In this video, we’ll extend your knowledge of the trig unit circle. Watch, absorb, and then be ready for questions on your next test.

# Transcript: Intro to the Unit Circle – II

Now we can continue our discussion of unit circle trigonometry. So in the previous video, we discussed the basic definitions of the elementary trig functions sign and cosign from the unit circle. And just a brief reminder here, what we did was of course. We put our socket to a triangle inside the unit circle so that the hypotenuse to the unit circle is the radius and of course the radius of the unit circle is 1.

That means that the adjacent side winds up equaling the x, the opposite side winds up equaling y. And we can define cosine as sine. Cosine of theta equals x and sine of theta equals y.

In other words sine and cosine are defined in terms of. If we have a radius at an angle theta where does that radius intersect the unit circle? And that the coordinates of that point x, y where the radius equals the unit circle provide the definition for sine and cosine.

## Basic Unit Circle Definition

And so that’s the basic unit circle definition of sine and cosine. In the last video we talked about how to find the values of sine, and cosine, simply on the axes. From the positive and negative x-axis, and the positive and negative y-axis. In this video, we can talk about getting into the quadrants. When we start looking at slanting angles that aren’t on the axis.

So first let’s talk about the quadrants. The quadrant of theta tells us the positive or negative sign of sine and cosine, precisely because we know the positive and negative signs of x and y in the different quadrants. Let’s think about this. In the first quadrant, x and y are both positive. And so that means that sin and cos are both positive there.

That makes sense. All the outputs are first quadrant outputs, all of those are positive. Now, when we move to the second quadrant In the second quadrant, remember that y is still positive but now cosine is negative. And so that means sine is positive and cosine is negative in the second quadrant. When we move to the third quadrant both y and x are negative so both sine and cosine are, and are negative.

When we move to the fourth quadrant, that now we have that the y is negative but the x is positive. And so that means that the sign is negative and the cosine is positive there. It’s very easy to tell the positive or negative signs of these two functions a different quadrant purely from the quadrant. The quadrant itself gives us the information that’s really important.

We can also look at tangent so tangent is sine over cosine of that is y over x. And so that’s gonna be positive wherever y and x have the same sign. Well, y and x are both positive in the first quadrant. And they’re both negative in the third quadrant.

So positive over positive is positive, but negative over negative is also positive. And so tangent is positive when sine and cosine have the same sign, either they’re both positive or both negative. And this happens in the first and the third quadrant.

In the other two quadrants, x and y have opposite signs. One is positive, and one is negative. So you have a negative over a positive or a positive over a negative that produces a negative fraction. And so tangent is negative, when sine and cosine have opposite signs in the second and fourth quadrants. Notice that sine, cosine, and tangent, all three are positive in quadrant one.

That’s the only quadrant in which all three are positive, and then in each other quadrant. Two of them are negative and only one of them is positive. And so as we go around all of our positive sin is positive, tangent is positive, cosine is positive. Some people like to remember that as all students take calculus.

If that helps you remember it that’s great. Now how does this help us? Well now we’re on our way to figuring out the values of sine, cosine and tangent for angles that are not on the axis, angles that are in the quadrants. And it’s very important to realize that we have the information of a positive or a negative purely from the quadrants.

So now suppose we need the values of sin of 150 degrees and cosine of 150 degrees. This is a quadrant two angle. So let’s think about what we know first. From the quadrant we know right away that of course we’re in the second quadrant, Y is positive, X is negative. So that means that the sine of under 50° will be positive, the cosine will be negative.

We easily have the positive and negative sines, now we need to get the numerical values. So let’s talk about that. Notice that 150°, Equals 180 minus 30. And so that means that one way to get to the angle of 150 degrees would be to go a full 180, and then back up 30. Let’s think about that. And so that means we could draw a little triangle in Q2.

And we see what we get that little triangle in QII there with the 30 degree angle. That is congruent to the first quadrant triangle that would have a 30 degree angle. What that means is, the lengths of those sides are gonna be exactly the same as the lengths of the sides of the 30 degree angle. And of course the lengths of the side determine the values to sin and cosine.

So it means that at least the absolute value of sin will be the same for 150 and 30. The absolute value of cosine will be the same for 150 and 30. So what are the values of sin and cosine of 30 degrees? Well we talked about this a little bit a couple videos back. And so you may remember that we figured out there that the sin(30°) = 1/2, the cos(30°) = root 3/2.

Well now we’re golden, because sine of sine and cosine of 150 degrees will have those same absolute values. So now we know the absolute values, all we need to know is the positive and negative signs. But we already have that figured out we get that from the quadrants. And so the sine is still gonna be positive and the cosine is gonna be negative in the second quadrant and so that tells us right away. Just take these absolute values and adjust the positive and negative signs.

So, sine of 150 degrees is gonna be one-half cosine of 150 degrees is gonna be negative root 3 over 2. And that’s how we find those values.

## Reference Angles

To solve that we used a reference angle. This is a really important idea. This is one of the most important ideas in trigonometry. The idea of a reference angle for any angle greater than 90 degrees.

So we wrap around into the second quadrant, the third quadrant, and the fourth quadrant. What we wanna do is we imagine a segment. That goes from where that point ends up, dropped to the x-axis. You’ll be drawing to either the positive or negative x-axis, and you’d make a little triangle and that little triangle would have an acute angle in it. That acute angle is the reference angle.

And it turns out that that triangle you make is always gonna become congruent to the triangle that you would have in the first quadrant with the reference angle. And so all the values of the trig function In any quadrant, will be equal to, they’ll have the same absolute value as the same trig functions of the reference angle. Of course the reference angle will have positive outputs. So the only thing you’ll have to figure out is what their sine or cosine should be negative in another quadrant.

And again, we can figure that out from the quadrant itself. So any trig function of any large angle has the same absolute value as that trig function of the reference angle. And then we just find the positive or negative from the quadrant and we’re done. So we get the positive or the negative from the quadrant. We get the numerical value from the reference setting.

So let’s stick with the 30 degrees and let’s think about what would have a reference angle of 30 degrees. We’ve already done it in the second quadrant. What about in the third quadrant? So in the second quadrant it’s 150 degree angle that has the reference of 30. In the third quadrant we have to go all the way to 180 degrees and then we have to go 30 beyond that so would be at 210 degrees.

And so 210 degrees has a reference angle 30. So it has exactly the same absolute values as sine and cosine of 30 degrees, but now both of them have to be negative. Now we could say that the sine of 210 degrees is negative one-half. The cosine of 210 degrees is negative root 3 over 2. So again we’re using the same absolute values. And just putting a negative sign in front of both of them well now let’s go to QIV.

Well in QIV, we have to wrap all the way around to 360 degrees, and then back up 30 degrees so that would put us at 330 degrees. And so that’s the angle in QIV that has a reference angle of 30 degrees. So again, same absolutely value as the sine of 30 degrees and the cosine of 30 degrees. In QIV we know that x is positive and y is negative.

So sine is negative and cosine is positive. Those same values with the negative sine sin(330°) = 1/2, cos(330°) = positive root 3/2. And so notice that we get to recycle the SOHCAHTOA values in the first quadrant. We get to recycle those in each quadrant just by flipping around the positive and negative signs. That’s the beauty of reference angle.

### Reference Angle: Powerful Tool

So the test tends not to ask us directly to evaluate sine and cosine for angles such as this. Nevertheless, knowing and understanding reference angle is an extremely powerful problem solving tool. And again, I’ll put in another advertisement here. If you’re moving on to calculus, you will have to do evaluations like this once you’re in calculus.

And so this is really an important idea to understand if you’re thinking about taking calculus later on.

## Practice Problem

Here’s a practice problem, test-like practice problem. Pause the video and then we’ll talk about this. Okay. If the tangent of y, some angle y, equals 2 root 2, and we know that y is between 180 and 270 degrees. So in other words, it is a third quadrant angle which of the following is the value of cosine?

And so I’ll just point out here third quadrant, we know both x and y are negative, so right away we know the cosine is going to be negative. And in fact all five answer choices are negative, so we’re gonna have to pick a negative anyway. Let’s think about how we’d find that numerical value. So we’re gonna use the reference angle.

Let’s just suppose we build an acute triangle where angle has a tangent of two root two. And so we’re gonna let R be the reference angle of Y. So R and Y, they’re two, R is an acute angle, Y is a third quadrant angle, but R is the reference angle of Y. And we know that if the tangent is two over two, we could just make the opposite equal to two over two, and the adjacent equal to one.

This is actually a triangle that would not fit inside the unit circle, it’s a little bigger. That doesn’t matter, the ratios will still be the same. And so we need to find d. And so we can do that just with the Pythagorean Theorem. So ordinary Pythagorean Theorem, d squared = 1 squared + (2 root 2) squared.

Well that’s 1 squared, and then 2 root 2, that’s 2 squared, which is 4 times root two squared which is two and so that’s four root two which is eight and then eight plus one is nine, so that’s d squared. D squared has to be positive because it’s a length so it’s a square root of nine, which is three, so now we know d. So now inside that little reference triangle, we can find the cosine.

The cosine(R) is one-third. Well, now we’re golden. We know that the cosine(Y) will have exactly the same absolute value and we just have to adjust the positive for a negative sign. And so of course again third quadrant we need something negative. So we take this same absolute value and we make it negative and so the cosine of y would be -1/3.

So we go back to our answers and we choose -1/3.

## Summary

In summary, we can figure out the positive or negative sign from the quadrant of the angle. For any angle greater than 90 degrees we know that the sine and cosine of that angle will have the same absolute values as the sine of the reference angle and the cosine of the reference angle.

And so we could figure out the little SOHCAHTOA of the reference angle. And you could figure this out either from the properties of special triangles, you might have to use the Pythagorean theorem, as we used in the practice problem. But we can figure out everything in that acute triangle and then we can adjust the positive or negative signs from the quadrant.

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