Trigonometric Functions
Amplitude, Period and Phase Shift
How a, b and c in y = asin(bx + c) stretch, compress and shift the wave, and a four-step method for sketching any of them.
What this page covers
- Read amplitude, period and phase shift from an equation
- Sketch a transformed sine or cosine curve
- Find where a shifted wave starts its cycle
- Handle a negative amplitude correctly
The three parameters
| Parameter | Name | Effect | Found from |
|---|---|---|---|
| a | Amplitude | Vertical stretch; a negative value flips the curve | |a| |
| b | Angular frequency | Horizontal compression | Period = 2pib |
| c | Phase | Horizontal shift | Starts where bx + c = 0 |
The three act independently, so they can be dealt with one at a time. The source works through them in exactly that order across pages 5 to 8.
Amplitude
Multiplying by a stretches the wave vertically. The range becomes [-|a|, |a|] instead of [-1, 1].
| Function | Amplitude | Range | Note |
|---|---|---|---|
| y = sin t | 1 | [-1, 1] | The basic wave |
| y = 2sin t | 2 | [-2, 2] | Twice as tall |
| y = -3sin t | 3 | [-3, 3] | Three times as tall, and inverted |
| y = 17sin 3t | 17 | [-17, 17] | The source's largest example |
The source writes 'amplitude = -2' for y = -2cos 2t and sketches it correctly, but amplitude is properly a magnitude and cannot be negative. The right description is amplitude 2 together with a reflection in the horizontal axis. The graph the source draws is right; only the label is loose.
For y = -3sin t the source's sketch shows the curve going down first, which is the reflection. Where y = 3sin t would rise to 3, this falls to -3.
Period
Multiplying the input by b compresses the wave horizontally, so it completes a cycle in less input. The source finds the period by asking when the argument reaches 2pi.
The source's own reasoning
For y = sin 2t: the function repeats when 2t = 2pi.
For y = sin 3t: 3t = 2pi, so t = 2pi3.
| Function | b | Period |
|---|---|---|
| y = sin t | 1 | 2pi |
| y = sin 2t | 2 | π |
| y = sin 3t | 3 | 2pi3 |
| y = 17sin 3t | 3 | 2pi3 |
| y = 1.5cos(3t - π) | 3 | 2pi3 |
| y = -2cos 2t | 2 | π |
y = 17sin 3t has amplitude 17 and period 2pi3. Changing a never affects the period, and changing b never affects the amplitude.
Phase shift
Adding c inside the function moves the wave sideways. The source's method is to find where the cycle starts, which is where the whole argument is zero.
Worked example — y = sin(t - π4)
The wave starts at t = π4 and has period 2pi, so it completes its cycle at π4 + 2pi = 9pi4. The source writes starts at t = π4; period = 2pi.
Worked example — y = 3sin(2t + π2)
| Parameter | Working | Value |
|---|---|---|
| Amplitude | |3| | 3 |
| Period | 2pi2 | π |
| Start | 2t + π2 = 0 | t = -π4 |
Source, Week 7, page 7. The cycle runs from -π4 to -π4 + π = 3pi4.
One complete cycle, from where the argument is zero to one period later.
Worked example — y = 1.5cos(3t - π)
A cosine curve starts at its maximum, so this one peaks at t = π3 with value 1.5.
For y = sin(2t + π2) the shift is -π4, not -π2. The b divides it. Finding the start by solving bx + c = 0 avoids the error entirely, which is why the source uses that method rather than a formula.
The sketching method
- Amplitude. |a| gives the height; a negative a flips the curve.
- Period. Solve bx = 2pi, giving 2pib.
- Start. Solve bx + c = 0, giving x = -cb.
- Divide the period into four and plot the five key points from the start — the source states this step explicitly.
- Draw a smooth curve and repeat it left and right.
| Position | Sine curve | Cosine curve |
|---|---|---|
| Start | 0 | Maximum |
| Quarter period | Maximum | 0 |
| Half period | 0 | Minimum |
| Three quarters | Minimum | 0 |
| Full period | 0 | Maximum |
A negative amplitude swaps every maximum with its minimum. For y = -2cos 2t the source's sketch starts at -2 rather than +2, which is the reflection working through.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| Period = 2pi b | = 2pib | A larger b compresses, so the period shrinks |
| Shift = -c | = -cb | Solve bx + c = 0 |
| Shifting the wrong way | Solve for the start explicitly | +c inside means a shift left |
| Calling a negative a a negative amplitude | Amplitude is |a| | The minus is a reflection |
| Starting a cosine curve at zero | It starts at its maximum | cos 0 = 1 |
| Thinking a affects the period | They are independent | 17sin 3t has period 2pi3 |
Frequently asked questions
How do I find the period?
Set the whole argument equal to 2pi and solve. For sin 3t that gives 3t = 2pi, so t = 2pi3. The source uses exactly this reasoning rather than quoting a formula.
Can the amplitude be negative?
Amplitude is properly |a|, so it is never negative. A negative a flips the curve upside down, which is a reflection rather than a change of amplitude. The source writes 'amplitude = -2' for y = -2cos 2t, which is best read as amplitude 2 with a reflection.
Where does the curve start?
Where the argument is zero. For sin(2t + π2) that is t = -π4, which the source computes directly.
Why divide the period into four?
Because the five key points of a wave sit at quarter-period intervals. The source gives the instruction explicitly: divide the period into 4.
Source. Handwritten teaching notes, Week 7, pages 5-8.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.