# AMPLITUDE PERIOD MAXIMUM AND MINIMUM OF SINE FUNCTIONS

## Amplitude Period Maximum and Minimum of Sine Functions

A periodic function is one which repeats itself over and over in horizontal direction.

What is period ?

The period of a periodic function is the length of one repetition or cycle

What is amplitude ?

The amplitude is the vertical distance between a maximum point and the principal axis.

What is maximum and minimum ?

• The maximum points occurs at the top of the crest.
• The minimum points occurs at the bottom of a through.

What is principal axis or man line ?

The graph oscillates about horizontal line called the principal axis or mean line.

• The amplitude is |A|
• Period is 2π/B, for B > 0

For the following sine functions find

i) amplitude

ii) period

iii)  maximum and minimum of y-values and corresponding x-values

iv) Graph two full periods of each function.

Problem 1 :

f(x) = 4 sin x

Solution :

f(x) = 4 sin x

f(x) = A sin B(x - C) + D

Comparing the given function with function above, we get

Amplitude (A) = 4, B = 1, period = 2π/1 ==> 2π

Midline y = 0

Length of 2 periods = 2 (2π) = 4π

 f(x) = 4 sin xx = 0f(0) = 4 sin 0f(0) = 0 f(x) = 4 sin xx = π/2= 4 sin π/2f(π/2) = 4 f(x) = 4 sin xx = πf(π) = 4 sin πf(π) = 0 f(x) = 4 sin xx = 3π/2f(3π/2) = 4 sin 3π/2f(3π/2) = -4 f(x) = 4 sin xx = 2πf(2π) = 4 sin 2πf(2π) = 0

(0, 0) (π/2, 4) (π, 0) (3π/2, -4) (2π, 0)

Maximum = 4 at x = π/2

Minimum = -4 at x = 3π/2

Problem 2 :

f(x) = 3 sin x + 2

Solution :

f(x) = 3 sin x + 2

f(x) = A sin B(x - C) + D

Comparing the given function with function above, we get

Amplitude (A) = 3, B = 1, period = 2π/1 ==> 2π

Midline y = 2, vertical translation of 2 units up.

Length of 2 periods = 2 (2π) = 4π

 f(x) = 3 sin x + 2x = 0f(0) = 3 sin 0 + 2f(0) = 2 f(x) = 3 sin x + 2x = π/2= 3 sin π/2 + 2f(π/2) = 3 + 2f(π/2) = 5 f(x) = 3 sin x + 2x = π= 3 sin π + 2f(π) = 0 + 2f(π) = 2 f(x) = 3 sin x + 2x = 3π/2= 3 sin 3π/2 + 2f(3π/2) = -3 + 2f(3π/2) = -1

x = 2π

f(x) = 3 sin 2π + 2

f( 2π) = 3(0) + 2

= 2

The points are :

(0, 2)  (π/2, 5) (π, 2) (3π/2, -1) and (2π, 2)

Maximum = 5 at x = π/2

Minimum = -1 at x = 3π/2

Problem 3 :

f(x) = sin (5x) + 2

Solution :

f(x) = sin (5x) + 2

f(x) = A sin B(x - C) + D

Comparing the given function with function above, we get

Amplitude (A) = 1, B = 5, period = 2π/5

Dividing 2π/5 by 4, we get 2π/20 ==> π/10

So, x values will be x = 0, π/10, 2π/10, 3π/10, 4π/10, .......

Midline y = 2, vertical translation of 2 units up.

 f(x) = sin (5x) + 2x = 0f(0) = sin (5x) + 2f(0) = 2 f(x) = sin (5x) + 2x = π/10f(π/10) = sin (5π/10) + 2= sin (π/2) + 2= 3 f(x) = sin (5x) + 2x = 2π/10f(2π/10) = sin (10π/10) + 2= sin (π) + 2= 2 f(x) = sin (5x) + 2x = 3π/10f(3π/10) = sin (15π/10) + 2f(3π/10) = sin (3π/2) + 2= -1 + 2= 1

f(x) = sin (5x) + 2

x = 4π/10

f(4π/10) = sin (20π/10) + 2

f(4π/10) = sin (2π) + 2

= 0 + 2

= 2

The points are :

(0, 2)  (π/10, 3) (2π/10, 2) (3π/10, 1) and (4π/10, 2)

Maximum = 3 at x = π/10

Minimum = 1 at x = 3π/10

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