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So we're going to pick up where we left
off in our last video.
We were talking about inverse functions.
And in particular, in this video, we're going to look at
the graphs of the sine, cosine, and tangent functions
and see that when we look at their graphs, their graphs
will not allow them, as they exist right now, to have an
inverse function because they don't pass the
horizontal line test.
But if I limit the domain of the sine function and some of
the other trigonometric functions, I will be able to
get an inverse function.
The question is, how do we agree in mathematics to limit
those graphs?
And in mathematics, mathematicians have agreed
that when we're dealing with the sine function, we want to
limit the graph so that it looks like this.
The graph will go now between negative pi over 2 to
positive pi over 2.
When my ratios are positive, my graph is above the x-axis.
When my ratios are negative, my graph is below the x-axis.
It would appear that if I'm dealing with an inverse sine
function and my ratio is positive, I need to report an
angle between 0 and pi over 2.
And I'm going to write that down.
So if the ratio is positive, the angle must be an angle
between 0 and pi over 2 or an angle in quadrant I.
If the ratio is negative, and you can see here I'm producing
negative ratios, then the angle that produces those
ratios is an angle between minus pi over 2 and 0.
We get these negative angles when we
proceed to go clockwise.
So I say that if the ratio is negative and the angle must be
between minus pi over 2 and 0, an angle between minus pi over
2 and 0 is an angle in quadrant IV.
Let's look at a couple of examples now having to do with
these inverse sine functions.
In the first example, I've been given a positive ratio.
So I know my angle is going to be an angle from quadrant I
that produced this ratio.
Now I have to think about the chart where I list the ratios
for 30, 45, and 60 degree angles.
And I say, what angle in that chart produces the ratio
positive square root of 3 over 2?
And as I think about that, I say that's a 60 degree angle.
Oh but wait a minute, I remember that when we're
reporting the angles for our inverse trig functions, we're
supposed to report our angles in radian measure.
So instead of saying 60 degrees, I'm going
to say pi over 3.
That's the equivalent of a 60 degree angle.
Now in the second problem here, I have a ratio, which I
remember, the square root of 2 over 2.
But it's a negative ratio.
First of all, I'm going to ignore that negative sign
temporarily.
I'm thinking about what angle produces this ratio, the
square root of 2 over 2, when I'm dealing
with the sine function.
And I believe that that's a 45 degree angle, or pi over 4
angle for radian measure.
But now it's negative.
And that means my angle has got to be an angle somewhere
between minus pi over 2 and 0.
And, therefore, I say that angle is a negative angle.
And it's a negative pi over 4.
That's an example of how I would answer these.
These are known ratios.
I know these ratios from my chart for 30, 45, and 60
degree angles.
And I'm following the cropping of this graph.
That's for the sine function.
Things get a little different when we get
to the cosine function.
Here's one period of the cosine function.
That's what it looks like going from 0 to 2 pi.
But, obviously, here we go again.
This is not going to pass the horizontal line test, is it?
So yet again, I'm going to chop my graph down so that I'm
going to force it to pass the horizontal line test.
And when I do that, I'm doing what members of the
mathematics community have agreed on.
We're going to crop that graph so that the graph now only
goes between 0 and pi.
Again, my ratios are positive.
Where?
When my angle is between 0 and 2 pi--
that's an angle in quadrant I--
and here, when my graph falls below the x-axis.
Those are negative ratios.
And I get negative ratios between pi over 2 and pi.
That would be an angle in quadrant II.
This is going to cause me to use reference angles if my
ratio is negative.
OK, let's take a look at a couple of examples here for
the cosine, inverse cosine function.
In the first example of the inverse cosine, I have a
positive ratio.
And all I'm doing right now is thinking, what angle for
cosine produces the ratio positive 1/2?
And it's a 60 degree angle.
I get that from my chart.
But I'm not going to say 60 degrees.
I'm going to say the radian equivalent of 60 degrees,
which, again, is pi over 3.
Moving onto example two, I see that I have the same ratio.
But my angle is negative.
Now this gets to be a little tricky, because the angle's
got to be an angle who's terminal side falls in
quadrant II or an angle between positive
pi over 2 and pi.
In order to get an angle in quadrant II, I need to use a
reference angle.
If this were positive, the angle that produces that
positive ratio is pi over 3.
But now pi over 3 is the reference
angle in quadrant II.
So I take pi.
And I subtract 1 pi over 3 from pi.
And that gives me 2 pi over 3.
And 2 pi over 3 is an angle between pi over 2 and pi.
It's an angle which will produce a
negative ratio for cosine.
And it is here, located on my cropped graph.
So it is legitimate.
Finally, let's look at tangent.
And if I were to draw the graph of the tangent
function-- and, again, here I've started to the left of
the origin.
You can see that's a negative pi over 2.
I have several repetitions of this graph here, present in
this diagram.
Obviously, here again, we're failing the
horizontal line test.
So what am I going to do with the tangent function?
I'm going to do what millions of other mathematicians do.
And I'm going to limit the graph of the tangent function
to existing between minus pi over 2 to positive pi over 2.
This is a graph that will pass the horizontal
line test quite nicely.
And it gives me the graph of a function that has an inverse.
Now if the ratio is positive, that is, produces this part of
my graph above the x-axis, the angle that produces it is
going to have to be an angle between 0 and pi over 2.
If, however, the ratio is negative, below the x-axis,
that is going to be produced by an angle between minus pi
over 2 and 0, or an angle in quadrant IV.
Let's look at a couple of examples of this--
the inverse tangent of this positive ratio of the
square root of 3.
Thinking about my chart for 30, 45, and 60 degree angles,
I realize that a 60 degree angle produces
this ratio for tangent.
But I'm not going to write 60 degrees here.
I'm going to write the radian equivalent of 60 degrees,
which is pi over 3.
Now in the next problem, my ratio is negative.
And my answer's going to have to be an angle between minus
pi over 2 and 0.
When I think about that I say, OK, let's just look at the
ratio as if it were positive.
What angle produces the ratio 1 for tangent?
And it is a 45 degree angle, or pi over 4.
Now I'm going to use that same angle.
But the same angle is going to be down in quadrant IV.
And I'm going to say that's a minus pi over 4.
And that's the angle that produces that ratio, negative
1 for tangent.
And so, in summary, we could say the following.
That if we're talking about the sine, the inverse sine,
and the inverse tangent functions and we have negative
ratio, that our angle is a negative angle between 0 and
minus pi over 2.
That if we're talking about the inverse cosine function
and we have a negative ratio, our angle is positive and from
quadrant II.
And it's between positive pi over 2 and pi.
If our ratio was positive, all three of these are going to
come from an angle whose terminal side is in quadrant
I. That's it.
Good luck with this assignment.