Graphing Sine Functions

Graphing Sine Functions

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

Graph: One Cycle

One cycle of y = sin x

This is one cycle of y = sin x.

The cycle starts from (0, 0).

As x changes from 0 to 2π,
sin x moves 0 → 1 → 0 → -1 → 0.

This table shows the coordinates of the blue points.

Amplitude, Period

y = sin x: (amplitude) = 1, (period) = 2pi

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 2π, the red cycle repeats.

So the period of y = sin x is 2π.

sin x = sin (x - 2pi)

For every 2π, the red cycle repeats.

So if y = sin x is under the translation
(x, y) → (x + 2π, y),

its image, y = sin (x - 2π),
will exactly cover y = sin x.

Translation of a function

So sin x = sin (x - 2π).

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

y = a sin x: (amplitude) = a, (period) = 2pi

y = (1/2) sin x

Amplitude: 1/2
Period: 2π

y = 1 sin x

Amplitude: 1
Period: 2π

y = 2 sin x

Amplitude: 2
Period: 2π

So, for y = a sin x,

Amplitude: a
Period: 2π

y = sin bx: (amplitude) = 1, (period) = 2pi/|b|

y = sin [1/2]x

Amplitude: 1
Period: 2π / |1/2|

y = sin 1x

Amplitude: 1
Period: 2π / |1|

y = sin 2x

Amplitude: 1
Period: 2π / |2|

So, for y = sin bx,

Amplitude: 1
Period: 2π / |b|

Formula: y = a sin bx

y = a sin bx. Amplitude: a, Period: 2pi/|b|

By combining these properties,
you can get this formula.

y = a sin bx

Amplitude: a
Period: 2π / |b|

Example 1

Sketch one cycle of the given function's graph. y = 3 sin 2x

Roughly draw one cycle of the function
with its amplitude and period.

Amplitude: 3
Period: 2π / |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

Sketch one cycle of the given function's graph. y = 3 sin (2x - pi) + 1

To find the amount of translation,
change 2x - 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 2π / |2| = π.

So the cycle ends at π/2 + π = 3π/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 (3π/2, 1).

Example 3

Sketch one cycle of the given function's graph. y = -2 sin x

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: 2π / |1| = 2π

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 = 2π.

Example 4

Find the values of a, b, c, and d. y = a sin [bx + c] + d (0 <= x <= 2pi)

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 5π/6.

So the period is 5π/6 - π/6 = 2π/3.

Recall the period formula: 2π / |b|.
So 2π/3 = 2π / |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.