# How to draw the graph of Sine square x?

- Step 1: Find the expression of sin
^{2}x
- Using the Double-Angle Formulas: cos 2x = cos
^{2}x - sin^{2}x , label it as equation(1)
- Using Pythagorean identities: sin
^{2}x + cos^{2}x = 1
- Substitute cos
^{2}x = 1 - sin^{2}x into equation(1), then
- cos 2x = 1 - sin
^{2}x - sin^{2}x
- cos 2x = 1 - 2sin
^{2}x .
- Express sin
^{2}x in the left side of the equation, then
- 2sin
^{2}x = 1 - cos 2x , divide the coefficient 2 in each item of the equation
- sin
^{2}x = 1/2 - (1/2)cos 2x
- thus, obtain the expression of sin
^{2}x

- 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 y
_{1} = 1/2 and y_{2}= -(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 = sin
^{2}x 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.