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.