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Find the analytical expression of the graph of the sine function y = A sin (Bx + C)

The figure below is a part of the graph of the sine function y = A sin(Bx + C). Given three points on the sine graph, find its analytical expression of this graph.

Given the graph of the graph of the sine function y = A sin(Bx + C), find the analytical expression of the graph.

Solution:

Let T be the period of the sine graph, the half period of the sine graph is:

T/2 = 5pi/9 - 2pi/9 = 3pi/9 = pi/3
so, T = 2pi/3
since T = 2pi/B
so, B = 2pi/T = 2pi ÷ 2pi/3 = 2pi × 3/2pi = 3
so, the analytical expression of the sine graph is y = A sin(3x + C)
Now we determine the initial phase C from the second point (2pi/9, 0)
Let 3x + C = t, when t = pi, sin t = 0
so, 3x + C = pi, when x = 2pi/9
3 × 2pi/9 + C = pi
2pi/3 + C = pi
so C = pi - 2pi/3 = 3pi/3 - 2pi/3 = pi/3
so, the analytical expression of this graph is: y = A sin (3x + pi/3)
Now we determine the amplitude A from the first point (0, (square root of 6)/2)
Note: any point on the graph of the sine function will satisfy the function y = A sin (Bx + C)
substitute x = 0, y = (square root of 6)/2 into the sine function
(square root of 6)/2 = A sin (3 × 0 + pi/3)
(square root of 6)/2 = A sin (pi/3)
(square root of 6)/2 = A (square root of 3)/2
multiply 2 in both side of the equation
(square root of 6) = A (square root of 3)
A = (square root of 6)/(square root of 3) = square root of 2

so, the analytical expression of this sine graph is y = (square root of 2) sin (3x + pi/3)