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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.