Introduction: In this lesson, the basic graphs of sine and cosine will be discussed and illustrated as they are shifted vertically. How the
equation changes and predicts the shift will be illustrated.
The Lesson: The graphs of
have as a domain, the possible values for x, all real numbers. We will use radian measure so that any real number can be used for x. This is because any real number can be associated with an arc length on a circle of radius 1, the definition of a radian measure. Therefore, when using a calculator, set the mode for radians, not degrees.
The graph of
is shown below. The WINDOW is X:
and Y: (– 2, 2, 1). This graph shows a maximum value of 1 at 0 and
, a minimum of – 1 at
and x-intercepts at
. 
The graph of
is shown below. 
The WINDOW is X:
and Y: (– 2, 2, 1). The only difference between this graph and the graph of
is that all points have been moved up ½ unit. This movement upward is called a vertical shift of ½. The maximum and minimum points are easy to determine. There is a max at (0, 1.5) and a min at
, for example. The x-intercepts are more difficult to determine and involve solving the equation
. This means that
and we have to evaluate
which is explained in the lesson on inverse operations. The three x-intercepts shown are located at
.
Let's Practice:- Describe the graph of
.
First we will look at the graph of
and then we will shift it vertically 3 units down.
The first graph below is of
while the second graph is of
. Notice that the vertical shift does not affect the amplitude which is 2. In this case there are no x-intercepts. The WINDOW for both graphs is X:
and Y: (– 6, 2, 1).
- Describe the graph of
.
First we have the graph of
. The amplitude is 3 and the graph has been reflected across the x-axis. The graph of
is shown on the same axes with a WINDOW of X:
and Y: (– 4, 6, 1). The graph shows an x-intercept of
for 

The next graph given below shows an x-intercept of 0.84 for
. 
This was determined using a calculator to solve the equation:
. 
We used 

to get approximately 0.84.