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# Graph of the Sine Function and Transformation

### Graph of the sine function y = sin x

### Graph of the sine function y = A sin (Bx + C)

# Two methods to draw the graph of the sine function

### Draw the graph of y = sin x using the five points method

### Difference among the graphs of y = sin x, y = (1/2) sin x and y = 2sin x

# Graph of Sine Function Transformation

### Draw the graph of y = sin 3x using transformation method

### Draw the graph of y = sin (x - 3pi/4) using the transformation method

### Draw the graph of y = sin (x + 3pi/4) using the transformation method

The figure above is the graph of y = sin x. For y = sin x, there is a constant T, when x is any value, there is sin (x + T) = sin x. The constant T is called the period of the function y = sin x. Look the figure above, sin 0 = sin (0 + 2pi), sin pi/2 = sin (pi/2 + 2pi), sin pi = sin (pi + 2pi), sin 3pi/2 = sin (3pi/2 + 2pi), sin 2pi = sin (2pi + 2pi), so, the period T of y = sin x is 2pi.

The general form of the sine function is y = A sin (Bx + C), in which A > 0 and B > 0. The coefficient A is the amplitude of the curve, let T be the period of the sine function, then T = 2pi/B and C is the initial phase.

In the figure above, the blue curve is the graph of y = sin x, its period is T = 2pi/B = 2pi/1 = 2pi. Its amplitude A = 1 and initial phase C = 0. The yellow curve is the graph of y = sin (x + 2pi/3). Its amplitude A = 1, period T = 2pi/B = 2pi/1 = 2pi and Initial phase C = 2pi/3. Therefore, we can left-shift the graph of y = sin x by 2pi/3 units to get the graph of y = sin (x + 2pi/3).

The graph of y = A sin (Bx + C) is called the curve of the sine function. There are two methods to draw the graph of the sine function. The method one is the five-point method, which selects five points in a period of the sine function. The five points include the maximum point of the function, the minimum point of the function, the zero points of the function. Draw the graph of the sine function in one period and then extend the graph to both direction for every 2pi/B interval to get the graph of the sine function y = A sin (Bx + C). The method two is the transformation method, which is transform the graph of y = sin x to the graph of y = A sin (Bx + C).

For the graph of y = sin x, its period is T = 2pi/B = 2pi/1 = 2pi. To draw the graph of y = sin x, we divide the interval 2pi by five. So, the x-coordinate of the first point is x = 0. The x-coordinate of the fifth point is x = 2pi. The x-coordinate of the third point is x = pi. The x-coordinate of the second point is x = pi/2. The x-coordinate of the fourth point is x = 3pi/2. For each x-coordinate, find its corresponding y-coordinate. When x = 0, y = sin 0 = 0, when x = pi/2, y = sin pi/2 = 1, when x = pi, y = sin pi = 0, when x = 3pi/2, y = sin 3pi/2 = -1, and when x = 2pi, y = sin 2pi = 0. Therefore, the five important points in one period are (0, 0), (pi/2, 1), (pi, 0), (3pi/2, -1), and (2pi, 0).

If you are interested in the details on how to draw the graph of the sine function using the five points method, please watch the video by click the link, how to draw the graph of y = sin x?

In the figure above, the blue curve is the graph of y = sin x, the yellow curve is the graph of y = 2 sin x and the grey curve is the graph of y = (1/2) sin x. The three curves have the same period but different Amplitudes. For example, the amplitude of y = sin x is one, the amplitude of y = 2 sin x is two and the amplitude of y = (1/2) sin x is one-half. Because the three curves have the same coefficient B, so they have the same period, T = 2pi/B = 2pi/1 = 2pi. Because the three curves have the same coefficient C, which is 0, so their initial phase is zero.

Step 1. Draw the graph of y = sin x, its period is T = 2pi/B = 2pi/1 = 2pi. Its five important points in one period are (0, 0), (pi/2, 1), (pi, 0), (3pi/2, -1) and (2pi, 0).

Step 2. Draw the graph of y = sin 3x

For the sine function y = sin 3x. Its coefficient B = 3, so its period is T = 2pi/3.

So, for each point on y = sin x, shrink the x-coordinate to one-third and keep the y-coordinate the same to get the graph of y = sin 3x.

Note: the period of y = sin x is 2pi. The period of y = sin 3x is 2pi/3. So, three period of y = sin 3x is 2pi. That is, from x = 0 to x = 2pi, the graph of y = sin 3x repeat three times. So, the coefficient B is the frequency of repeat times of the curve. Therefore, in the interval of x from 0 to 2pi, the graph y = sin x has only one period but the graph of y = sin 3x has three period.

In the figure above, the blue curve is the graph of y = sin x. The yellow curve is the graph of y = sin (x - 3pi/4). For sine function y = sin (x - 3pi/4), its initial phase is C = -3pi/4, we can right-shift the graph of y = sin x by 3pi/4 units to get the graph of y = sin (x - 3pi/4).

In the figure above, the blue curve is the graph of y = sin x. The yellow curve is the graph of y = sin (x + 3pi/4). For the sine function y = sin (x + 3pi/4), its initial phase is C = 3pi/4, we can left-shift the graph of y = sin x by 3pi/4 units to get the graph of y = sin (x + 3pi/4).

**Example**

Draw the graph of y = sin (2x - pi/3)

Solution

**Method one - shift first then shrink**

Step 1. Draw the graph of y = sin x

Step 2. Right-shift the graph of y = sin x by pi/3 units to get the graph of y = sin (x - pi/3)

Step 3. For each point on y = sin (x - pi/3), shrink the x-coordinate to half and keep the y-coordinate the same to get the graph of y = sin (2x -pi/3)

In the figure above, the grey curve is the graph of y = sin (2x - pi/3) and the yellow curve is the graph of y = sin (x - pi/3). For each point on the grey curve, its x-coordinate is one-half the x-coordinate of the yellow curve.

**Method two - shrink first then shift**

Step 1. Draw the graph of y = sin x

Step 2. Draw the graph of y = sin 2x by shrink the x-coordinate of the graph y = sin x to one-half and keep the y-coordinate the same to get the graph of y = sin 2x.

Step 3. Draw the graph of y = sin (2x - pi/3) by shift the graph of y = sin 2x right pi/6 units to get the graph of y = sin (2x - pi/3).

Look the figure above, the yellow curve is the graph of y = sin 2x. The grey curve is the graph of y = sin (2x - pi/3). By shift the graph of y = sin 2x right pi/6 units, we get the graph of y = sin (2x - pi/3).

**Method 3 - Five points**

The five important points for y = sin x in one period are (0, 0), (pi/2, 1), (pi, 0), (3pi/2, -1) and (2pi, 0).

Now we will find the five important points for y = sin (2x - pi/3)

The first point, when 2x - pi/3 = 0, y = 0.

- 2x - pi/3 = 0
- 2x = pi/3
- x = pi/6
- so, the first point is (pi/6, 0)

The second point, when 2x - pi/3 = pi/2, y = 1

- 2x - pi/3 = pi/2
- 2x = pi/3 + pi/2
- 2x = 2pi/6 + 3pi/6 = 5pi/6
- x = 5pi/12
- so, the second point is (5pi/12, 1)

The third point, when 2x - pi/3 = pi, y = 0

- 2x - pi/3 = pi
- 2x = pi/3 + pi
- 2x = pi/3 + 3pi/3 = 4pi/3
- x = 4pi/6 = 2pi/3
- so, the third point is (2pi/3, 0)

The fourth point, when 2x - pi/3 = 3pi/2, y = -1

- 2x - pi/3 = 3pi/2
- 2x = pi/3 + 3pi/2
- 2x = 2pi/6 + 9pi/6 = 11pi/6
- x = 11pi/12
- so, the fourth point is (11pi/12, -1)

The fifth point, when 2x - pi/3 = 2pi, y = 0

- 2x - pi/3 = 2pi
- 2x = pi/3 + 2pi
- 2x = pi/3 + 6pi/3 = 7pi/3
- x = 7pi/6
- so, the fifth point is (7pi/6, 0)

In the figure above, the five important points of y = sin (2x - pi/3) are (pi/6, 0), (5pi/12, 1), (2pi/3, 0), (11pi/12, -1), (7pi/6, 0). Extended the graph for each pi interval in both direction to get the graph of y = sin (2x -pi/3).