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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.

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)