# Graphing Sine Functions

How to graph sine functions: graph, amplitude, period, formula, examples, and their solutions.

## Graph: One Cycle

This is one cycle of *y* = sin *x*.

The cycle starts from (0, 0).

As *x* changes from 0 to 2*π*,

sin *x* moves 0 → 1 → 0 → -1 → 0.

This table shows the coordinates of the blue points.

## Amplitude, Period

This is the graph of *y* = sin *x*.

Amplitude means

'(max) - (middle)' or '(middle) - (min)'.

So the amplitude of *y* = sin *x* is 1.

[1 = 1 - 0 = 0 - (-1)]

For every 2*π*, the red cycle repeats.

So the period of *y* = sin *x* is 2*π*.

For every 2*π*, the red cycle repeats.

So if *y* = sin *x* is under the translation

(*x*, *y*) → (*x* + 2*π*, *y*),

its image, *y* = sin (*x* - 2*π*),

will exactly cover *y* = sin *x*.

Translation of a function

So sin *x* = sin (*x* - 2*π*).

This property is true for any periodic functions:*f*(*x*) = *f*(*x* - [period]).

*y* = (1/2) sin *x*

Amplitude: 1/2

Period: 2*π*

*y* = 1 sin *x*

Amplitude: 1

Period: 2*π*

*y* = 2 sin *x*

Amplitude: 2

Period: 2*π*

So, for *y* = *a* sin *x*,

Amplitude: *a*

Period: 2*π*

*y* = sin [1/2]*x*

Amplitude: 1

Period: 2*π* / |1/2|

*y* = sin 1*x*

Amplitude: 1

Period: 2*π* / |1|

*y* = sin 2*x*

Amplitude: 1

Period: 2*π* / |2|

So, for *y* = sin *bx*,

Amplitude: 1

Period: 2*π* / |*b*|

## Formula: *y* = *a* sin *bx*

By combining these properties,

you can get this formula.*y* = *a* sin *bx*

Amplitude: *a*

Period: 2*π* / |*b*|

## Example 1

Roughly draw one cycle of the function

with its amplitude and period.

Amplitude: 3

Period: 2*π* / |2| = *π*

Use these values to sketch one cycle.

Mark the maximum and minimum values.

(*y* = ±3)

Start from the origin.

And draw one cycle that ends at *x* = *π*.

## Example 2

To find the amount of translation,

change 2*x* - *π*

to 2(*x* - *π*/2).

So *y* = 3 sin [2(*x* - *π*/2)] + 1 is under the translation

(*x*, *y*) → (*x* + *π*/2, *y* + 1).

Translation of a function

Roughly draw one cycle of the function

with its details.

The middle value is 1.

And the amplitude is 3.

So (max) = 1 + 3 = 4,

(min) = 1 - 3 = -2.

The cycle starts at *π*/2.

And the period is 2*π* / |2| = *π*.

So the cycle ends at *π*/2 + *π* = 3*π*/2.

Use these values to sketch one cycle.

Mark the maximum and minimum values.

(*y* = 4, -2)

Start from (*π*/2, 1).

And draw one cycle that ends at (3*π*/2, 1).

## Example 3

Roughly draw one cycle of the function

with its amplitude and period.*y* = -2 sin *x* is -*y* = 2 sin *x*.

If you compare *y* = 2 sin *x* and -*y* = 2 sin *x*,

it shows (*x*, *y*) → (*x*, -*y*).

So -*y* = 2 sin *x* (white)

is the image of *y* = 2 sin *x* (gray)

under the reflection in the *x*-axis.

Amplitude: 3

Period: 2*π* / |1| = 2*π*

Use these values to sketch one cycle.

Mark the maximum and minimum values.

(*y* = ±2)

Start from the origin.

And draw the fliped cycle

that ends at *x* = 2*π*.

## Example 4

Find one sine cycle from the given graph.

And find *a* and *b* from the cycle.

The middle value is 2.

The maximum value is 4.

And the minimum value is 0.

So the amplitude, *a*, is 2.

The cycle starts at *π*/6.

And it ends at 5*π*/6.

So the period is 5*π*/6 - *π*/6 = 2*π*/3.

Recall the period formula: 2*π* / |*b*|.

So 2*π*/3 = 2*π* / |*b*|.

So *b* = 3.

(The sine cycle is not flipped: no reflection.

So you don't have to think about *b* = -3.)

*a* = 2, *b* = 3.

And the cycle is under the translation

(*x*, *y*) → (*x* + *π*/6, *y* + 2).

So the given graph shows *y* = 2 sin [3(*x* - *π*/6)] + 2.

Solve [3(*x* - *π*/6)] to find *c*.

So *a* = 2, *b* = 3, *c* = -*π*/2, and *d* = 2.