In this discussion, we are going to graph the sine function as well as explain the graph in detail.
We will also discuss the standard equation of the sine function.
Here is the outline to this article:
Graph the Parent Graph of the Sine Function
- Find the Max and Min Values given the Graph
- Find the Horizontal Midline Given the Graph
- Find the Phase Shift Given a Graph
- Find the Vertical Shift Given a Graph
- Find the Cycle given the Graph
- Find the Period given the Graph
- Find the x-scale Given the Standard Equation
The Standard Equation of the Sine Function
- Find the Amplitude Given the Standard Equation
- Find the Period Given the Standard Equation
- Find the Phase Shift Given the Standard Equation
- Find the Vertical Shift Given the Standard Equation
- Find the Cycle Given the Standard Equation
- Find the x-scale Given the Standard Equation
It is assumed that you already understand the unit circle.
Thus, we begin with what you already know.
Recall the unit circle:
We have terminal points in the form (x,y);
and sine takes the y coordinate.
Thus,
Starting from zero and going counterclockwise,
sin(0) = 0
sin (pi/4) = sqrt(2)/2 = .07
sin(pi/2) = 1
sin (3pi/4) = sqrt(2)/2
sin(pi) = 0
sin (5pi/4) = -sqrt(2)/2
sin(3pi/2) = -1
sin (7pi/4) = – sqrt(2)/2
sin(2pi) = 0
Starting from zero and going clockwise,
sin(0) = 0
sin (-pi/4) = -sqrt(2)/2
sin(-pi/2) = -1
sin (-3pi/4) = -sqrt(2)/2
sin(-pi) = 0
sin (-5pi/4) = sqrt(2)/2
sin(-3pi/2) = 1
sin (-7pi/4) = sqrt(2)/2
sin(-2pi) = 0
We draw the graph of sine, connecting the dots. Once you understand the shape of the sine graph, we only need to use the quadrantal angles, which are the angles where its terminal points are along the x and/or y axis.
The quadrantal angles on the unit circle are:
0, pi/2, pi, 3pi/2, and 2pi.
Find the Max and Min Values Given the Graph
The max and min values are the graph’s high and low values respectively.
On our graph, the high value is 1 and the low value is -1.
Thus, the max is 1 and the min is -1.
Find the Horizontal Midline Given the Graph
The horizontal midline of the sine graph is the horizontal line that passes through the middle (between the max and min values) of the sine graph.
We see that the horizontal midline of y=sinx is the x-axis (y=0).
To find the horizontal midline algebraically,
add the max and min values together and divide by 2.
In our graph, our max value is 1 and our min value is -1.
Thus, the horizontal midline is
[(1) + (-1)]/2 = 0/2 = 0
The Phase Shift Given the graph
The phase shift, or the horizontal shift, of the graph is when the graph shifts to the right or to the left.
Since our graph begins at zero, we know that the phase shift is
zero.
The Vertical Shift Given the graph
The vertical shift of the graph is when the graph shifts up or down. Since our horizontal midline is at y=0, we know that the vertical shift is zero.
We begin our graph at the phase shift and the horizontal midline.
Thus, we begin our graph at (0,0).
The Amplitude of Sine Given the Graph
The amplitude tells us how high or how low the graph will go from the midline.
To find the amplitude, take max – min and divide by 2.
In our graph, we have a max of 1 and a min of -1.
Thus, the amplitude = (1- (-1)) / 2 = (1+1) / 2 = 2/2 =1
The Cycle of Sine Given the Graph
The cycle of sine shows one complete period.
Beginning at 0 and ending at 2pi is one cycle.
Beginning at -2p and ending at 0 is one cycle.
The Period of Sine Given the Graph
The period of sine is the length of one cycle.
We see that from 0 to 2pi, we have one cycle.
Thus, period = Length of one cycle = End – Beginning = 2pi -0 = 2pi
Similarly, we see that from -2p to 0 is one cycle.
Thus, 0 – (-2pi) = 0 +2pi = 2pi
In our picture, we graphed the sine function over two periods.
The x-scale of Sine Given the Graph
The x-scale is the increment value along the x-axis.
We see that, from our graph, we have an x-scale of pi/2 because the distance between each point is pi/2.
pi/2 – 0 = pi/2
pi – pi/2 = pi/2
3pi/2 – pi/2 = pi/2
2pi – pi/2 = pi/2
The Standard Equation of Sine
The standard equation of the sine function is of the form:
y = Asin[b(x-c)] + d.
If we were to write the original sine function in standard form, we have
y = Asin[b(x-c)] + d.
y = sin[1(x-0)] + 0, where
|A| = 1, b=1, c=0, d=0
Let us look at each part in detail.
|A| is the amplitude. This tells us how high and how low our graph will go from the midline.
It is also the vertical stretch/compress.
If |A| is greater than zero, but less than 1, the graph compresses vertically.
If |A| is greater than 1, the graph stretches vertically.
If A is less than zero, the graph reflects the x-axis of the graph of
y = |A|sin[b(x-c)] + d.
We use b to algebraically find the period.
period = 2pi/|b|
b is also the horizontal stretch/compress.
If |b| is greater than 0 but less than 1, the graph stretches horizontally.
If |b| is greater than 1, the graph compresses horizontally.
If b is less than zero, the graph reflects the y-axis of the graph of
y = |A|sin[|b|(x-c)] + d.
Or, since sine is an odd function, we can use the fact that
y = Asin[-b(x-c)] + d = -Asin[b(x-c)] + d
and pull out the negative sign
then reflect the x-axis of |A|sin[b(x-c)] + d
c is the phase shift, or horizontal shift; it tells us how many units to shift right and left.
If c is grater than zero, we shift right.
If c is less than zero, we shift left.
Note, that in the form
y = Asin[b(x-c)] + d , to find c,
set x-c=0 and solve for x to get x = c.
In the form
y = Asin[b(x+c)] + d,
set (x+c)=0 and solve for x to get x = -c.
Notice that we flip the signs from the standard equation when finding c.
In our equation, we see that c = 0.
d is the vertical shift; it tells us how many units to shift up and down.
If d is greater than 0, we shift up
If d is less than zero, we shift down.
The horizontal midline begins at y = d.
In our equation, our d is 0. Thus, our horizontal midline is y=0.
We begin our graph at the phase shift and the horizontal midline.
Thus, we begin our graph at (c, d).
The Cycle Given an Equation
The Cycle is one complete period.
The cycle begins at “c” and ends at “c + the period”
The cycle is on the interval [c, c + the period] = [c, c + 2pi/b]
The Period Given an Equation
The period is the length of one cycle.
To algebraically compute the period given an equation,
we use the value of b in the following equation:
2pi/b
We compute the period of the original sine graph as follows:
2pi/b = 2pi/1 =2pi
The x-scale of Sine Given an Equation
The x-scale is the increment values along the x-axis.
To algebraically compute the x-scale of the original sine function, we say,
x-scale = period/4 = 2pi/4 = pi/2
Always begin your graph at x=c and y = d = horizontal midline, then add increments of the x-scale and follow the sine pattern.
For the original sine graph, our c is 0.
So, we begin at 0, then add increments of
pi/2.
The values are as follows:
0
0 + pi/2 = pi/2
pi/2 + pi/2 = pi
pi + pi/2 = 3pi/2
3pi/2 + pi/2 = 2pi
We see that the values above are on the parent graph for sine.
Now you are ready to graph the sine function with multiple transformations!
Thank you for reading! I hope that I can assist you on your math journey!
Sincerely,
MathAngel369
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