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This video is provided as supplementary material
for courses taught at Howard Community College, and in this video
I want to do a few more examples of finding the trigonometric functions values
given point on the terminal side of the angle.
So here's the first problem. We're given a point (2, 3),
we're told it's on the terminal side of an angle, and we have to find the
trigonometric function values for it. So to understand this,
I'll draw a simple graph. and label the point
(2, 3). And using that point
I'm going to draw a line from the point
to the origin, and say that the angle that
that forms -- this angle -- is theta. I'm going to draw another line
vertically down from that point to the x-axis,
and that will give me a side
opposite that angle theta . The length of that side
is going to be the y-value for the point.
So the length of that side is 3. For the side
adjacent to angle theta, I'm going to have a length of 2,
that's the x-value for the point. So now I know the
opposite side... The opposite is 3
and the adjacent is 2.
Using that information and the Pythagorean
theorem,I can find this diagonal,
which is the hypotenuse. So the hypotenuse
is going to equal the square root
of the opposite side squared -- that's going to be 3-squared
plus
the adjacent side squared. That's
2-squared. That's going to give me
the square root of 3-squared, which is 9, plus 2-squared, which is 4.
And that will equal √13,
which can't be reduced. So now we know that the hypotenuse
has a length of 13.
With those two sides and the hypotenuse,
I can find trigonometric function values.
So to find the sine of theta, all I have to do
is take the opposite side, which is 3,
and put that over the hypotenuse,
which is √13. Now
you might be told that you're not supposed to leave radical signs in the
denominator.
Then what you have to do is multiply
this fraction 3 over √13
by the fraction √13
over √13. That second fraction just equals 1,
but when we multiply, we get
3 times √13, and in the denominator we have √13 time √13,
so that's just 13. And now we've got the sine of theta
as a fraction without a radical sign in the denominator.
For the cosine of theta
we're going to get the opposite side, which is 2,
over the √13. I will do the same trick to get rid of the radical
sign.
We'll multiply by √13 over √13.
So this will be 2√13
over 13. For the tangent,
we just take the opposite,
which is 3, over the adjacent, which is 2.
So we have 3 over 2. For the reciprocal functions,
cosecant, secant,
and cotangent, we're just going to take the values for the sine, cosine
and tangent,
and invert them for their reciprocals, and that will be the values for the other
three functions. Okay, let's do one more.
This one has some fractions and square roots in it.
So I've got the point
(-1/4, -√3/4). So,
(-1/4, -√3/4). So,
as before, we know that
the opposite side is going to have a length that's the same as the y-value.
So that's -√3 / 4.
And the adjacent side
is going to be -1/4.
So that would put this terminal point in this third
quadrant.
The hypotenuse is going to equal
the square root of...
(let's put this in parentheses)
-√3 /4 squared plus
-√3 /4 squared plus
-1/4 squared.
This simplifies nicely when we square everything.
We're going to get -√3 / 4 squared.
We're going to get -√3 / 4 squared.
is going to be 3
over 16.
plus -1/4 squared
will be 1/16.
That's going to equal √d(4/16),
which is the same as 2/4. I could reduce this down to 1/2,
but since my sides
both have denominators that are 4, I'm going to leave this as 2/4.
You'll see why in a second.
So I've got the two sides,
the opposite side and the adjacent side, and the hypotenuse
and now I want to find the sine.
The sine of theta is going to be
that opposite side, that's
-√3 / 4
over
2/4. Now let's remember something about fractions so we can
simplify this.
I've got a fraction divided by a fraction. When that happens,
I take the first fraction -√3 / 4 and
that's divided by the second fraction,
2 over 4, and I can rewrite that
as that same first fraction, -√3 / 4
times the reciprocal of the second fraction,
which is 4/2. And then those 4's are gong to cancel.
And I get -√3 /2.
Okay.
So this sine of theta can be reduced very simply if I just disregard
in each the fractions the denominators, the fraction in the numerator
and the fractions in the nominator. I just end up with
-√3 / 2.
The same kind of thing will happen when I do the cosine.
So the cosine of theta is going to equal to
-1/4,
that's the adjacent, over the hypotenuse, which is 2/4.
I'm going to cancel out those 4's.
and I just get -1/2.
And for the tangent,
I'm going to have
-√3 /4
over -1/4.
I can cancel out the 4's.
I've got two negative signs, so that becomes positive,
and all I've got here is just √3.
And once again,
given this information, you can fairly easily get your reciprocal function
values -- the cosecant,
secant and cotangent.
I hope this helps. Take care. I'll see you next time.