This material appears in sections 4.5-4.6.
For trig functions there are some important quantities that give us information about the shape of the graphs, and we should figure out how the various transformations effect these quatities.
-periodic:
The same is true for secant and cosecant.
Tangent and cotangent are
-periodic:
The period of a trig function is the minimum a such that for all
we have
.
If you have seen waves in other contexts, like a Physics class, you are also probably familiar with the notion of "frequency" of a wave. Period and frequency contain the same information about a wave.
They are inverses of each other. Frequency tells you number of cycles per second while period tells you number of seconds per cycle.
has a phase shift of
.
You may have guessed that the phase shift was $c$, and sometimes it is defined that way, but it is important to notice that
does NOT look like the graph of
shifted horizontally by $c$.
To shift
horizontally by $c$ we would replace $x$ with
, giving:
Here are some examles:
The black curve is the graph of
.
The red curve is the graph of
.
The green curve is the graph of
.
The red curve is obtained by shifting the black curve to the right by
. It makes sense to say that the graph of the red function has a phase shift of
.
The red curve is the graph of
, it does not have an obvious
in it, we ge the
by taking
.
The green curve is obtained by shifting the black curve to the right by
. The green curve is the graph of
and we get the
by taking the
in the formula and dividing by 2.
would look like the graph of
, but would be stretched vertically by a factor of 2 and then the whole graph would be shifted up 4.
These transformations only change the graph vertically, so phase and period are unaffected.
Amplitude is affected by the multiplication, but not by translation.
has amplitude 1, so
has amplitude 2, the crests are twice as high and the troughs are twice as low.
also has amplitude 2. Now the crests go up to 6, but the troughs only go down to 2 so the difference is 4, and half of that difference is the amplitude.
Important: You can not determine amplitude just by looking at the maximum height in a graph, you have to consider maximum change. The amplitude of
is 2, even though the maximum height of the graph is 6!
. Multiplication by
causes the
curve to oscillate faster (for
).
This changes the period.
I recommend that you DO NOT try to memorize how the period changes. Figure it out.
has a period of
. This means that
will complete one cycle as its input varies from 0 to 2π. The input is now
.
when
.
when
. So
completes one cycle as
goes from 0 to
.
Thus the period of
is
.
We can achieve a horizontal shift by replacing
with
. This requires some care if there is also a multiplication.
Consider
again. To shift right by
replace
by
and get
.
It is important to understand that
is NOT
shifted by
. As we just said, to shift by
you would need
.
is only a shift by
. See the discussion of Phase above for more examples of phase shift.
by using what we learned about transformations of trig functions. We know that:
So given a formula we can write down the amplitude, vertical shift, phase shift, and period of the function, and from this information we can draw a graph, which we will do below. Conversely, given a graph we can figure out the amplitude, period, phase shift, and vertical shift, and from this information we can figure out a formula for the graph.
Step 1: Draw horizontal lines at the maximum and minimum heights.
The distance between these two lines is 4, so the amplitude is 2.
Step 2: Draw a horizontal line midway between the max line and the min line.
This is the equilibrium line. Here it is at height 1, which tells us that the vertical shift is 1.
Step 3: Draw vertical lines marking one complete cycle of the curve, beginning where the curve crosses the equilibrium line going up.
In this figure the starting line is at
and the finish line is at
. (It would be hard to tell from the picture exactly where these lines should go, but suppose we know that our curve crosses the equilibrium line at exactly
and
.) The distance between these two lines is the period, so in this case the period is
Also, the position of the starting line tells us the phase shift. Here the phase shift is
.
Now we know:
All that remains is to find
and
.
From the period we have:
From the phase shift we know
and since
that means
.
Now we plug these values in to the formula
So a formula matching this graph is:
From the formula we can write down:
Now we want to draw a graph with these properties.
Step 1: Draw a horizontal line at the height of the vertical shift, in this case at height -1. This is the equilibrium line:
Step 2: Draw horizontal lines above and below the equilibrium line at distance equal to the amplitude.
In this case the amplitude is
, so we draw horizontal lines at heights
and at
.
These are the max and min lines.
Step 3: Draw a vertical starting line at
coordinate equal to the phase shift, in this case at
.
Step 4: Draw a vertical finish line at
-coordinate equal to the phase shift plus the period, so at
.
Step 5: So far we've drawn a box which should contain one complete cycle of the sine function. Divide the box into four boxes of equal width.
Step 6: Draw in the pieces of the sine curve.
Step 7: We've drawn one cycle of the curve. Now use periodicity to make copies of this cycle.
Author: Christopher Cashen <nil >
Date: 2008/12/10 20:21:00