Hello.
Well, welcome to the next presentation in the
trigonometry modules.
Just to start off a little bit, let's review what
we've done so far.
In the last couple modules, we learned the definitions-- or at
least, I guess, we could call it a partial definition-- of
the sine, the cosine, and the tangent functions.
And the mnemonic we used to memorize that was sohcahtoa.
Let me write that down.
Sohcahtoa.
And what that told us is, let's say we had a right triangle.
Let me draw a right triangle.
This is a right angle here.
This is the hypotenuse.
Let me label the hypotenuse, h.
Let me label this-- and so we want to figure out, we want to
use this angle right here.
Theta, we'll call this theta.
Whatever.
Then this is the adjacent side, and this is the opposite side.
And that's an o.
So soh tells us that sine is equal to opposite
over hypotenuse.
Cosine is equal to adjacent over hypotenuse.
And tangent is equal to opposite over adjacent.
And I think, by this point-- and especially if you did some
of the exercises on the Khan Academy-- that should be second
nature and should make a lot of sense to you.
But this definition, using a right triangle like this,
actually breaks down at certain points.
Actually, at a lot of points.
For example, what happens as this angle right here
approaches 90 degrees?
You can't have two 90 degree angles in a right
triangle, can you?
Then it would be like a rectangle or something.
But you could actually probably figure out what happens as
it approaches 90 degrees.
But the definition, this definition, breaks
down for that.
Also, what happens if this angle is negative?
Or what happens if this angle is more than 90 degrees?
Or what happens if it's 800 degrees?
Or you know, 8 pi radians?
Not that 800 and 8 pi radians are the same thing.
But obviously, this definition starts to break down.
Because we couldn't even draw a right triangle that
has those properties.
So now I'm going to introduce you to an extension
of this definition.
It's really the same thing, but it allows the sine, the cosine,
and the tangent functions to be defined for angles greater than
or equal to pi over 2, or 90 degrees, or less than 0.
So let's draw a unit circle.
So this is just the coordinate axis, and here is a
circle of radius 1.
And let's make-- let me see.
Let me make sure I'm using the correct pen tool.
OK.
So let's call this right here-- so this is theta.
This is an angle, right?
Between the x-axis and this line I just drew here.
And this is a radius, right?
And we said that this has a radius 1.
So the length of this line is 1, right?
Because it just goes from the origin to the
outside of the circle.
So it has a radius of 1.
And now I'm going to draw a right triangle again.
Let me just drop a line from here.
So there I have a right triangle.
So if we use the old definition we learned before.
Let's just focus on sine for now.
So sine is equal to opposite over hypotenuse.
Let's apply that to this right triangle right here.
This is the right angle.
So what's the opposite angle of this?
What's the opposite side from this angle?
I'm going to change to yellow.
It's this side, right?
This is the opposite side.
And what's the hypotenuse?
The hypotenuse is just this radius, right?
And let's just say that this point, where it intersects
the circle-- let's call this point right here x comma y.
So what's the height of this opposite side?
Well, it's y, right?
Because it's just the height of that point.
This is of height y.
So sine of this angle right here, sine of theta, is
going to equal the opposite side-- which is this yellow
side, which is just the y-coordinate-- is going to
equal y over the hypotenuse.
The hypotenuse is this pink side here.
And what's the length of the hypotenuse?
Well, it's the radius of this unit circle.
So it's 1.
And y divided by 1?
Well, that's just y.
So we see that sine of theta is equal to y.
Let's do the same thing for cosine of theta.
Well, we know that cosine is equal to adjacent
over hypotenuse.
Well, what's the adjacent side here?
I'm running out of colors.
The adjacent side is this bottom side, right here.
So that would equal-- so if I said-- I'm running
out of space, too.
Cosine of theta would equal this gray side-- which is
the adjacent side-- and what is that?
What is this length?
What is the length of this side?
Well, it's just x, right?
If this is the point x, y then this distance here is x and we
already learned this distance-- or we already observed--
that this distance is y.
So this distance being just x, we know that the
adjacent length is x.
So we say cosine of theta is equal to x over the hypotenuse.
And once again, the hypotenuse is 1.
So cosine of theta is equal to x.
I know what you're thinking.
Sal, that's very nice and cute.
Cosine of theta equals x, sine of theta equals y.
But how is this really different from what we
were doing before?
Well, if I define it this way, now all of a sudden when the
angle becomes 90 degrees, now I can actually define
what sine of theta is.
Sine of theta now is just y.
Is just the y-coordinate, which is 1.
If theta is equal to-- I'm going to make sure it's very
messy right here-- if theta is equal to 90 degrees,
or pi over 2 radians.
This is pi over 2.
This angle right here.
And similar, cosine of pi over 2 is 0.
Because the x-coordinate right here is 0.
Let me do it with a couple more examples.
Oh, I'm forgetting the tangent function.
And you could probably figure out now, what is the
definition now we can use for the tangent function?
Well, going back-- let's use this green theta here.
Because it's kind of a normal angle.
So in this green angle here, tangent is
opposite over adjacent.
So tangent now, we can define as y/x.
And remember, these y's and x's that we're using are the point
on the unit circle where the angle that's defined by this--
by whatever-- where the radius that is subtended by this
angle, or I guess the arc, intersects-- actually,
I'm getting confused with terminology.
It's where this line intersects the circumference.
The coordinate of that-- the sine of theta is equal to y.
The cosine of theta is equal to x.
And the tangent of theta is equal to y/x.
Let's do a couple of examples and hopefully this'll make a
little bit more sense to you.
Let me try to really fast draw a new unit circle.
So that's my unit circle.
And here's the coordinate axis.
It's one of them.
And here is the other one.
So if we use the angle-- let's use the angle pi over 2, right?
Theta equals pi over 2.
Well, pi over 2 is right here.
It's a 90 degree angle, if you wanted to use degrees.
And now, we just figure out where it intersects
the unit circle.
And once again, this is a unit circle, so it
has a radius of 1.
So we can see that sine of pi over 2 equals the y-coordinate
where it intersects the unit circle.
So that's just 1.
What's cosine of pi over 2?
Well, it's just the x-coordinate, where you
intersect the unit circle.
And the x-coordinate here is 0.
And what's the tangent of pi over 2?
This is interesting.
The tangent of pi over 2.
Well, the tangent we defined now as y/x.
So the y-coordinate, this is the point 0, 1, right?
The y-coordinate is 1.
So it equals 1/0.
So this is undefined.
So still, we don't have a tangent function that can
define itself at certain points.
But in the next module, we're actually going to graph this.
And you'll see that it approaches infinity.
And similarly, we could try to find the functions for when
theta equals pi, right?
That's like 180 degrees.
That's this point right here.
So sine of pi.
What's the y-coordinate at this point?
Well, this point is negative 1 comma 0.
So the y-coordinate is 0.
What's the x-coordinate?
Cosine of pi.
That's negative 1.
And of course, what's the tangent of pi radians?
It's y/x.
So it's 0 over negative 1, which equals 0.
Hopefully this makes sense.
Now in the next module I'll actually graph these points.
And you'll see how it all comes together and why it is useful
to define the sine, the cosine, the tangent functions this way.
See you soon.
Bye.