Find the analytical expression of the sine graph

Question:

The graph is a part of y = A sin (Bx + C), find its analytical expression.

Solution:

Find the amplitude A of the sine graph

A = [3 - (-3)]/2 = (3 + 3)/2 = 6/2 = 3.

Find the coefficient B of the sine function y = A sin (Bx + C)

T is the minimum period of the sine graph.

From the given sine graph, quarter period T/4 = 5/8 - 1/4 = 5/8 - 2/8 = (5 - 2)/8 = 3/8

The minimum period T = 4 × 3/8 = 3/2

The relationship between the minimum period T and the coefficient B is: T = 2pi/B

So, B = 2pi/T

B = 2pi ÷ 3/2 = 2pi × 2/3 = 4pi/3

So far, the sine function is: y = 3 sin (4pix/3 + C). We need to determine the initial phase C. Look the given graph, when x = 5/8, y = 0. Set a middle variable t, let t = 4pix/3 + C, then y = sin t, when t = pi, y = 0.

so, (4pi/3) × (5/8) + C = pi

5pi/6 + C = pi

C = pi - 5pi/6 = 6pi/6 - 5pi/6 = (6pi - 5pi)/6 = pi/6

so, y = 3sin (4pix/3 + pi/6)

Therefore, the analytical expression of the given sine graph is y = 3sin (4pix/3 + pi/6). Looking for more detail? Please watch the video.