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This video is provided as supplementary material
for courses taught at Howard Community College, and in the video
I want to show how to find the six trigonometric function values
when you're given a point on the terminal side of the angle.
So let's start with a fairly simple problem to get the concept in this down,
and then we'll go to a slightly harder one. So you might be given a problem that
says something like this:
Find the trigonometric functions given the point
(3, 4) on the terminal side of the angle.
So to understand the concept, I'm going to draw
a simple graph. Here's the x-axis and y-axis,
And I'll put a point onthat graph that we'll call
(3, 4). Now using that point, I can create a right triangle.
I'm going to connect that point with a straight line down to the origin.
I'm also going to connect the point with a vertical line
down to the x-axis. So this angle that I've created
at the origin is an angle we'll call theta.
We can say that the side opposite to theta
has a length of 4,
the y-value that I was given. So the
opposite side has a length of 4.
We can see that the adjacent side
has a length of 3. That's the x-value I was given.
If I want to find the hypotenuse
I can use the Pythagorean theorem.
The hypotenuse is going to be equal to
the square root of the opposite side
squared -- so that's 4 squared --
plus the adjacent side squared --
that's 3 squared. This is basically going to be a 3-4-5
right triangle. So 4 squared is 16
and 3 squared is 9. So that's the square root of 16 plus 9.
16 plus 9 is 25, and the square root of 25 is 5.
So now I know that the hypotenuse equals 5.
Once I've got the lengths of the two sides and
the hypotenuse of a right triangle, I can figure out all the trigonometric
functions values. So the sine
of theta is going to equal
the opposite over the hypotenuse, or 4 over 5.
The cosine of theta
is the adjacent over the hypotenuse. That's 3 over 5.
And the tangent of theta
is going to be the opposite over the adjacent, or 4 over 3.
For the reciprocal functions,
I've got the cosecant of theta
is the reciprocal of the sine, so that's 5 over 4.
The secant is the reciprocal of the cosine.
That's 5 over 3.
And the cotangent is going to be
3 over 4. So that's the basic concept.
Let's do a slightly harder one. This problem says
Find the trig functions given the point (-1, √3)
(-1, √3) on the terminal side of the angle.
So we can draw a quick
picture of this if we want to. We're gonna have something at
(-1, √3).
So that's in the second quadrant.
So we know the
two sides that we're looking for.
The opposite side is going to be the y-value.
So the opposite side is the y-value,
and that's going to be the √3. Don't be scared by the fact
that it's a square root.
The adjacent side
is going to be the x-value. That's -1, and don't be scared by the fact
that it's negative.
And now we want to find the hypotenuse.
Well, once again, to find the hypotenuse,
what we're going to do is take the square root
of the opposite side squared plus the adjacent side squared.
So we want the √3 squared
So we want the √3 squared
plus -1 squared.
Well √3 squared
is just 3. And -1 squared is a positive 1.
So now I've got √4
which just equals 2. So the hypotenuse
has a length of 2.
So once again, we've got the two sides and hypotenuse,
so we can just fill in
the values for each function. So the opposite over the hypotenuse
is going to be√3/2.
The cosine
will be the adjacent, -1,
over 2.
And the tangent will be the opposite over the adjacent,
or √3 over -1.
I'll put the negative sign in the numerator
And then we're just going to take the reciprocals of each of these
to get the other three functions.
So I'll have 2 over √3,
and you might want to convert that into something without a radical sign
in the denominator. I"ve going to have
the secant of theta.
That's going to equal
2 over -1. I'll just make that -2.
And I've got the cotangent
of theta. That equals
-1 over √3, and once again
you might want to change this into something without a radical sign
in the denominator.
In other words, multiply the numerator and denominator by
√3....
Let just do that quickly. So this is going to be -1
times √3, or just -√3,
over √3 times √3, which is just 3.
So that's the basic concept. I'll do another video
with some harder examples so you can get more practice
and understand this a little better. Okay, take care.
I'll see you soon.