back to Trigonometry video lesson index

How to find initial phase of a sine function

Question:

The graph of y = sin (5x + C) + 1 is symmetry to the line x = pi/4. Find the initial phase C of the sine graph.

Solution:

We are given the graph of the sine function y = sin (5x + C) + 1 is symmetry to the line x = pi/4. What is the property when the sine function y = sin x is symmetry to a line?

Look the graph of the y = sin x function,

how to find the initial phase of the sine graph?

when x = pi/2, y = 1. The graph is symmetry to the line x = pi/2.

when x = 3pi/2, y = -1. The graph is symmetry to the line x = 3pi/2

when x = 5pi/2, y = 1. The graph is symmetry to the line x = 5pi/3

when x = 7pi/2, y = -1. The graph is symmetry to the line x = 7pi/2

Now we will find when y is either positive one or negative one, what is the x?

pi/2 = pi/2 + 0 × pi
3pi/2 = pi/2 + 1 × pi
5pi/2 = pi/2 + 2 × pi
7pi/2 = pi/2 + 3 × pi

so, when y is either positive one or negative one, x = pi/2 + k × pi, in which k is an integer.

Because the graph of y = sin (5x + C) + 1 is symmetry to the line x = pi/4, so we get,

sin (5 × pi/4 + C) = +- 1
5 × pi/4 + C = pi/2 + k pi
C = pi/2 - 5 × pi/4 + k pi
= 2pi/4 - 5pi/4 + k pi
= - 3pi/4 + k pi

The initial phase is: C = - 3pi/4 + k pi, (k is an integer.)

when k = 0, C = - 3pi/4, y = sin (5x - 3pi/4) + 1

when k = 1, C = - 3pi/4 + pi = - 3pi/4 + 4pi/4 = pi/4, y = sin (5x + pi/4) + 1

when k = 2, C = - 3pi/4 + 2pi = - 3pi/4 + 8pi/4 = 5pi/4, y = sin (5x - 5pi/4) + 1

how to find the initial phase of the fine graph?

when k is an even number, the graph is the same as blue curve. The graph is symmetry to the line x = pi/4. In x = pi/4, y gets the maximum 2.

when k is an odd number, the graph is the same as the orange curve. The graph is symmetry to the line x = pi/4. In x = pi/4, y gets the minimum value 0. Watch the video for more details.