Hello.
I'm now going to use the actual Khan Academy website to
do some more problems.
And this time we're going to go the other way around.
We're going to look at the graph of a trigonometric
function and we're going to try to figure out the equation.
So let's start with the problem we have in front of us.
So we have a-- well we don't know if it's a sine curve or a
cosine curve, but I guess it's fair to say that it's
one of the two.
Actually, let's answer that question first.
What do you think this is?
Do you think that this is a sine curve or a cosine curve?
What's the difference?
Or what's the easiest difference to differentiate
between the two?
Well, what was sine of 0?
Let me get a little chalkboard in here.
Let me bring it right here.
Let me make sure my pen tool is correct.
All right.
What is sine of 0?
And it could be 0 degrees or 0 radians.
Well if you remember from a couple of the other modules,
or even if you want to use a calculator, sine of 0 is 0.
And what is cosine of 0?
Well cosine of 0, if you remember from the last modules
or you want to use a calculator-- although you
shouldn't have to use a calculator for cosine of 0--
you might remember that it was 1.
So this graph that we have here, it's definitely when x is
0, when the x-axis is 0, the function definitely isn't 0.
In fact, in this case it's 1 1/2.
So this tells us that this isn't a sine curve-- although
later with shifts we'll learn that it could be a shifted sine
curve-- that this isn't a sine curve, that this is
a cosine curve.
And then you might ask, well Sal if this is a cosine curve
why is f of 0 equal to 1 1/2, or 3/2, instead of 1?
Because I just said here that cosine of 0 is 1.
Well that's because there must be some type of a coefficient
here, let's call it A, that is changing the amplitude
of this cosine curve.
And if you remember from the last module, what
do you think this A is?
Well that A is just literally the amplitude of the curve.
And what's the amplitude of this curve?
Well, the amplitude of this curve, if we just see how much
it moves above and below the x-axis, well it's that 3/2,
or that 1 1/2, we've been talking about.
See it moves up 3/2, and it moves down 3/2.
So let me just write that.
So we know that this is 3/2 cosine of, well, something x.
Right?
We know f of x is equal to 3/2 cosine of something x.
And we could use the formula that we learned in the previous
video, that it equals 2pi over the period x.
So now we just have to look at the graph and try to figure
out what the period of the graph is.
Well how many radians does it take for the graph to
start repeating again?
Let me click on the hint button, and maybe
this'll help us.
When I click hint-- there; drew the period.
And you could have figured it out on your own.
If you just go from any point and then follow the curve back
to the same point again, you'll see how long it's period is.
And the hint on the Khan icon actually told us
that the period is 4pi.
And you could just start from any point to any other point.
You could have gone from this point and then gone down,
gone back up, come down.
And then you would have seen that this distance is also 4pi.
So we know that the period is 4pi.
And then I could click hint again and it'll tell us stuff
that we already figured out.
The amplitude, we already figured out, was 1 1/2.
But let's just use a period, because we already knew
what the amplitude is.
So the period here we already figured out was 4pi.
So let's just write that in our equation.
So f of x is equal to 3/2 cosine of 2pi divided by
the period-- the period in this case is 4pi-- x.
That equals 3/2 cosine of 1/2 x.
Now if you ever forget this formula, which frankly
I always do forget it.
I've actually never memorized it.
I just try to think about what the period would be.
The way I think about it is the coefficient on the x-term,
that's a measure of how many cycles does the graph
do within 2pi radians.
Let me see if I can explain that within the context
of this problem.
So this problem, if we start at 0 and then 2pi is here, how
many cycles do we complete by the time we get to 2pi?
We start here, we go back here, and then we're at 2pi.
Well we only got halfway done through a cycle.
So that's the coefficient on the x-term.
And that's how I remember it.
So I could just say, well that's 3/2 cosine of
how many cycles do I complete in 2pi radians?
Well I only complete half a cycle.
3/2 times the cosine of 1/2 x.
So that's our f of x.
Let's do another problem.
All right.
OK.
This one's interesting.
So the first thing, just by inspection we can figure out
what this amplitude is.
This is pretty easy.
Right?
How much does it move above and below the x-axis?
Well it only goes 1, so we know that the coefficient, or the
multiplier times the sine or the cosine function--
whichever this is-- is 1.
So let's write that down.
Let's write down that the amplitude is equal to 1.
Now let's try to figure out if this is a sine
or a cosine function.
In the last problem we said sine of 0 is 0
and cosine of 0 is 1.
Well f of 0 of this function, whichever it is, is 0.
So we know this is a non-shifted sine function.
So there.
We have another piece of information.
It's a sine function.
So the last thing we have to figure out, we can either
figure out the period or we could use the method that I
just showed you where we say, well how many times does it
cycle within 2pi radians?
So let's do it that way.
And then we immediately know the coefficient.
Let's see.
Well actually, this graph board doesn't even get
all the way to 2pi.
But let's see.
It goes one cycle, two cycles.
And I did two cycles in only pi radians, right?
Because I'm only at x equals pi here.
So if I did two cycles in pi radians, then we must be able
to do four cycles in 2pi radians.
Or we could actually start here.
Actually, this is better.
Right?
Going from negative pi to pi.
That's 2pi radians.
So we finish one cycle, two cycles, three
cycles, four cycles.
So then we know what the coefficient on the x-term is.
So we know that it is sine of 4x.
So the answer here is f of x is equal to 1-- because that's the
amplitude-- the amplitude times sine of 4x.
I think we have time for one more.
And I want you to-- don't just mechanically do whatever
I'm telling you.
I want you to think about why counting the number of cycles
within 2pi radians, why that makes sense to you.
Then think back to the unit circle.
Or think back why that formula, the 2pi divided by the period,
is also the coefficient.
Think about why that makes sense.
If you realize why it makes sense you'll never have to
memorize it, and then 20 years later when you're doing it like
I'm doing it right now you won't be confused.
You'll be able to re-derive the formulas.
Let's do one more.
All right.
So what's the amplitude here?
Well, let's see.
The amplitude is 1/2.
So let me delete the old stuff that I was writing before.
For some reason it's not deleting.
OK.
Hope I don't confuse you.
So the amplitude, let's just call it Amplitude,
is equal to 1/2.
And how many cycles does it complete within 2pi radians?
Let's see.
If we start here it looks like it completes only half a cycle.
Right?
Because it takes 4pi radians to complete the entire cycle.
So it only completes half a cycle.
So we could think of it either two ways.
We could say that the period is equal to 4pi, because that's
how long it takes to complete one cycle, or we could say
it can only complete half a cycle within 2pi radians.
The last thing we have to figure out: Is it a sine
or a cosine function?
Well, f of 0 is 0.
Right?
So it's a non-shifted sine function.
So then we're done.
We have f of x is equal to 1/2-- figured out it's a sine
function-- sine of what?
How many cycles did it complete in 2pi radians?
It only completed half a cycle.
Let me not cover the problem.
It only completes half a cycle.
So it's 1/2 sine of 1/2 x.
Or we could use the formula f of x equals goes 1/2 sine of
2pi divided by the period x.
Right?
Because 2pi divided by the period is equal to 2pi over
4pi, which also equals 1/2.
I think that'll give you a sense of how to do
these problems now.
And I encourage you to practice them on the
Khan Academy website.
Have fun.