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# How to draw the graph of Sine square x?

Step 1: Find the expression of sin2x
Using the Double-Angle Formulas: cos 2x = cos2x - sin2x , label it as equation(1)
Using Pythagorean identities: sin2x + cos2x = 1
Substitute cos2x = 1 - sin2x into equation(1), then
cos 2x = 1 - sin2x - sin2x
cos 2x = 1 - 2sin2x .
Express sin2x in the left side of the equation, then
2sin2x = 1 - cos 2x , divide the coefficient 2 in each item of the equation
sin2x = 1/2 - (1/2)cos 2x
thus, obtain the expression of sin2x
Step 2: Draw the graph of y = (1/2)cos 2x
Note: For the period function y = A cos (Bx + C), Its amplitude is A. period T = 2 pi/B and phase shift is C.
In this case, its amplitude A = 1/2, period T = 2 pi/2 = pi and phase shift C = 0
Step 3: Draw the graph of y = - (1/2)cos 2x
Reflect the graph of y = (1/2)cos2x in x-axis to get the graph of y = - (1/2)cos2x
Step 4: Draw two graphes such as y1 = 1/2 and y2= -(1/2)cos2x in order to add them.
When x = 0, y=1/2 + (-1/2)cos2x = 1/2 + (- 1/2) = 0
When x = pi/4, y=1/2 + (-1/2)cos2x = 1/2 + 0 = 1/2
When x = pi/2, y=1/2 + (-1/2)cos2x = 1/2 + 1/2 = 1
When x = 3pi/4, y=1/2 + (-1/2)cos2x = 1/2 + 0 = 1/2
When x = pi, y=1/2 + (-1/2)cos2x = 1/2 + (- 1/2) = 0
When x = 5pi/4, y=1/2 + (-1/2)cos2x = 1/2 + 0 = 1/2
When x = 3pi/2, y=1/2 + (-1/2)cos2x = 1/2 + 1/2 = 1
When x = 7pi/4. y=1/2 + (-1/2)cos2x = 1/2 + 0 = 1/2
When x = 2pi, y=1/2 + (-1/2)cos2x = 1/2 + (- 1/2) = 0
Therefore, the graph of y = 1/2 + [- (1/2)cos 2x] is :
Summary:
The graph of y = sin2x is the sum of the graph y = 1/2 and the graph y = (-1/2)cos2x
The sum of the graph y = 1/2 and the graph y = (-1/2)cos2x is the graph of y = (-1/2)cos2x move up 1/2 unit.