Video Transcript
In this video, we will learn how to
use the fact that the quadrant where an angle lies determines the sign of its
trigonometric functions sine, cosine, and tangent to solve equations. Let’s start by thinking about what
we know about angles on a coordinate plane.
On a coordinate plane, the center
is the origin. For standard angles measured in the
coordinate plane, the positive 𝑥-axis will be the initial side, the starting point
of our measurement. This positive 𝑥-axis and the
beginning represents zero degrees. By the time we get to the positive
𝑦-axis, we’ve gone 90 degrees. The negative 𝑥-axis will be 180
degrees. The negative 𝑦-axis will be 270
degrees. And all the way back around for a
full turn is 360 degrees. In radians, that would be zero, 𝜋
over two, 𝜋, three 𝜋 over two, and then two 𝜋.
The ray that represents where we
stop measuring our angle is called the terminal side. And the angle and standard position
will be located between the initial and terminal side. But we want to specifically
consider angles that fall on the unit circle. The unit circle is centered at the
origin and has a radius of one unit. For some coordinate 𝑥, 𝑦 in the
first quadrant, the lengths 𝑥 and 𝑦 become the legs of a right triangle. And because we know that the radius
of this circle is one, the hypotenuse of this right triangle will be one.
If we label the side lengths here
𝑥 and 𝑦, we can take the Pythagorean theorem, which tells us 𝑎 squared plus 𝑏
squared equals 𝑐 squared, in the unit circle 𝑥 squared plus 𝑦 squared equals
one. We can also say that the equation
of the unit circle must be 𝑥 squared plus 𝑦 squared equals one. But what we wanna do now is
consider a bit of trigonometry with regards to the unit circle sine, cosine, and
tangent. We know the sine of an angle
represents the opposite side length over the hypotenuse, the cosine of an angle 𝜃
represents the adjacent side length over the hypotenuse, and the tangent
relationship of an angle is the opposite side length over the adjacent side
length.
For our angle 𝜃 inside the unit
circle, the opposite side length will always be the 𝑦-value, it’s the vertical
leg. The sin of 𝜃 is then 𝑦 over one
since in the unit circle the hypotenuse is always one. sin of 𝜃 is, therefore, equal to
𝑦 in the unit circle. Similarly, the adjacent side will
be the horizontal leg, the 𝑥-value. And that means in the unit circle,
the cos of 𝜃 will be equal to 𝑥. And this means that the tan of
angle 𝜃 will be equal to 𝑦 over 𝑥.
Before we look at some examples,
there’s one final thing we need to consider. And that is the sine of angles in
quadrants. If we keep our unit circle, and we
look at our 𝑥, 𝑦 in the first quadrant, in quadrant one, all 𝑥-values are
positive and all 𝑦-values are positive. This means that the sin of 𝜃 would
be positive since 𝑦 is positive, the cos of 𝜃 would be positive since 𝑥 is
positive, and since 𝑥 and 𝑦 are both positive, the tan of 𝜃 will also be
positive. This means that in quadrant one,
all trig relationships are positive.
But as we move to quadrant two and
create a right angle from the point negative 𝑥, 𝑦, we’re dealing with negative
𝑥-values and positive 𝑦-values. The 𝑦-value is positive in the
second quadrant, making the sine value positive. But the 𝑥-value is negative. And that means in quadrant two, the
cosine will be negative. With a positive value and a
negative value, The tangent then becomes negative. So, we can generalize and say in
quadrant two, the sine is positive but the cosine and tangent will be negative.
Moving around to quadrant three, in
quadrant three, the 𝑥-value and the 𝑦-value will be negative. This means the sine will be
negative, the cosine will be negative. However, with tangent, negative 𝑦
over negative 𝑥 will be positive. And so, in quadrant three, sine and
cosine are negative, while tangent is positive. And finally, in quadrant four, the
𝑥-values will be positive and the 𝑦-values will be negative. This gives us a negative sine
value, a positive cosine value, and a negative tangent value.
One way that you can remember these
sine values is by using the CAST diagram that looks like this. The CAST diagram indicates which
trig value will be positive in which quadrant. In quadrant one, all are
positive. In quadrant two, sine is
positive. In quadrant three, tangent is
positive. And in quadrant four, cosine is
positive. Now, we’re ready to consider some
examples.
Find sin of 𝜃, given 𝜃 is in
standard position and its terminal side passes through the point three-fifths,
negative four-fifths.
It’s probably helpful here to
sketch a coordinate plane. On this coordinate plane, we want
to plot the point three-fifths, negative four-fifths, which is here. If we know that the terminal side
of our angle goes through this point and that our angle is in standard position,
then the initial side will be the ray that begins at the origin and goes through the
positive 𝑥-axis. That means the angle we’re
interested in is the space from the initial side to the terminal side. However, to calculate this, we’ll
form a right angle with the 𝑥-axis.
When we do that, we end up with a
right-angled triangle. And we’ll use the angle created
with the 𝑥-axis to solve. We can solve this using
right-triangle trigonometry, where we have a right triangle with side lengths
three-fifths and four-fifths. We remember that sin of 𝜃 will be
the opposite over the hypotenuse. However, we don’t currently know
the hypotenuse. We don’t know the distance from the
origin to the point three-fifths, negative four-fifths. We can use the Pythagorean theorem
to find that, which means three-fifths squared plus four-fifths squared equals 𝑐
squared. Nine twenty-fifths plus sixteen
twenty-fifths equals 𝑐 squared. 25 over 25 equals one. And if 𝑐 squared equals one, then
𝑐 must be equal to one.
Now that we know the hypotenuse is
equal to one, we can say something else about this angle in our coordinate
plane. And that is that our point falls on
the unit circle. The unit circle has its center at
the origin and has a radius of one. We’re looking for the sin of
𝜃. It’s equal to the opposite over the
hypotenuse. Also in a unit circle, the sin of
𝜃 is equal to its 𝑦-coordinate. The 𝑦-coordinate for this point is
negative four-fifths. And so, we say that the sin of 𝜃
equals negative four-fifths. And if we wanted to check this, we
could remember based on our CAST diagram that for angles falling in the fourth
quadrant, the cosine is positive, but the sine and tangent will be negative, which
confirms that the sin of our angle 𝜃 is equal to negative four-fifths.
Let’s consider another example.
Suppose 𝑃 is a point on the unit
circle corresponding to the angle four 𝜋 over three. Is there another point on the unit
circle representing an angle in the interval zero to two 𝜋 that has the same
tangent value? If yes, give the angle.
First, we might wanna sketch a
coordinate plane and then add a unit circle, which is a circle with the center at
the origin and a radius of one. From there, we might also want to
label our coordinate plane with radians beginning at zero, 𝜋 over two, 𝜋, three 𝜋
over two, and two 𝜋. Because we have the interval zero
to two 𝜋, we know we’re only interested in one full turn. Our point 𝑃 is on the unit circle
and corresponds to the angle four 𝜋 over three. This means our first job is to find
out where the angle four 𝜋 over three would land.
I know for 𝜋 over three is greater
than 𝜋, but it’s probably worth comparing to find out if four 𝜋 over three is
greater than or less than three 𝜋 over two. If we give these fractions common
denominators, four 𝜋 over two becomes eight 𝜋 over six and three 𝜋 over two
becomes nine 𝜋 over six. Since eight 𝜋 over six is less
than nine 𝜋 over six, we can say that four 𝜋 over three is less than three 𝜋 over
two. And that means point 𝑃 is going to
fall in our third quadrant and that four 𝜋 over three would be this angle.
Since we know that our angle falls
in the third quadrant, we can use the CAST diagram, which will tell us that the
tangent of the angle in the third quadrant is going to be positive. In order for us to find another
point inside the unit circle that has the same tangent value, we’ll be looking for
the other place where the tangent value could be positive. And that will be in the first
quadrant. In the first quadrant, all trig
values will be positive.
But in order for us to find what
the value of that angle would be in the first quadrant, we need to break up our
angle four 𝜋 over three into smaller parts. We could say that four 𝜋 over
three is equal to 𝜋 plus 𝜋 over three, the distance from zero to 𝜋 and then an
additional 𝜋 over three. The right triangle created inside
the unit circle four 𝜋 over three in the third quadrant would look like this.
And in the first quadrant, there
would be some point such that we would be dealing with the angle of 𝜋 over
three. In the first quadrant, that would
have 𝑥, 𝑦 coordinates. And in the third quadrant, it would
have negative 𝑥, negative 𝑦 coordinates. And we know that in a unit circle,
the tan of 𝜃 will be equal to 𝑦 over 𝑥. And we would say that the tan of
four-thirds 𝜋 is equal to negative 𝑦 over negative 𝑥 and the tan of 𝜋 over three
is equal to 𝑦 over 𝑥. But we simplify negative 𝑦 over
negative 𝑥 to just 𝑦 𝑥. And so, we’ve shown that yes, there
is another angle in this interval that has the same tangent value. And it’s the angle 𝜋 over
three.
In our next example, we’ll consider
an application of the unit circle.
Consider a windmill with the blades
of length one meter. The position of the top 𝑃 of a
given blade is given by the coordinates 𝑎, 𝑏, which depends upon the angle 𝜃 as
shown. Express 𝑎 and 𝑏 as functions of
the measure of angle 𝜃 in radians. If angle 𝜃 at a certain time is
five-thirds 𝜋, what will it be after the blade has completed half a rotation?
Because the length of the blades
are one meter and in our diagram, this represents the radius of the windmill, we can
use our knowledge of the unit circle to help us solve this problem. The distance from the center of the
windmill to point 𝑃 would be one. And we’re told that the point 𝑃 is
located at 𝑎, 𝑏. And so, we can create a right
triangle with the 𝑥-axis and say that its side lengths are 𝑎 and 𝑏,
respectively. We use the distance from the
terminal side to the 𝑥-axis to calculate the angle 𝜃. We know that in the unit circle, we
can represent the angle as a sine and cosine relationship. For our angle 𝜃, the opposite side
length would be 𝑏 and its hypotenuse is one meter, so we have sin of 𝜃 equals 𝑏
over one. And we can represent 𝑏 as sin of
𝜃.
Similarly, if we look at cosine, we
end up with the side length 𝑎 over one, which means we can say that the distance 𝑎
must be equal to the cos of the angle 𝜃. And without more information, this
is as far as we can go with these two functions. We can say that 𝑏 equals sin of 𝜃
and 𝑎 equals cos of 𝜃.
Part two of our questions says if
the angle of 𝜃 at a certain time is five 𝜋 over three, what will it be after the
blade has completed a half a rotation? First of all, we’ve already been
told that we’re operating in radians, and so it might be helpful to label our
coordinate plane. Beginning at the 𝑥-axis, zero
radians, then 𝜋 over two radians, 𝜋 radians, three 𝜋 over two radians, and a full
turn, which is two 𝜋 radians. In a system like this, a full turn
is two 𝜋. And therefore, a half turn is going
to be equal to 𝜋 radians.
If we start with the angle 𝜃 five
𝜋 over three radians and we add a half a rotation, we’re adding 𝜋 to that
angle. And just like adding any fractions,
we need a common denominator. We can write 𝜋 as three 𝜋 over
three. We add five 𝜋 plus three 𝜋 to get
eight 𝜋. And the denominator wouldn’t
change. The angle 𝜃 after a half rotation
would be eight 𝜋 over three.
Before we finish, let’s review our
key points. The unit circle is a circle with
the center at the origin and a radius of one unit. For any point on the unit circle
𝑥, 𝑦, the sin of the angle 𝜃 created by that point will be equal to 𝑦 and the
cos of angle 𝜃 will be equal to 𝑥. Additionally, 𝑥 squared plus 𝑦
squared must equal one. We also saw that the CAST diagram
can be used to identify the signs of trigonometric functions in each quadrant.
