11	-0.707107
2·sin(2(x − π/4)) + 1	1	3	1	-1	1
Midline y = 1	1	1	1	1	1

Interactive Sandbox

Interactive Grapher

Tap or hover over the graph to read off values.

Try these examples:

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

Expression Calculator

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History

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Practice Problems

18 problems

Problem 1 of 18

Easy

What is the amplitude of

y = -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 $h(t) = -20\cos(\frac{\pi}{15}t) + 22$ meters after $t$ minutes, and the period is $\frac{2\pi}{\pi/15} = 30$ .

1Try It Yourself

The height of the seat is $h(t) = -20\cos(\frac{\pi}{15}t) + 22$ meters, with $t$ 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 $d(t) = 1.5\sin(\frac{\pi}{6}(t - 3)) + 4$ meters, $t$ hours after midnight, has midline 4 m, amplitude 1.5 m and period $\frac{2\pi}{\pi/6} = 12$ hours. The depth swings between 2.5 m and 5.5 m.

2Try It Yourself

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

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

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, $L(t) = 4\sin(\frac{2\pi}{365}(t - 80)) + 12$ hours, where $t$ 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 $L(t) = 4\sin(\frac{2\pi}{365}(t - 80)) + 12$ hours, where $t$ 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 $y = A\sin(B(x - C)) + D$ : $|A|$ is the amplitude, $\frac{2\pi}{B}$ the period, $C$ the phase shift and $y = D$ the midline
2Stretch first, then shift: inside the bracket $B$ before $C$ , outside $A$ before $D$
3Always factor out $B$ before reading the phase shift: $\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2}))$
4 $x - C$ shifts right by $C$ , $x + C$ shifts left by $C$
5Five key points: start at $x = C$ , step $= \frac{\text{period}}{4}$ . Sine: midline, max, midline, min, midline; cosine: max, midline, min, midline, max; $A < 0$ swaps max and min
6The domain is all real numbers and the range is $[D - |A|,\ D + |A|]$

Frequently Asked Questions

Because

B = 2

multiplies the whole bracket. Factor it out:

\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2}))

. The cycle starts where the bracket is

0

, at

x = \frac{\pi}{2}

, so the shift is

\frac{\pi}{2}

Glossary

Amplitude: Distance from the midline to a maximum or minimum: $|A|$
Period: Horizontal length of one full cycle: $\frac{2\pi}{B}$
Phase shift: Horizontal shift of the graph: $C$ in $x - C$ (right if positive, left if negative)
Midline: The horizontal line $y = D$ 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 $A$

Formula Card

General form

y = A\sin(B(x - C)) + D

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

Amplitude

|A|

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

Period

\frac{2\pi}{B}

Length of one cycle

Phase shift and midline

C

and

y = D

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

Key point step

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

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

Range

[D - |A|,\ D + |A|]

From the minimum to the maximum

Today's challengeOne problem, three tries

Transformations of Trig Graphs

Definition

Order: stretch first, then shift

The five key points method

Worked Examples

Common Mistakes

Interactive Visual

Function graph

Function graph

Table of values

Interactive Sandbox

Interactive Grapher

Expression Calculator

History

Practice Problems

Why It Matters

Real World Applications

Ferris Wheel Height

Tides in a Harbor

Hours of Daylight

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

General form

Amplitude

Period

Phase shift and midline

Key point step

Range

More in This Topic

Graphing the Sine Function

Graphing Cotangent

Graphing the Cosine Function

Amplitude and Period