Video Transcript
Trigonometric Functionsโ Values
with Reference Angles
In this video, we will learn how to
find the reference angle and principal angle of an angle in standard position, and
we will learn how to use these to evaluate trigonometric functions.
To do this, we first need to recall
how we evaluate the trigonometric functions. Thatโs the sine of an angle and the
cosine of an angle. Now we can evaluate the
trigonometric functions by using right triangles. However, to extend these functions
even further, we need to use the unit circle. And thatโs the circle of radius one
centered at the origin. We can then use this to evaluate
trigonometric functions. For example, letโs sketch the angle
150 degrees in standard position. Well, remember, to draw an angle in
standard position, we draw it from the positive ๐ฅ-axis measured counterclockwise if
the angle is positive and clockwise if the angle is negative.
We then know that the point of
intersection between the terminal side of our angle and the unit circle tells the
value of the sine and cosine evaluated at this angle. The ๐ฅ-coordinate tells us the
cosine of this angle, and the ๐ฆ-coordinate tells us the sine of this angle. So our point of intersection will
have coordinates cos 150 degrees, sin of 150 degrees. We want to determine these values,
so we need to find the values for these coordinates. To do this, we first recall that
the angles in a straight line sum to 180 degrees. This means we can find the value of
the following angle. This angle is 30 degrees, since it
needs to add to 150 degrees to make 180 degrees.
If we then drop a vertical line
from the point of intersection to our ๐ฅ-axis, we construct a right triangle. And to help us with our
calculations, it might help us to sketch this right triangle on its own. First, we can see the hypotenuse of
this right triangle is a radius of the circle. And remember, since this is the
unit circle, its radius is equal to one. So the hypotenuse of our triangle
is one. Next, we can see that the base and
height of this right triangle are given by the ๐ฅ- and ๐ฆ-coordinate of the point of
intersection.
But remember, lengths are not
allowed to be negative. Since weโre in the second quadrant,
our ๐ฅ-coordinate is going to be negative, while our ๐ฆ-coordinate will be
positive. So to be extra safe, weโll take the
absolute value of both of these coordinates. Remember, that just means we take
the nonnegative value. We can then determine values for
these expressions by applying trigonometry to our right triangle. If we label our sides relative to
the angle 30 degrees, then we see that the absolute value of sin of 150 degrees is
the opposite side to our angle and the absolute value of the cos of 150 degrees is
the adjacent side of our angle.
Weโre now ready to find expressions
for these values. First, recall that the sine of an
acute angle is equal to the ratio of the length of the opposite side and the
hypotenuse in a right triangle. In our right triangle, the opposite
side had length absolute value 150 degrees and the hypotenuse had length one. So the quotient of these two values
is equal to the sin of 30 degrees. We know that dividing by one
doesnโt change the value and 30 degrees is a special angle. We know how to evaluate the sin of
30 degrees. Itโs equal to one-half. We can actually prove this by using
equilateral triangles.
But weโre not done yet. Remember, weโve only shown this is
equal to the absolute value of the sin of 150 degrees. What weโve actually found here is
the length in our right triangle. So weโve found that the length of
this side in our right triangle is one-half. The sin of 150 degrees is the
๐ฆ-coordinate of our point of intersection. Since we can see this is in the
second quadrant, we know this is positive. So we take the positive value. Weโve shown the sin of 150 degrees
is equal to one-half.
We can do the same to find the cos
of 150 degrees. This time, we need to take the
ratio of the adjacent side and the hypotenuse. Using trigonometry, we have the cos
of 30 degrees is equal to the absolute value of the cos of 150 degrees divided by
one. Once again, we know dividing by one
doesnโt change a value. And we can evaluate the cos of 30
degrees. Itโs equal to the square root of
three divided by two. But remember, we do need to be
careful. What we found is the length of the
base of our right triangle, whereas the cos of 150 degrees is the ๐ฅ-coordinate of
the point of intersection. We can see that this is
negative. Therefore, weโre going to need to
take the negative value. Hence, weโve shown the cos of 150
degrees is equal to negative the square root of three divided by two.
We can follow the exact same
procedure to evaluate the trigonometric functions of many different values. For example, if we were asked to
evaluate the sin of negative 405 degrees, we would start by sketching negative 405
degrees in standard position, where we remember since the angle is negative, we need
to measure our angle clockwise. We know that one full revolution
clockwise is negative 360 degrees. We need to then go 45 degrees
further clockwise. This gives us the following
angle. And itโs worth reiterating here the
terminal side of an angle in standard position is not unique. For example, we could measure an
angle clockwise of exactly the same terminal side as negative 405 degrees.
One way of determining this value
is to add integer multiples of 360. We see negative 405 plus two times
360 is 315 degrees. Both of these angles will have the
same terminal side when sketched in standard position. We want to use this to determine
the sin of negative 405 degrees. We recall this is given by the
coordinates of the point of intersection between the terminal side and the unit
circle. The ๐ฅ-coordinate will be cos of
negative 405 degrees, and the ๐ฆ-coordinate will be sin of negative 405.
degrees. We want to determine the value of
this ๐ฆ-coordinate in the same way, by using trigonometry.
We drop a perpendicular from the
point of intersection to the ๐ฅ-axis, giving us a right triangle. To help us evaluate this, weโll
sketch the triangle outside of the circle. Once again, we note the hypotenuse
of this right triangle is a radius of the circle. And since this is the unit circle,
its radius is one. So the hypotenuse of our right
triangle is one. The height of this right triangle
is going to be the absolute value of the ๐ฆ-coordinate of the point of
intersection. Thatโs the absolute value of the
sin of negative 405 degrees. Itโs worth noting we could add the
base into our diagram; however, itโs not needed to answer this question.
We want to apply trigonometry to
find a value for this expression. However, to do this, weโre going to
need to know one of the angles. There are a few different ways of
doing this. Letโs consider only the value of
315 degrees. Since the terminal side of these
two angles are the same, we can use either of these two values. And itโs often easier to work with
positive values between zero and 360 degrees. We can then see one of the missing
angles in our triangle adds to make a full revolution with the angle 315
degrees. If we call this angle ๐, then we
have ๐ plus 315 degrees makes a full revolution. In other words, this is equal to
360 degrees.
We can then solve for ๐ by
subtracting 315 degrees from both sides. We get that ๐ is equal to 45
degrees. We can then add this into our right
triangle. We can then use right-triangle
trigonometry to find an expression for the sin of negative 405 degrees. We start by labeling the sides
relative to the angle of 45 degrees. And we see we have the length of
the opposite side and the length of the hypotenuse. So weโre going to need to use the
sine ratio. Applying trigonometry, we get the
sin of 45 degrees is equal to the absolute value of the sin of negative 405 degrees
divided by one.
We can simplify this. First, the sin of 45 degrees is
equal to the square root of two divided by two. Next, we know dividing by one wonโt
change the value. Finally, remember that this is
telling us the height of our right triangle; however, weโre trying to find the value
of the ๐ฆ-coordinate of the point of intersection. By applying trigonometry, we showed
that the height of this right triangle was root two over two. However, we can see weโre in the
fourth quadrant, so the ๐ฆ-coordinate is negative. So we need to take the negative of
this value. We have the sin of negative 405
degrees is equal to negative root two divided by two.
In this example, to evaluate the
sin of negative 405 degrees, we started by finding an equivalent positive angle
between zero and 360 degrees. And then we found the value of ๐,
which is the acute angle the terminal side makes with the ๐ฅ-axis. We can actually evaluate the sine
or cosine of any angle in exactly the same way. We first find an equivalent angle
in standard position between zero and 360 degrees and then find the acute angle that
the terminal side makes with the ๐ฅ-axis. And we give names to both of these
types of angles. First, if ๐ is an angle in
standard position, then the counterclockwise angle between the initial and terminal
side of angle ๐, which is less than a full turn, is called the principal angle of
๐.
For example, we showed that 315
degrees had the same terminal side as negative 405 degrees in standard position. This means that 315 degrees is the
principal angle of negative 405 degrees. And we use this angle to help us
evaluate the trigonometric functions at negative 405 degrees. Similarly, if ๐ is an angle in
standard position, then we call the acute angle that the terminal side of angle ๐
makes with the ๐ฅ-axis the reference angle of ๐. For example, we showed when
negative 405 degrees was drawn in standard position, its terminal side made an angle
of 45 degrees with the ๐ฅ-axis. So we could say the reference angle
of negative 405 degrees is 45 degrees. We used both of these to help us
evaluate trigonometric functions at this value.
There are four different
possibilities for the principal angle and the reference value. And these are related to which of
the four quadrants the terminal side of our angle lies in. First, if the terminal side lies in
the first quadrant, then the principal angle will just be an acute angle and the
reference angle will have the same measure. If the terminal side is in the
second quadrant, then the principal angle will be obtuse and the reference angle
will be 180 degrees minus the principal angle. If the terminal side is in the
third quadrant, then we can see that our principal angle will be between 180 and 270
degrees. However, the reference angle is
only going to be the following acute angle. This is the principal angle minus
180 degrees. Finally, if the terminal side is in
the fourth quadrant, then our principal angle will be between 270 and 360 degrees
and the reference angle will be the angle needed to make this a full turn. It will be 360 degrees minus the
principal angle.
Letโs now see an example of
determining a principal angle.
Given the angle negative two ๐ by
three, find the principal angle.
Weโre asked to determine the
principal angle of negative two ๐ by three. We start by recalling that the
principal angle will be an angle in standard position with the same terminal side
and our angle should be between zero and two ๐. Therefore, to find the principal
angle, weโll start by sketching negative two ๐ by three in standard position. First, since our angle is negative,
this means weโre going to need to measure our angle clockwise from the positive
๐ฅ-axis. Next, it might help us to measure
each of the right angles. Going clockwise from the positive
๐ฅ-axis, thatโs negative ๐ by two, negative ๐, and negative three ๐ by two.
Next, either by using the
coefficients or calculating, we can see that negative two ๐ by three lies between
negative ๐ by two and negative ๐. In other words, its terminal side
lies in the third quadrant. We want to find the principal
angle. Remember, this is measured in
standard position and itโs between zero and two ๐. So itโs measured clockwise. And it needs to have the same
terminal side. This gives us the following
sketch. And together with the measure of
negative two ๐ by three, we can see this makes a full revolution. In other words, the principal angle
plus two ๐ by three must be equal to two ๐.
And we can then solve for the
principal angle by subtracting two ๐ by three from both sides of the equation. We get that the principal angle
will be two ๐ minus two ๐ by three, which we can evaluate by noting that two ๐ is
equal to six ๐ by three. We subtract two ๐ by three from
this to get four ๐ by three, which is our final answer. Therefore, we were able to show the
principal angle of negative two ๐ by three is four ๐ by three.
Letโs now see an example where we
use the principal and reference angles to evaluate a trigonometric function without
a calculator.
Find the value of the sin of 11๐
by six.
In this question, weโre asked to
evaluate the sin of on angle in radians. We can notice that 11๐ by six is
not an acute angle, so we canโt do this just by using right-triangle
trigonometry. So instead, we need to recall that
we can evaluate trigonometric functions by sketching the angle in standard position
and then using the intersection with the unit circle. To do this, we need to find the
coordinates of the point of intersection between the terminal side of our angle in
standard position and the unit circle. So weโll start by sketching our
angle in standard position.
Remember, this is measured
counterclockwise from the positive ๐ฅ-axis. To determine which side the
terminal side lies in, it can help to add the right angles to our diagram, where we
note that one full revolution gives us two ๐. If we were to evaluate 11 over six,
we would see itโs equal to 1.83 recurring. So itโs between three over two and
two. Multiplying this inequality through
by ๐, we see that our angle must lie in the fourth quadrant. This gives us the following
sketch.
Remember, the coordinates of the
point of intersection between the terminal side and the unit circle will help us
evaluate the trigonometric function at this angle, in this case, the coordinates of
the point of intersection are the cos of 11๐ by six and the sin of 11๐ by six. This means weโre asked to determine
the ๐ฆ-coordinate of this point of intersection. To do this, weโll drop a
perpendicular line from the point of intersection to our ๐ฅ-axis. Then the length of this line can be
used to find the value of the sin of 11๐ by six.
To help us determine this value,
letโs isolate the following right triangle from our diagram. We want to determine the height of
this right triangle. And to do this, we can notice we
can find the hypotenuse of our triangle. We can see itโs a radius of the
circle. And this is the unit circle, so all
of the radii have length one. Next, we know the height of our
triangle was given by sin of 11๐ by six. But we do need to be careful
because this ๐ฆ-coordinate is negative. So instead, weโll write this as the
absolute value of the sin of 11๐ by six to make this a positive value.
Finally, we can determine one of
the angles in our right triangle. We can see that, together with 11๐
by six, this angle makes a full revolution, which means this angle has measure two
๐ minus 11๐ by six. And if we evaluate this, we get ๐
over six. Itโs worth noting that this angle
has a name. Itโs called the reference angle of
11๐ by six. Itโs the acute angle that the
terminal side makes with the ๐ฅ-axis. We now have an angle and one side
in our right triangle, and we want to determine the other side in our right
triangle. So we can do this by using
trigonometry.
Remember, the sine of an angle in a
right triangle is the ratio of the length of the opposite side and the hypotenuse
side in our right triangle. In our case, this tells us the sin
of ๐ by six is equal to the absolute value of the sin of 11๐ by six all divided by
one. And we can evaluate this
expression. First, dividing by one doesnโt
change the value. Next, ๐ by six is a special
angle. We know the sin of ๐ by six is
equal to one-half. In fact, we can prove this by using
equilateral triangles. Therefore, weโve shown that the
height of our right triangle is equal to one-half.
But weโre not done yet. Remember, weโre asked to evaluate
the sin of 11๐ by six, which is the ๐ฆ-coordinate of the point of intersection
between the terminal side and the unit circle. This is why itโs important we
remember our absolute value sign. We can see in the diagram that this
point of intersection has a negative ๐ฆ-coordinate. This means we need to take the
negative value of our answer, giving us that the sin of 11๐ by six is equal to
negative one-half, which is our final answer. Therefore, by sketching the angle
11๐ by six in standard position and then finding its reference angle, which was ๐
by six, and then applying trigonometry, we were able to show the sin of 11๐ by six
is equal to negative one-half.
Before we finish the video, there
is one more thing worth pointing out about this method. We can use this method to evaluate
the reciprocal trigonometric functions as well, for example, if we were asked to
evaluate the csc of 11๐ by six. We would start by recalling the
cosecant of an angle is exactly equal to one divided by the sine of that angle. And we know how to find the sine of
an angle; we find the reference angle and then use trigonometry. We already found the sin of 11๐ by
six to be equal to negative one-half. We can then substitute this value
in to evaluate the csc of 11๐ by six. Itโs equal to one divided by
negative one-half, which, of course, simplifies to give us negative two.
Letโs now go over some of the key
points of this video. First, if ๐ is an angle in
standard position, then the counterclockwise angle between the initial and terminal
side of angle ๐, which is less than a full turn, is called the principal angle of
๐. This is useful because the
principal angle has the same terminal side, so evaluating the trigonometric
functions at the principal angle will not change their value. Next, the acute angle between the
terminal side of ๐ and the ๐ฅ-axis is called its reference angle. Itโs worth pointing out here we
assume the angle is not quadrantal because otherwise it wonโt have a reference
angle; weโll just use the right angle instead. Finally, we saw that we can
evaluate equivalent arguments for trigonometric functions by sketching the argument
in standard position and then using its principal and reference angles.