# 3. Graphs of *y *=* a* sin(*bx* + *c*) and *y* = *a* cos(*bx* + *c*)

by M. Bourne

### Phase shift Interactives...

Later on this page:

In this section, we meet the following 2 graph types:

y=asin(bx+c)

and

y=acos(bx+c)

Both ** b** and

**in these graphs affect the**

*c***phase shift**(or

**displacement**), given by:

`text(Phase shift)=(-c)/b`

The **phase shift** is the amount that the curve is moved
in a horizontal direction from its normal position. The displacement will be to the **left** if the phase shift is negative, and to the **right** if the phase shift is positive.

There is nothing magic about this formula. We are just solving the expression in brackets for zero; `bx + c = 0`.

### Example 1

### Need Graph Paper?

Sketch the curve

y= sin(2x+ 1)

Answer

Let's first graph `y = sin 2x` (without the "+ 1") to get an idea of what phase shift means.

The amplitude is 1 and the period is `(2π)/2 = π.`

I have drawn a little more than 2 cycles, starting from `x = 0`.

The graph of `y=sin(2x)` for `0 ≤ x ≤ 6.5`

Now let's consider the phase shift. Using the formula above, we will need to shift our curve by:

Phase shift `=-c/b=-1/2=-0.5`

This means we have to shift the curve to the **left** (because the phase shift is negative) by `0.5`. Here is the answer (in green). I have kept the original *y* = sin 2*x * (in dotted gray) so you can see what's happening.

The graph of `y=sin(2x+1)` for `-0.5 ≤ x ≤ 6.7`

Note that the curve passes through:

- (−0.5, 0) on the
*x*-axis (because we shifted the curve to the left by 0.5) and - sin 1 = 0.84 on the
*y*-axis. This figure is obtained by substituting*x*= 0 into*y*= sin(2*x*+ 1); (in radians, of course).

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### Example 2

Sketch

`y=12cos(2x-pi/8)`

Answer

In this example, the

amplitude`a = 12`, and

`b = 2`, so theperiod= `(2π)/b = (2π)/2 = π`, and

`c=-pi/8`

so the **phase shift** is

`-c/b=-(-pi"/"8)/2=pi/16`

This means we need to move the cosine graph to the
**right** of its normal position (because the displacement is
positive in this example) by `pi/16`.

Without phase shift, the cosine curve will be *y* = 12 cos 2*x* and its graph is as follows:

The graph of `y=12cos(2x)` for `0 ≤ x ≤ 2pi`

Now, let's shift it to the right by *π*/16 ≈ 0.1963:

The graph of `y=12cos(2x-pi/8)` for `0 ≤ x ≤ 2pi`

You can see the original cosine curve (in dashed gray) and we have shifted it by `pi/16` to the right to give the required graph of `y = 12 cos (2x − π/8)`, in green.

## Phase shift Interactive

In the following interactive, use the slider to change the value of *c*, which displaces the curve. Observe the
"*c*" and "displacement" values and how they change when you
move the curve left and right.

The example you see is `y=sin(pi t)`. This has **period** given by `(2 pi)/b = (2π)/π = 2`.

You can also see the **cosine** case by choosing it at the top.

Choose graph type:

Sine Cosine

Guide curve:

On Off

Graph: `y=sin(pi t - 0)`

Displacement = `(-c)/b = 0/pi = 0`

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## Phase Angle or Phase Shift?

**Phase angle** is not always defined the same as **phase
shift**.

The phase **angle** for the sine curve *y *=* a* sin(*bx *+* c*) is usually taken to be the value of *c* and the phase **shift** is usually given by `-c/b`, as we saw above.

**Reminder: **In the last section, we saw how to express sine curves in terms of frequency.

**Example: **Electronics engineers separate the terms "phase angle" and "phase shift", and they use a mix of radians and degrees. We may have a current expressed as follows:

I= 50 sin (2π(100)t+ 30°)

This means the amplitude is `50\ "A"`, the frequency is `100\ "Hz"` and the phase angle is`30°`.

See an application of phase angle at An Application to AC Circuits in the complex numbers chapter.

Also, see a discussion on this issue at Phase shift or phase angle? in the math blog.

To keep things simple for now, we will mostly use the term **phase shift** in this chapter.

## Amplitude, period and phase shift graph applet

The following interactive will help you to explore the three key concepts when drawing trigonometric graphs - amplitude, period and phase shift.

Use the sliders under the graph to vary each of the **amplitude**, **period** and **phase shift** of the graph.

The `x`-axis has an integer scale (it's radians, of course), and multiples of `pi` are indicated with a red stroke.

You can also change the function to cosine. Hopefully you can see the concepts work the same for both sine and cosine curves.

Choose graph type:

Sine Cosine

Guide curve:

On Off

Graph: `y=a sin(bx + c)=` `sin(x)`

Period = `(2pi)/b = 6.28/1 = 6.28`

Displacement = `(-c)/b = 0/1 = 0`

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### Exercises

Sketch the graph of the following.

1. `y = sin(2x + π/6)`

Answer

`a = 1;`

`"period" = (2π)/2 = π`

`"displacement" = -c/b = -(π/6)/2 = -π/12`

So we need to move the sine graph to the **left** (because
we have a negative number) of its normal position by `pi/12`.

The graph of `y=sin(2x+pi/6)` for `0 ≤ x ≤ pi`

The curve `y=sin(2x)` is shown as grey dashed, and the answer, shifted by `pi/12` to the left, is in dark green.

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2. `y = 3 sin(x + π/4)`

Answer

`a = 3`;

`"period" = 2π`;

`"displacement" = -c/b = -(π/4)/1 = -π/4`

The graph of `y=3sin(x+pi/4)` for `0 ≤ x ≤ 2pi`

The curve `y=3sin(x)` is shown as grey dashed, and the answer, shifted by `pi/4` to the left, is in dark green.

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3. `y = 2 cos(x - π/8)`

Answer

`a = 2`;

`"period" = 2π`;

`"displacement" = -c/b = -(-π/8)/1 = π/8`

This means we need to move *y* = 2 cos *x* to the
**right** by ` π/8.`

The graph of `y=2 cos(x-pi/8)` for `0 ≤ x ≤ 2pi`

The curve `y=2cos(x)` is shown as grey dashed, and the answer after shifting by `pi/8` to the right, is in dark green.

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4. `y = -cos(2x - π)`

Answer

Let's take this one a step at a time.

We'll start by simply drawing the graph of `y = cos\ 2x`.

`a = 1`; `"period" = (2π)/2 = π.`

The graph of `y= cos(2x)` for `0 ≤ x ≤ 2pi`

Now, let's shift the curve (we are now considering the "minus π" part of the question.

`"displacement" = -c/b = -(-π)/2 = π/2`

So we have to shift every point on the curve to the right (because phase shift is a positive number) by `pi/2`. This give us *y* = cos(2*x* - *π*), the blue dotted curve.

The graph of `y= cos(2x-pi)` for `0 ≤ x ≤ 2pi`

Now we need to consider the minus out the front of the expression *y* = −cos(2*x* - *π*). The minus will just give us a mirror image in the *x*-axis, since every positive value becomes negative and every negative value becomes positive. In other words, the minus turns it upside down.

The answer for the graph of *y* = −cos(2*x* - *π*) is the green curve:

The graph of `y= -cos(2x-pi)` for `0 ≤ x ≤ 2pi`

But wait, this is what we started with! It looks the same as `y = cos\ 2x`.

This example shows an interesting thing about phase shift and periodic functions. If you shift far enough, you can easily obtain equivalent sine or cosine expressions.

You will get a better understanding of why this works in Sums and Differences of Angles, which we meet later.

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