Graphing trig functions

Analyzing the amplitude and periods of the sine and cosine functions.

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Closed Caption:


In this presentation we're
going to learn how to graph
trig functions without
having to kind of
graph point by point.
And hopefully after this
presentation you can also look
at a trig function and be able
to figure out the actual
analytic definition of
the function as well.
So let's start.
Let's say f of x.
Let me make sure I'm using
all the right tools.
So let's say that f of x is
equal to 2 sine of 1/2 x.
So when we look at this, a
couple interesting things here.
How is this different than just
the regular sine function?
Well, here we're multiplying
the whole function by 2,
and also the coefficient
on the x-term is 1/2.
And if you've seen some of the
videos I've made, you'll know
that this term affects the
amplitude and this term affects
the period, or the inverse of
the period, which
is the frequency.
Either way.
It depends whether you're
talking about one or the
inverse of the other one.
So let's start with
the amplitude.
This 2 tells us that the
amplitude of this function
is going to be 2.
Because if it was just a 1
there the amplitude would be 1.
So it's going to
be 2 times that.
So let's draw a little dotted
line up here at y equals 2.
And then another dotted line
at y equals negative 2.

So we know this is
the amplitude.
We know that the function is
going to somehow oscillate
between these two points, but
we have to figure out how fast
is it going to oscillate
between the two points,
or what's its period.
And I'll give you a
little formula here.

The function is equal to the
amplitude times, let's say,
sine, but it would also
work with cosine.
The amplitude of the function
times sine of 2pi divided
by the period of the
function, times x.
This right here is a "p."
So it might not be completely
obvious where this comes from.
But what I want you to do is
maybe after this video or
maybe in future videos we'll
experiment when we see what
happens when we change this
coefficient on the x-term.
And I think it'll start
to make sense to you why
this equation holds.
But let's just take this as
kind of an act of faith right
now, that 2pi divided by the
period is the coefficient on x.
So if we say that 2pi divided
by the period is equal to the
coefficient, which is 1/2.
I know this is extremely messy.
And this is separate from this.
So 2pi divided by the
period is equal to 1/2.
Or we could say 1/2 the
period is equal to 2pi.
Or, the period is equal to 4pi.
So we know the amplitude
is equal to 2 and the
period is equal to 4pi.
And once again, how did
we figure out that the
period is equal to 4pi?
We used this formula: 2pi
divided by the period is the
coefficient on the x-term.
So we set 2pi divided by the
period equal to 1/2, and then
we solved that the
period is 4pi.
So where do we start?
Well, what is f of 0?
Well, when x is equal to
0 this whole term is 0.
So what's sine of 0?
Sine of 0 is 0,
if you remember.
I guess you could use a
calculator, but that's
something you should remember.
Or you could re-look at the
unit circle to remind yourself.
Sine of 0 is 0.
And then 0 times 2 is 0.
So f of 0 is 0.
Right?
We'll draw it right there.
And we know that it
has a period of 4pi.
That means that the function
is going to repeat after 4pi.
So if we go out it should
repeat back out here, at 4pi.

And now we can just kind
of draw the function.
And this will take a little bit
of practice, but-- actually I'm
going to draw it, and then we
can explore it a little
bit more as well.
So the function's going
to look like this.

Oh, boy.
This is more difficult
than I thought.
And it'll keep going in
this direction as well.

And notice, the period here you
could do it from here to here.
This distance is 4pi.
That's how long it takes for
the function to repeat, or
to go through one cycle.
Or you could also, if you
want, you could measure this
distance to this distance.
This would also be 4pi.
And that's the period
of the function.
And then, of course, the
amplitude of the function,
which is this right here, is 2.
Here's the amplitude.
And then the period of 4pi we
figured out from this equation.
Another way we could have
thought about it, let's say
that-- let me erase some of the
stuff-- let's say I didn't
have this stuff right here.
Let's say I didn't know
what the function was.
Let me get rid of
all of this stuff.
And all I saw was this
graph, and I asked you
to go the other way.
Using this graph, try to figure
out what the function is.
Then we would just see,
how long does it take for
the function to repeat?
Well, it takes 4pi radians for
the function to repeat, so
you'd be able to just visually
realize that the period
of this function is 4pi.
And then you would say,
well what's the amplitude?
The amplitude is easy.
You would just see how
high it goes up or down.
And it goes up 2, right?
When you're doing the amplitude
you don't do the whole swing,
you just do how much it
swings in the positive
or negative direction.
So the amplitude is 2.
I'm using the wrong color.
The period is 4pi.

And then your question
would be, well this is
an oscillating, this is
a periodic function.
Is it a sine or is it
a cosine function?
Well, cosine function, assuming
we're not doing any shifting--
and in a future module I will
shift along the x-axis-- but
assuming we're not doing any
shifting, cosine of 0 is 1.
Right?
And sine of 0 is 0.
And what's this function at 0?
Well, it's 0.
Right?
So this is going to
be a sine function.
So we would use
this formula here.
f of x is equal to the
amplitude times the sine of 2pi
divided by the period times x.
So we would know that the
function is f of x is equal to
the amplitude times sine of 2pi
over the period-- 4pi-- x.
And, of course,
these cancel out.
And then this cancels out and
becomes 2 sine of 1/2 x.
I know this is a little
difficult to read.
My apologies.
And I'll ask a question.
What would this
function look like?
f of x equals 2
cosine of 1/2 x.
Well, it's going to look the
same but we're going to
start at a different point.
What's cosine of 0?
When x is equal to 0 this
whole term is equal to 0.
Cosine of 0, we
learned before, is 1.
So f of 0 is equal to 2.
Let me write that. f
of 0 is equal to 2.
Let me do this in a
different color.
Let me draw the cosine function
in a different color.
We would start here.
f of 0 is equal to 2, but
everything else is the same.
The amplitude is the same
and the period is the same.
So now it's going
to look like this.
I hope I don't mess this up.
This is difficult.
So now the function is
going to look like this.
And you're going to go down
here, and you're going
to rise up again here.
And on this side you're
going to do the same thing.
And keep going.
So notice, the cosine
and the sine functions
look awfully similar.
And the way to differentiate
them is what they do-- well,
what they do in general.
But the easiest way is, what
happens when you input
a 0 into the function?
What happens at the y-axis, or
when x is equal to 0, or when
the angle that you input
into it is equal 0?
Unless we're doing shifting--
and don't worry about shifting
for now, I'll do that in future
modules-- sine of 0 is 0
while cosine of 0 would be 1.
And since we're multiplying it
times this factor right here,
times this number right
here, the 1 becomes a 2.
And so this is the
graph of cosine of x.
This is this graph
of sine of x.
And this is a little bit of
a preview for shifting.
Notice that the pink graph,
or cosine of x, is very
similar to the green graph.
And it's just shifted this
way by-- well, in this
case it's shifted by pi.
Right?
And this actually has something
to do with the period
of the coefficient.
In general, cosine of x is
actually sine of x shifted
to the left by pi/2.
But I don't want to
confuse you too much.
That's all the time I
have for this video.
I will now do another video
with a couple of more
examples like this.