Trigonometric Functions on the Unit Circle: Videos & Practice Problems

The unit circle is essential for understanding trigonometric functions, where angles correspond to specific x and y coordinates. The sine and cosine of an angle are represented by the y and x values, respectively, while tangent is the ratio of sine to cosine. For angles outside the first quadrant, reference angles simplify calculations. Reciprocal functions like cosecant, secant, and cotangent are derived from sine, cosine, and tangent. Key equations include the unit circle equation $x^2+y^2=1$.

Introduction to the Unit Circle

Welcome back, everyone. Here we're going to take our first look at what's called the unit circle, which is just a circle of radius one where all of the angles going around our circle correspond to specific x y coordinates on our graph. Now, the unit circle is going to be massively important in this course because you're going to need it to solve a bunch of different trigonometry problems. And the first time you see the unit circle, it may be a bit overwhelming because there is a ton of information here, and you might not know any of it yet. But that's totally okay because here I'm going to completely break down the unit circle for you throughout this chapter, and soon you'll be able to look at a completely blank unit circle and fill in all of that missing information completely on your own.

So let's go ahead and get started with what the unit circle even is. Now like I mentioned, the unit circle is a circle of radius one, and like any circle, it goes around from zero to 360 degrees. Now this might be really familiar to you but we want to look specifically in the context of this unit circle so we're gonna go around and fill in all of these angle measures. Now remember, the convention when looking at angle measures is to start measuring from the x-axis from zero degrees here. And going a quarter of the way around my circle, I reach 90 degrees and then halfway around my circle I am at 180 degrees and then two seventy degrees at the bottom of my circle here until I reach a full rotation of 360 degrees.

Now we don't just want to know these angle measures in degrees because it's also important to know our angle measures in radians, especially when working with the unit circle. So remember, a full rotation around three sixty degrees is also equal to two pi radians. So looking at our unit circle here, zero degrees is zero radians. And then going up to 90 degrees, this is π2radians. Then halfway around our circle, we reach π radians.

And then at the bottom here, we have 3π2 radians until we reach a full rotation, of course, of two pi radians at the end here. Now the angle measures are not the only important thing here, and this might not have been any new information. But remember, all of these angle measures on our unit circle specifically correspond to an x y coordinate. So looking at our graph here at zero degrees and noticing the scale of my graph, I see that zero degrees is located at this point one zero. Then up to 90 degrees, this is the point zero one.

Over at one hundred and eighty degrees is the point negative one zero. And then down here at the bottom of my circle this is the point zero negative one. So all of these angle measures correspond to these ordered pairs as well. Now knowing that the radius of our unit circle is one and seeing all of these points on our graph, you may have also noticed that our center is right here at the origin so our unit circle is centered at the point zero zero. Now with this knowledge in mind, we can also represent our unit circle using an equation.

Now here we have the general equation of a circle, but knowing that our center is at zero zero, this h and this k go away. And we also know that our radius is one. So this gives us our our equation for the unit circle as x squared plus y squared equals one. So that means any x y point on my unit circle, I can plug back into this equation and it would give me a true statement. So all of these points that I have on here, I could plug into that equation and get a true statement because I know that they're all on the unit circle.

But you might actually be given a completely random point and be asked to verify whether it's on the unit circle or not because these are not the only four points on my graph. Right? I have all of these points in between them. So let's go ahead and take a look at our first example here. In this first example, we're asked to identify which points are which points are on the unit circle and then label them on the graph.

So the first point that we're given here is this point one one, and we wanna verify if this is on the unit circle. Now this is actually a point that's going to be really easy to just go ahead and plot, so let's go ahead and just do that first. So looking at my graph, going up to one one, here is the location of that point, one one. And I can clearly see that this is not on my unit circle. It's a little bit outside of it, so I can already say that this is not on the unit circle.

But if we want to perform an extra check here we can also go ahead and just plug this point into our unit circle equation. So doing that I get one squared plus one squared and I want to know is that equal to one because if it is then it's on the unit circle. Now one squared is just one so this is one plus one And is this equal to one? Well, one plus one is equal to two, so this is definitely not a true statement. Two is not equal to one, so this verifies what we already saw that this point is not on our unit circle.

Let's look at one other point here. We have the point half root three over two. Now this seems like a rather random point and it's definitely not one that I can just plot on my graph easily like one one because I don't know exactly where root three over two is. So let's just go ahead and plug this right into our equation to verify whether it's on the unit circle or not. So plugging in our x and y values here, I have one half squared plus root three over two squared, And we're verifying is this equal to one.

So squaring these values, remember, with fractions, we wanna square both the numerator and the denominator. So this one half squared is going to be one fourth. And then squaring root three over two, I know that it will cancel out that square root in the numerator giving me a three and then squaring that denominator, I get a four again. And we're verifying is this equal to one. Now adding these two fractions together, this gives me four over four.

And is this equal to one? Well, four over four is indeed one. So one equals one. This is definitely a true statement, which tells me that this point, half root three over two, is actually on our unit circle. And we can go ahead and plot it there as well.

So looking at our unit circle on our coordinate system over here, if I go over to the point 0.5 on that x axis, I know that wherever it's touching the unit circle that is this point half root three over two. So my point is right here half root three over two. But remember, every single point on our unit circle also corresponds to an angle. And it just so happens that this point corresponds to the angle 68 degrees. Now we'll continue to see this angle and this point throughout this chapter, but for now since we know what the unit circle is and how to verify if a point is on it, let's get a bit more practice.

Thanks for watching and let me know if you have questions.

Identify the quadrant that the given angle is located in.
$\frac{7\pi}{4}$ radians

• A. Quadrant I
• B. Quadrant II
• C. Quadrant III
• D. Quadrant IV

Answer: (D)

Identify the quadrant that the given angle is located in.
$\frac{\pi}{7}$ radians

• A. Quadrant I
• B. Quadrant II
• C. Quadrant III
• D. Quadrant IV

Answer: (A)

Identify the quadrant that the given angle is located in.
$\frac{2\pi}{3}$ radians

• A. Quadrant I
• B. Quadrant II
• C. Quadrant III
• D. Quadrant IV

Answer: (B)

Identify the quadrant that the given angle is located in.
$\frac{6\pi}{5}$ radians

• A. Quadrant I
• B. Quadrant II
• C. Quadrant III
• D. Quadrant IV

Answer: (C)

Test whether the point is on the unit circle by plugging it into the equation, $x^2+y^2=1$.
$\left(\frac{-\sqrt2}{2},\frac{-\sqrt2}{2}\right)$

• A. On Unit Circle
• B. NOT on Unit Circle
• C. More information is required

Answer: (A)

Sine, Cosine, & Tangent on the Unit Circle

Hey, everyone. We just learned about the unit circle, and we saw that every angle corresponds to a specific xy coordinate on that unit circle. Like our angle of zero degrees corresponds to the point one zero. But how exactly are these two things related mathematically? Well, they're related using trigonometric functions, which specifically relate angles to points on our unit circle.

Now you may have worked with these trig functions, sine, cosine, and tangent before using right triangles. So it may seem really strange here that we're now working with these trig functions using a circle, but it's actually the exact same thing because we can think of all of these angles as forming a right triangle. And it's actually going to be even easier to find trig values of an angle because they're just going to be equal to the x and y coordinates on our unit circle. And I'm gonna show you exactly how that works here. So let's go ahead and get started.

Now looking at our angle of 53 degrees here, I see that it corresponds to this point three-fifths four-fifths on that unit circle. Now if I trace my x and my y values here, I have formed a right triangle. And this right triangle has a hypotenuse of one because that's the radius of the unit circle. Then it has a base of my x value and a height of that y value that we see in this coordinate right here. Now looking at this triangle, if I asked you to find the sine of 53 degrees and the cosine of 53 degrees using this triangle, you would actually find that the sine is just equal to that y value or the height of our triangle and the cosine is going to be equal to the x value or the base of our triangle.

And this is actually how it's going to work for every single angle on our unit circle. The sine of an angle is always the y value or the height of our corresponding triangle, and the cosine of an angle is always the x value or the base of our corresponding triangle. Now you're not going to want to have to set up a triangle and work this out algebraically every single time. So I like to remember this by just looking at my ordered pair. Now we know that ordered pairs go in the order xy, and I like to think about my trig functions as going in alphabetical order.

I know that c comes before s in the alphabet, so cosine comes before sine here. My cosine is my x value and my sine is my y value. Now like I said, this works for any angle on our unit circle, so let's come down to our zero-degree angle measure here. Now this can be kind of strange to think about as a triangle, but think about your triangle as being squished against that x-axis to give it a height of zero. Now when looking at these values here, one and zero, I know that the cosine of zero degrees is equal to one and the sine of zero degrees is equal to zero, which makes sense because the height of my triangle is zero here.

Now we don't want to forget about tangent because we still need to find the value of our tangent. Now you might have calculated the tangent previously by dividing sine over cosine. But since our sine and our cosine correspond to our y and x values here, we can find the tangent of an angle by simply dividing y over x. So coming back over here to the tangent of zero degrees, if I take my y value of zero and divide it by my x value of one, I end up getting that the tangent of zero degrees is simply zero. Now let's go ahead and look at the tangent of our other three quadrantal angles on our unit circle starting with 90 degrees.

So the tangent of 90 degrees, if I take my y value of one divided by my x value of zero, I end up with a fraction dividing by zero which is not something that we can do so this is actually an undefined value. Now if I come over here to 180 degrees I will end up with zero on the top of my fraction the same way I did with zero degrees. So the tangent of 180 degrees is also equal to zero. Now let's come down here to the tangent of 270 degrees. We will end up dividing by zero the way we did with 90 degrees.

So this is again an undefined value. But these are not the only four points on our unit circle. Right? We know that we have all of these other points actually forming the rest of our circle. So let's go ahead and take a look at the trig values of some other points along this circle.

Now looking at two seventeen degrees here, first I come down to this third quadrant looking at my angle of 217 degrees. Now we can go ahead and draw our triangle here in order to visualize this a little bit more. And then looking at this ordered pair, I remember, okay, xy, cs telling me which is my cosine and which is my sine. So looking at this two sine of 217 degrees, I know that that's my y value. So the sine of 217 degrees is negative three-fifths.

Now it may seem strange here to have a negative value for a trig function, but this is totally fine because looking at the triangle that is on this graph, we see that we are in the negative y values because this is attached to a coordinate system. Right? So it's totally fine to have a negative trig value. Now our cosine here is our x value. Remember negative four-fifths which is also a negative value which makes sense looking at the triangle on our graph here.

And then finally we have the tangent of two seventeen degrees. Now remember for the tangent of our angle, we want to take our y value and divide it by our x value. So here we would take negative three over five and divide it by negative four over five. Now whenever we divide fractions, we're just really multiplying by the reciprocal. So this would be negative three-fifths times negative five fourths, the reciprocal here.

Now those fives are going to cancel, and the negatives will cancel as well, leaving me with an answer of three over four. Now looking at this, you may have noticed that because these have the exact same denominator, we have effectively just divided their numerators. So you can think about that as kind of a shortcut here as we continue to work with dividing fractions. So this tells me the tangent of 217 degrees degrees is three over four. Now let's move on to our final example here.

We have an angle measure of 60 degrees in that first quadrant. Now looking at my 60-degree angle, I can go ahead and draw that triangle if it makes you happy, and then we can go ahead and identify our trig values. Now remember looking at our ordered pair xy cs my cosine value is that one-half that I can go ahead and fill in and then my sine value is that y square root of three over two. Now for the tangent we again take y over x and looking at these numbers they again have the same exact denominator of two so effectively we're just going to be dividing their numerators. So if I take the numerator of each of them, square root of three over one, that just gives me a value of the square root of three and I'm done here.

The tangent of 60 degrees is the square root of three. Now that we've seen how trig values relate angles to their ordered pair on the unit circle, let's get a bit more practice together. Thanks for watching, and I'll see you in the next one.

Find the sine, cosine, and tangent of each angle using the unit circle.
$\theta=-1.18$ rad, $\left(\frac{5}{13},-\frac{12}{13}\right)$

• A. $\sin\theta=-\frac{12}{13},\cos\theta=\frac{5}{13},\tan\theta=\frac{12}{5}$
• B. $\sin\theta=-\frac{12}{13},\cos\theta=\frac{5}{13},\tan\theta=-\frac{12}{5}$
• C. $\sin\theta=\frac{12}{13},\cos\theta=\frac{5}{13},\tan\theta=\frac{5}{12}$
• D. $\sin\theta=\frac{5}{13},\cos\theta=\frac{-12}{13},\tan\theta=\frac{5}{12}$

Answer: (B)

Find the sine, cosine, and tangent of each angle using the unit circle.
$\theta=225\degree,\left(-\frac{\sqrt2}{2},-\frac{\sqrt2}{2}\right)$

• A. $\sin\theta=-\frac{\sqrt2}{2},\cos\theta=-\frac{\sqrt2}{2},\tan\theta=2$
• B. $\sin\theta=\frac{\sqrt2}{2},\cos\theta=-\frac{\sqrt2}{2},\tan\theta=-1$
• C. $\sin\theta=-\frac{\sqrt2}{2},\cos\theta=-\frac{\sqrt2}{2},\tan\theta=1$
• D. $\sin\theta=\frac{\sqrt2}{2},\cos\theta=\frac{\sqrt2}{2},\tan\theta=\frac{\sqrt2}{1}$

Answer: (C)

Sine, Cosine, & Tangent of 30°, 45°, & 60°

Hey everyone. We just learned how trig functions relate angles to their corresponding point on the unit circle, where x and y values are cosine and sine values of that angle, respectively. Now sometimes you won't be given this ordered pair, and you'll just be asked to find the sine or cosine of some angle with no other information. And that's where memorization comes in. Now memorization and math can be really tricky, but here I'm going to walk you through two different ways to memorize the trig values of the three most common angles, thirty, forty-five, and 60 degrees.

And with this knowledge, you'll be able to solve pretty much any trig problem that gets thrown your way. So let's go ahead and get started here with what you may hear referred to as the one two three rule. Now no matter how you choose to memorize these values, you're always going to start in the same way with the square root over two. Because all of these trig values are the square root of something over two and our job is just to memorize what that something is. So memorizing that with the one, two, three rule, you may be wondering why it's called that.

And it's because we're going to start with our x values and count one two three going clockwise. Then for our y values, we're going to count one two three going counterclockwise. Now what exactly do I mean by that? Well, we're going to start in this upper left corner with the x value of 60 degrees, and we're going to start counting from one, going one, two, three clockwise around our unit circle for our x values. Then we're going to go back counterclockwise and count one, two, three back up for our y values.

Now we're done here. These are all of our trig values. We've gone one, two, three clockwise, one, two, three counterclockwise, and we're done. Now we can do a bit more simplification here because we know that the square root of one is just one. So this x value or this cosine of 60 is really just one half, and the sine value of 30 degrees or the y value is also just one half.

Now remember that these values, these x and y values, also represent the base and height of the corresponding triangle. And we also don't want to forget about our tangent value. Remember that the tangent of any angle can be found by simply dividing sine by cosine. So once we have those sine and cosine values, we can find our tangent pretty easily. So looking at 30 degrees here, if I take the my sine value one half and divide it by my cosine value to get tangent, here I'm really just effectively dividing those numerators because they have the same exact denominators.

So for my tangent, I get one over the square root of three. Or with my denominator rationalized, rationalized, I get the square root of three over three as my tangent. Now you can find the tangent value of these other two angles the same exact way, and you can feel free to pause here and try that on your own. Now this was the one two three method, but remember that I promised you two different methods of memorizing this. So let's take a look at one more, a bit more visual hands-on approach to memorizing these values referred to as the left hand rule.

So for our left hand rule, you're going to take, you guessed it, your left hand and put it in front of your face like this. Now to you, your hand will look something like this. And I want you to consider your pinky as being at zero degrees and your thumb as being at 90 degrees, effectively making your left hand the first quadrant of the unit circle. So your other three fingers are going to represent those three common angles. So 30 degrees, 45 degrees, and 60 degrees.

Now really imagine your left hand as being that first quadrant of the unit circle. And from here, we can find our trig values by simply counting on our fingers. So let's go ahead and focus in on the angle 30 degrees. Now to do that, you're going to look at your hand and you're going to take that finger closest to your pinky that represents 30 degrees and you're going to fold it inward. Now we're going to count the number of fingers that are above that folded in finger and below that folded in finger in order to find our trig values.

So for our sine and cosine of 30 degrees, we're going to start in that same way. Remember, the square root of something over two and our number of fingers is going to tell us what that something is. So with our finger folded in here, we're going to count the number of fingers that are above that folded in 30 degree finger and put that under our square root in order to get the cosine of 30 degrees. So the cosine of 30 degrees here, counting those fingers, I have three fingers above. So this is the square root of three over two.

The cosine of any angle is the square root of your fingers above divided by two. Now for our sine, we're instead going to look at the number of fingers below, in this case, just one, our pinky. So we get the sine of 30 degrees as being the square root of one over two or just one half. Now for our tangent, remember that we can always just take the sine divided by the cosine. Or here we can also rely on counting our fingers again.

So for the tangent of an angle, we're going to take the square root of the fingers below that folded in finger and divide it by the square root of the fingers above. So here the fingers below again were one. So the square root of one over the square root of three. So the tangent of 30 degrees, we get one over the square root of three. Or with that denominator rationalized, we end up with the square root of three over three.

Now this left hand rule will work for any angle of the first quadrant, any finger you can fold in, and use this in order to find your trig values. Now that we've seen these trig values of these common angles, let's get a bit more practice in that first quadrant. Thanks for watching, and I'll see you in the next one.

Sine, Cosine, & Tangent of 30°, 45°, & 60° Example 1

A common problem you may be faced with is filling in an entirely blank unit circle, and being able to do that starts with the first quadrant. So here we're going to fill in all of the missing information in the first quadrant of this unit circle. Now the way that I do this might not be the way that you choose to do this, and that's totally okay. Always find what works best for you. And here I'm going to walk you through my thought process of filling in this first quadrant.

Now feel free to pause here and try this on your own before jumping back in with me. Now the way that I like to start with this unit circle is to start with my x and my y axes because this is the easiest information for me to remember. So starting with zero degrees, I know that this point is right here on my x-axis at one zero. Then going up to this other axis, my y-axis, this is at 90 degrees, which I also know is π2 radians and is located at the point on the y-axis zero one. So from here, now that I have my information on each axis, I like to move on to my degree angle measures because this is also something that's rather easy for me to remember.

I know that I start with 30 degrees, then 45 degrees, and then 60 degrees. Now here is where it usually gets a bit trickier because I'm going to move on to our radian angle measures, which can be a bit more difficult to remember because you're probably more familiar with degrees than radians. Now you could choose to use a formula to convert these degrees into radians, or you can kind of reason it out using the information you already have on your unit circle, which is what I'm going to do here. So here I'm going to start with 45 degrees. Now I know that 45 degrees is halfway between 90 degrees.

So that means that that radian angle measure needs to be halfway between zero and π2. And half of π2, if I multiply those together, that gives me a value of π4. So that radian angle measure for 45 degrees is π4. Then moving on to 30 degrees. 30 degrees I know is one third of 90 degrees, so I know that that radian angle measure has to be one third of π2.

Now multiplying that out that gives me a value of π6 which is what my radian angle measure is here, π6. Now moving on to 60 degrees. I know that 60 degrees is two times my 30 degree angle measure. So that means my radian angle measure is going to be two times π6. So multiplying π6 times two gives me two π6 or simplifying that fraction, π3.

So that radian angle measure for 60 degrees is π3. Now you can also choose to just memorize what these radian values are if that's more your style. So here we would want to think about what are the because all of the numerators are the same. They're all pi. So we would think about the denominators.

Okay. I have zero radians here. Then I have six, four, three, two. So you can kind of think about it as counting down. Six, four, three, two for your denominators.

Of course, skipping five there. So it can be a little bit tricky to memorize, but remember you can always reason it out as we did here. Now that we have all of those angle measures filled in, we can move on to really the meat of the unit circle, the cosine, sine, and tangent values, all of our trig values, or in this case also our x and our y values. So let's start there with our x and y values or our cosine and sine values. Now to do this, remember we're always going to start the same exact way with the square root over two for every single one of these values.

It's always the square root of something over two, not the square root of two, the square root of something over two. And I'm going to do this using the one, two, three method because that's the easiest for me to remember, but feel free to use the left-hand method or the left hand rule or whatever else works. So let's go ahead and get started here. Now remember with the one two three rule, we're going to start with this x value up here and count one two three clockwise and then go back one, two, three counterclockwise and those are our cosine and sine values. Now remember we can simplify a bit here because the square root of one is just one.

So for both of these instances of square root of one that just becomes one. Now that we have our cosine and sine values, we can go ahead and find our tangent. Now remember, the tangent of any angle, we can just use our information that we already know because the tangent of an angle is the sine of that angle divided by the cosine of that angle. Now, knowing what we know about our x and our y values, this is also just equal to y over x, which we can find by looking at our unit circle right here. Now, looking at this first value, 30 degrees, identifying the tangent of 30 degrees, I have one half and root three over two as my sine and my cosine values.

So I wanna take my sine one-half and divide it by my cosine root three over two. Now because these have the same exact denominator, I am effectively just dividing those numerators. So really, this just gives me a value of one over root three. Now rationalizing this denominator, this gives me a value of root three over three as that tangent. Now moving on to 45 degrees.

I have root two over two and root two over two, the same exact values for sine and for cosine.

Reference Angles on the Unit Circle

Hey, everyone. We now know all of our trig values of our three common angles in quadrant one, but the unit circle has three other quadrants, all of which we'll need to know trig values in. Now, this can seem really overwhelming at first because there are a ton more angles here that we haven't even looked at yet. But you don't have to worry because for all of these angles not in quadrant one, we can simply use what's called a reference angle. Every single one of these angles corresponds to an angle in quadrant one that we already know absolutely everything about, allowing us to find our trig values really quickly and easily for all of these other angles.

But before we even think about trig values, we first need to know how to even find a reference angle. So here, I'm going to show you how to do just that. Let's go ahead and jump right in here. Now like I said, for angles that are not in quadrant one, they all correspond back to angles that are in quadrant one. So every single one of these angles here has a reference angle of thirty, forty-five, or 60 degrees.

Now how exactly do we find that reference angle? Well, let's take a look here at our 150 degree angle. We know that this angle measures 150 degrees away from our zero degree measure there. But what if instead we were to measure this angle from the other side of the x-axis here at 180 degrees? Well, this angle formed with this side of the x-axis from 150 degrees to 180 degrees is only 30 degrees.

And if I were to draw this triangle out, we actually see that it forms the exact same triangle as our 30 degree angle measure in quadrant one just flipped over to the other side. So our 150 degree angle has a reference angle of 30 degrees. And this is how it's going to work for all of these angles. We're always going to measure them to the nearest part of the x-axis and write this angle measure as a positive number. So this angle formed here is a positive 30 degrees and that represents its reference angle.

Now let's take a look at our remaining angles in this quadrant two starting with 135 degrees. Now we want to measure this again to the nearest part of the x-axis which still happens to be this 180 degrees over here. Now from 180 to 135 degrees this measures 45 degrees away so it has a reference angle of 45 degrees, which again we can see that it forms the same exact triangle just flipped in the opposite direction. Now our final angle in quadrant two is this 120 degrees, which when I take a look at this angle and measure it to the nearest part of the x-axis, again still 180 degrees, this is 60 degrees away from 120 down to 180. So this tells me that 120 degrees has a reference angle of 60 degrees and again we can take a look at that triangle and verify that those are the exact same triangle just flipped in the other direction.

Now we want to take a look at our remaining two quadrants, quadrant three and quadrant four. And we're going to do the exact same thing here, measuring to the nearest part of the x-axis to determine that reference angle. Now remember, we're going to write these as a positive number because it doesn't matter what quadrant it is. This is always going to have a positive reference angle. So let's first take a look at two hundred and ten degrees and three hundred and thirty degrees.

Now for 210 degrees here, looking at its nearest x-axis, the nearest x-axis is still that 180 degrees and I see that it does measure 30 degrees away from that. So it has a reference angle of 30 degrees. Now looking at 330 degrees, its closest x-axis is coming around to a full rotation at 360 degrees, which is also 30 degrees away. So it too has a reference angle of 30 degrees. Now let's take a look at two more angles here.

We have 225 degrees and 315 degrees. Now again measuring to their nearest point on the x-axis here looking at two hundred and twenty-five and three hundred and fifteen. These are directly in the center angle-wise in those quadrants the same way that 45 degrees is. So when we look at that angle measure, we see that those are both 45 degrees away from their nearest x-axis. So they both have a reference angle of 45 degrees.

Now for our two remaining angles here, we have 240 degrees and 300 degrees. Now again measuring to the nearest part of the x-axis as we have every other time, we see that these are both 60 degrees away from their nearest x-axis respectively, so they each have a reference angle of 60 degrees. Now you might see that I've color-coded these here and this is also a way that's going to allow us to kind of remember what all of our reference angles are. So whenever I'm remembering reference angles, I like to think of them as forming an x. When I look at all of my 30-degree angles or my angles that have a reference angle of 30 degrees, they form a perfect x with that reference angle 30.

Now the same thing is true of our 45-degree angles. Looking here, all of these angles that have that reference angle of 45 form a perfect x. Now the same thing is true of all of my angles that have a reference angle of 60 forming a taller thinner x. Now the other way that you can think about this is by looking at these radian angle measures because all of the ones that have a corresponding reference angle have the exact same denominator. So all of our 30-degree reference angles have a denominator of six when we look at this here.

Then for 45 degrees, they all have a denominator of four when looking at that radian angle measure. And finally, all of my 60-degree reference angles have a denominator of three. Now that's a couple of different ways that you can choose to think about that. And now we know what we need to know about finding reference angles. So thanks for watching, and I'll see you in the next one.

Identify the reference angle of each given angle.
120°

• A. 30°
• B. 45°
• C. 60°

Answer: (C)

Identify the reference angle of each given angle.
$\frac{7\pi}{4}$ rad

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$

Answer: (B)

Identify the reference angle of each given angle.
210°

• A. 30°
• B. 45°
• C. 60°

Answer: (A)

Trig Values in Quadrants II, III, & IV

Hey, everyone. We now know all of the trig values of our three common angles in quadrant one. And we also know that all of our other angles have a reference angle equal to one of those common angles. Now all that's left to find is the trig values of all of these angles. And learning that many more new trig values does not sound like fun.

But luckily, you don't have to do that at all because the trig values of all of these angles are the exact same as those of their reference angle with just one tiny difference that I'm going to walk you through right now. So let's go ahead and get started. Now I mentioned one tiny difference and that difference is the sign. So the sine, cosine, and tangent of each of these angles is the exact same value as that of their reference angle, but just with a different sign depending on what quadrant our angle is located in. So let's go ahead and just come right down here and look at our unit circle.

So in quadrant one, we know all of these trig values, the cosine, sine, and tangent of each of these angles, and we see that being in quadrant one, these are all positive values. But what about as we move over to quadrant two? Let's specifically take a look at our 150-degree angle here. Now remember that this has a reference angle equal to 30 degrees because that's the angle it makes with the nearest X-axis. So with that reference angle of 30 degrees, we can further visualize this by seeing that they form the exact same triangle just flipped.

So it makes sense that the base and height of these triangles would be the exact same, which are cosine and sine values. So we can take the cosine, sine, and tangent here, our trig values, and just copy them over here because they're the exact same value. Now doing that, I end up with \( \frac{\sqrt{3}}{2} \) for that cosine or that X value, \( \frac{1}{2} \) for that Y value, that sine, and \( \frac{\sqrt{3}}{3} \) for tangent. Now the only thing that remains to consider here is the sign of each of these values because we're no longer located in quadrant one. Now looking at our relationship to the X and Y axes, I see that my X values over here should be negative, whereas my Y values remain positive.

So I want to reflect that in these trig values. So my X value is negative, \( -\frac{\sqrt{3}}{2} \). My Y value is positive. Now when looking at the tangent, we just consider how we actually find the tangent. Right?

So the tangent is the sine over the cosine or Y over X. So knowing that our Y value is positive and our X value is negative, if I were to divide a positive value by a negative value, that would give me a negative value. So here my tangent should also be negative. Now looking at these values altogether, only one of them is positive and this will be true for all of our trig values in quadrant two. Only the sine value, our Y value here, is going to be positive for all trig values in quadrant two.

Now let's move on to quadrant three and specifically 225 degrees. Now I know that the reference angle of 225 degrees is 45 degrees in quadrant one, so I can take my trig values here and simply copy them over here for the trig values of 225 degrees. So here I can copy these in with the cosine and sine both being \( \frac{\sqrt{2}}{2} \), my X and Y values, and then the tangent as one. Now all that we need to consider here is remember the sign. So based on where we are in the X and Y axes, we see that these both should be negative values.

My X and my Y value are both negative. Now thinking about the tangent, in order to find the tangent, I would be dividing a negative by a negative, which would actually give me a positive value. So my tangent here in quadrant three remains positive. Now looking at these values altogether, the only one that's positive is tangent, which will of course be true of all of my trig values in that quadrant three. So here in quadrant three, the tangent will always be positive and it's the only positive one.

Now finally, let's take a look over here at quadrant four. Now in quadrant four, we're going to look specifically at 300 degrees. Now we know that 300 degrees has a reference angle of 60. 60, so I can go ahead and take my 60-degree trig values and copy them right down here for the trig values of 300 degrees. Now doing that, my cosine and sine, I get \( \frac{1}{2} \) and \( \frac{\sqrt{3}}{2} \), literally just copying those values, then for the tangent \( \sqrt{3} \).

Now we need to consider of course the sign. Now the sign here looking in this quadrant, quadrant four, we know that we are in the negative Y values and positive X values so we need to reflect that on these trig values. So negative Y values, this negative \( \frac{\sqrt{3}}{2} \). And then for our tangent, we need to consider what we're doing. Of course, here we would be dividing a negative by a positive, which would give us a negative value.

So our tangent here will also be negative. Now finally looking at all three of these trig values as a whole, the only one that's positive is this cosine, this X value, \( \frac{1}{2} \). So here in quadrant four, the cosine will always be positive. Okay. We've looked at all of our quadrants here and we've seen the sign difference in each of them.

All of our trig values are the same, just with a little variation in sign depending on our location and on the unit circle. Now how can we remember this? We can remember this using a mnemonic device using the first letter of each function that's positive. So in quadrant one, all of them are positive. Quadrant two, the sine is positive.

Quadrant three, the tangent. And quadrant four, the cosine. ASTC. All students take calculus. Now this might not always be a true statement, but it's one that can help us remember which trig function is positive in our unit circle.

So now we have all of the information we need in order to completely fill in an entirely blank unit circle. So let's continue getting some more practice together. Thanks for watching, and I'll see you in the next one.

Trig Values in Quadrants II, III, & IV Example 2

Hey, everyone. In this problem, we're asked to use reference angles to complete the missing trig values in quadrants two, three, and four. Now that we know that all of these trig values are going to be the exact same as those in quadrant one, with just varying signs based on the quadrant, we can do this really quickly. So remember our mnemonic device here of how we remember which function is positive in each quadrant. "All students take calculus" tells us the first letter of each function that's positive within each quadrant.

Now let's first take a look at quadrant two. Looking at 135 degrees and 120 degrees. These have reference angles of 45 degrees and 60 degrees respectively, so we could just go ahead and copy those trig values over to this quadrant. For 135 degrees, I have 22, 22, and 1. Then for 120 degrees, I have a 12, 32, and the square root of three for that tangent.

Now in this quadrant, only the sine is positive based on that mnemonic device. So that means that both my cosine and tangent need to be negative here. So, my x values of both of these should be negative as well as this tangent here at the end. Now we're done with quadrant two. Let's move on to quadrant three down here.

Now looking at quadrant three, I have 210 degrees and 240 degrees as those missing trig values. Looking back up to quadrant one, I know that 210 degrees has a reference angle of 30 degrees, and 240 degrees has a reference angle of 60 degrees, so I can copy those trig values over. So for 210 degrees, I have 32 for my cosine, 12 for my sine, and 33 for that tangent. Then for 240 degrees or four pi over three radians, I have 12 for the cosine, 32 for the sine, and a 3 for that tangent. Now in this quadrant, only the tangent is positive, so that means both my sine and cosine need to be negative.

So x and y here are both negative when looking at these coordinates. Now finally, let's take a look at quadrant four over here. Now looking in this quadrant, my missing information is for 330 degrees and 315 degrees which have reference angles of 30 degrees and 45 degrees respectively. So let's go ahead and copy those trig values over here in this quadrant four. So for 330 degrees, I have 32, 12, and 33.

Then for that 315 degrees I have 22, 22, and finally 1 for that last tangent value. Now in quadrant four only the cosine is positive here, so that tells me that I need to go ahead and make both my sine and my tangent negative. So, my sine value and my tangent negative in both of these angles. Now we have completely filled in all of our information for the unit circle. Continue to practice this on your own and check-in with me as needed.

Thanks for watching, and I'll see you in the next one.

Identify what angle, $\theta$ , satisfies the following conditions.
$\sin\theta=\frac12$; $\tan\theta$ < $0$

• A. 30°
• B. 150°
• C. 60°
• D. 300°

Answer: (B)

Identify what angle, $\theta$ , satisfies the following conditions.
$\cos\theta=\frac{\sqrt3}{2}$; $\sin\theta$ < $0$

• A. 30°
• B. 60°
• C. 120°
• D. 330°

Answer: (D)

Identify what angle, $\theta$ , satisfies the following conditions.
$\tan\theta=-1$; $\cos\theta$ > $0$

• A. 45°
• B. 135°
• C. 315°
• D. 330°

Answer: (C)

Trig Values in Quadrants II, III, & IV Example 3

In this problem, we're given a completely blank unit circle and asked to fill in all of the missing information. Now this is a specific type of problem that you'll get better at with repetition, and it's definitely something that you should try on your own before jumping back in with me. Now here, I'm going to walk you through my thought process in filling in the unit circle and what works best for me might not be what works best for you. So, of course, try this on your own and figure out exactly what method works best for you for filling in the entire unit circle. Let's go ahead and get started here.

Looking at our unit circle, the first thing that I like to do is fill in all of the information for my four quadrantal angles because to me it's the simplest information on here. Looking at this first angle measure, I have zero degrees or zero radians, and I'm going to go ahead and fill in all of my angle measures for my three other quadrantal angles. Up here, I have 90 degrees, 180 degrees, 270 degrees, and remember, coming back around to a full rotation is 360 degrees. Often, a blank unit circle will only have one blank there, but it's important to remember that this is also a full rotation around. Let's also fill in those radian angle measures.

We have π/2 radians, π radians, and 3π/2 radians. Now we have all those angle measures filled in, in both degrees and radians, so let's consider our trig values of our quadrantal angles as well. Remember, on the coordinate system, we know that this point is located right here at (1, 0), and then we find our tangent value by simply dividing y over x. So doing that here, I have 0 over 1, which gives me a value of 0. Then up here for 90 degrees, this is located at the point (0,1), and dividing y over x here gives me 1 over 0, an undefined value.

Now over here with π radians, we're at (-1,0), and we again get zero for our tangent, and then down here at 270 degrees, we're at (0,-1), which is also an undefined tangent value. Okay, we have all of the information for our four quadrantal angles, so now let's focus on all of the other angle measures for all of our other quadrants. Let's start with quadrant one here because these are likely the easiest for you to remember, as they are for me. So here, we're looking at our three common angles. We have 30 degrees, 45 degrees, and 60 degrees.

I'm also going to go ahead and fill in my radian angle measures here. In this first quadrant, we have π/6, π/4, and π/3. Now here, let's go ahead and focus on all of our reference angles. All of these angles here have a reference angle of 30, so that's going to really help me to figure out what these angle measures are. Let's start here in quadrant two.

In quadrant two, I look at this reference angle in reference to the nearest part of the x-axis, which is at 180 degrees. So this is 30 degrees away from that 180, which tells me I can find this angle measure by simply taking 180 - 30 degrees. Now this gives me an angle measure of 150 degrees. Now for that radian angle measure, I can do the same exact thing but just with π. This is π/6 radians away, so I take π minus π/6 here, and that gives me my radian angle measure of 5π/6.

This will work for any angle in this quadrant because we can just do that for any angle. We just subtract our angle from 180 and from π. Now let's look in our quadrant three here and do the same exact thing. This is in the opposite direction of 180, so here I can actually take 180 and add my angle to it instead to get that angle measure. This is 30 degrees away because we're still looking at our 30-degree reference angle here.

If I take 180 and I add 30 degrees, that's going to give me an angle measure of 210 degrees. I can do the same thing with π, of course. I add π to my angle here, which is π/6, so π plus π/6 gives me an angle measure of 7π/6. Now let's move on to quadrant four. We look at this in reference to our final angle, our full rotation of 360 degrees, and this is, of course, still a 30-degree reference angle, so here I can take 360 degrees and I can subtract my angle here.

So 360 minus theta, my angle. Here, this is 30 degrees, so 360 minus 30 gives me 330 degrees. Of course, this is also 2π radians, so I can also take 2π and subtract my angle in radians to get the same radian angle measure here. So here, 2π minus π/6 will give me 11π/6. Looking at all of these radian angle measures, you can also choose to memorize these sorts of patterns that happen with your radian angle measures.

So, looking at all of these, I have π/6, 5π/6, 7π/6, and 11π/6. For angles that have the same reference angle, they always have the same denominator in that radian angle measure. These all have a denominator of six, and they count from one, five, seven, 11. You can remember that pattern if it's easier for you to memorize here. Now here, looking at this, I can apply this same line of thinking to all of my other reference angles.

So I want you to go ahead and try that on your own and then check back in with me. Okay. Now that we have all of those angle measures in there, we can move on to trig values. Let's start in quadrant one because we have some memory tools that will help us fill this in rather easily. Remember, in quadrant one, we always start the same way regardless of how we choose to memorize this.

We always start with the square root of two for every single one of these values. So I can go ahead and fill that in for every trig value in quadrant one. Now once I do that, I can go ahead and use a memory tool. Here, I'm going to use the one, two, three memory tool, but of course, you can use whatever works for you. So here, using the one, two, three memory tool, I'm going to start here with this value and I'm going to count one, two, three in the clockwise direction for those x-values, and then one, two, three back in the counterclockwise direction for those y-values.

So these are my trig values in quadrant one. We can, of course, simplify these square roots of one because they're just one. Now from here, we want to fill in our tangent values. You can choose to memorize your tangent values, but it's easier for me just to remember that this is y over x. So I can take these values that I already have and simply divide them.

Now because they have the same denominator, we're effectively just dividing those numerators. So here, in this first one, I divide my numerators of y over x, one over the square root of three. This is a correct value, but remember, it's going to be better if we rationalize our denominator here. So this will give me the square root of three over three for the tangent of 30 degrees. For 45 degrees, if I take y over x, those numerators, I get a value of one.

And then for 60 degrees, I get the square root of three over one, which is just the square root of three. Now we have all of our trig values in quadrant one, but we have these three other quadrants. Remember, all of our trig values are the exact same as those of their reference angles, and we only have to worry about the sign. So from here, I'm just going to go ahead and fill in all of these trig values, copying them over from quadrant one to their respective reference angle in every quadrant, and then we can worry about the sign after. I'm going to go ahead and fill all of those values in now.

All of our trig values are copied over to quadrants two, three, and four. But remember, the sign of these values may be different depending on what quadrant they're in. That's when we consider the sign. Remember a mnemonic device to remember the signs of these values. "All students take calculus" telling us the first letter of whatever trig function is positive in that quadrant. So in quadrant two, we have s. That tells us that only the sine is positive. So only my y-values here, root three over two, square root of two over two, and one-half should be positive. So all of my other values are negative, and I can go ahead and fill that in for all of my other values, putting a negative sign on each of them.

We can do the same exact thing in quadrants three and four. In quadrant three, only the tangent is positive, and in quadrant four, only the cosine is positive. So all of my other trig values must be negative. I'm going to go ahead and copy those negative signs in now based on that, and then we'll rejoin together. Okay.

You should have all of your trig values filled in now with negative signs on the correct values based on our mnemonic device, "all students take calculus." Take a chance to double-check all of your values here with mine. Now we have completely filled in all of the missing information on our unit circle. Again, this is a problem that gets easier with repetition, so feel free to try this as many times as you need until you have it down. Thanks for watching.

And of course, let me know if you have questions.

Coterminal Angles on the Unit Circle

Hey everyone! If I gave you a unit circle and I asked you to find the sine of 30 degrees, you'd probably locate 30 degrees on your unit circle and simply identify your y value. But what if instead I asked you to find the sine of 390 degrees? It doesn't sound quite so simple. But if we look at our circle here and we go a full rotation around from zero to 360 degrees and then just go an additional 30 degrees, I'm actually right here at 390 degrees.

So the trig values are going to be the exact same as those of 30 degrees, an angle that I already know. And that's because these angles are coterminal, so their trig values are going to be the exact same. Now the first time that you see an angle that's either really large or maybe even negative, it might seem a little bit strange. But here I'm going to walk you through exactly how to identify a coterminal angle on the unit circle and use that to really quickly and easily identify your trig values. So let's go ahead and get started here.

Now you might remember coterminal angles more formally as an angle with the same terminal side. And you might also remember finding them by adding or subtracting multiples of 360 degrees or two pi radians. So with my 390 degree angle, if I were to subtract a full rotation around a circle of 360 degrees, I would end up with my coterminal angle of 30 degrees. Now we're gonna do basically the same thing here, but it's gonna be even easier to visualize on our unit circle because we already have all of these angles here. So we can do this without having to set up any algebraic equation and just visually finding these angles.

So let's go ahead and jump right into this first example and find the tangent of three pi. Now the tangent of three pi. The first thing we want to do here is find the coterminal angle of three pi so that we can easily find our trig value that's already on our unit circle. So let's consider this by looking at our unit circle. If I go around my circle once from zero to two pi radians and then I go another additional pi radians, I've really gone three pi radians around my circle.

But I've ended right back here at pi, which is my coterminal angle. So the coterminal angle of three pi is simply pi. So to find the tangent, I just need to look at the tangent value of pi here, which happens to be zero. So the tangent of three pi is equal to zero and we're done here. Now let's move on to our next example here.

Here we're asked to find the cosine of negative pi over four. Now remember for a negative angle, that just tells us that we're going clockwise around our circle rather than counterclockwise. So let's again visualize this on our unit circle here. Now if I start at zero and I go negative pi over four radians, so this way, clockwise, negative pi over four, I'm gonna end up here at seven pi over four radians. So the coterminal angle of negative pi over four is seven pi over four.

Now that I know that, I can identify the cosine by just finding the cosine of that coterminal angle seven pi over four, which is of course my x value here. So the cosine of negative pi over four is 22 and I'm good to go here. Now we have one final example, the sine of 390 degrees. So we want to go ahead and identify this on our unit circle. Now you might remember this from earlier but let's just visualize this one more time.

So 390 degrees. If I go around my circle once, that's 360 degrees. And if I go another 30 degrees, that puts me at 390 degrees. So the coterminal angle here is, of course, 30 degrees. Now to find the sine of this angle we're just going to look at the sine of 30 degrees which is of course our y value in this case one half.

So the sine of 390 degrees is one half. Now that we know how to find trig values using coterminal angles, let's get some more practice together. Thanks for watching and let me know if you have questions.

For each expression, identify which coterminal angle to use & determine the exact value of the expression.
$\sin\frac{7\pi}{3}$

• A. $\frac12$
• B. $2$
• C. $\frac{\sqrt3}{2}$
• D. $\frac{2\sqrt3}{3}$

Answer: (C)

For each expression, identify which coterminal angle to use & determine the exact value of the expression.
$\tan765\degree$

• A. $-1$
• B. $1$
• C. $0$
• D. undefined

Answer: (B)

For each expression, identify which coterminal angle to use & determine the exact value of the expression.
$\cos\left(-\frac{10\pi}{4}\right)$

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{\sqrt2}{2}$

Answer: (A)

Secant, Cosecant, & Cotangent on the Unit Circle

Hey, everyone. Up to this point, we've been working with three trig functions, sine, cosine, and tangent, on our unit circle. But remember, there are actually three other trig functions, our reciprocal trig functions, cosecant, secant, and cotangent. Now you may be worried that we're going to have to add a bunch of new information on our unit circle to find these reciprocal trig functions, but you don't have to worry about that at all because we can find the values of cosecant, secant, and cotangent using what's already on the unit circle. So let's go ahead and jump right in here.

Now these functions, remember, are called the reciprocal trig functions because they are simply one over our three original trig functions, sine, cosine, and tangent. So for the cosecant of an angle, this is really just one over the sine of that angle. And on the unit circle, since we know that the sine of an angle is equal to that y value, the cosecant of the angle is just one over y. Then for the secant, since this is one over the cosine of that angle and we know that the cosine of any angle on our unit circle is just that x value, the secant of that angle is 1x. So looking at our unit circle, if I asked you to find the secant and cosecant of 60 degrees, you could simply take one over that x value and one over that y value and you're done.

Now let's not forget the tangent here because we also have the cotangent as one of our reciprocal trig functions. And when we find the cotangent of an angle, this is just one over the tangent. Now we know the tangent of an angle to be y over x on our unit circle, so we can simply flip that fraction to get the reciprocal telling us that the cotangent of an angle is simply xy. So again, if I asked you to find the cotangent of 60 degrees, you could go over to your unit circle, take your x value, divide it by your y value, and you'd be good to go. Now with that in mind, let's go ahead and work through some examples together.

So our first example here is the cosecant of pi over six. Now we know that the cosecant is one over the sine of that angle so this is one over the sine of pi over six which we also know to be the y value. So going up to our unit circle here at pi over six, looking at that y value of one half here for the cosecant of pi over six, I can take one over one half. Now whenever I take the reciprocal of a fraction, I'm really just flipping that fraction. So this gives me a value of two.

So the cosecant of pi over six is equal to two. Now let's look at the cotangent here. Now here we're asked to find the cotangent of pi over four. Now we know that the cotangent of any angle is just our x value over our y value. So here we think about that again, xy.

Now we can go up to our unit circle at pi over four, look at those x and y values and simply divide them. So here we have the square root of two over two divided by the square root of two over two. Now because these are the exact same value, we end up with an answer of one. The cotangent of pi over four is just one, which just so happens to be the exact same as the tangent of pi over four because those x and y values are the same. Now let's look at one final example here.

Here we have the secant of zero. Now the secant of zero, I know that this is just one over the x value at zero degrees, so I can go up here to my unit circle at zero degrees, identify that x value, and plug that in. So this is just going to be one over one which is of course just equal to one. So the secant of zero is just one, and we're done here. Now that we know how to find these reciprocal trig values from all of the information we already have on the unit circle, let's get some more practice.

Thanks for watching, and let me know if you have questions.

Evaluate each expression.
$\cot\left(\frac{11\pi}{6}\right)$

• A. $\frac12$
• B. $-\frac{\sqrt3}{3}$
• C. $-\sqrt3$
• D. $2$

Answer: (C)

Evaluate each expression.
$\csc225\degree$

• A. $1$
• B. $-\frac{\sqrt2}{2}$
• C. $\sqrt2$
• D. $-\sqrt2$

Answer: (D)

Evaluate each expression.
$\sec\left(\frac{\pi}{3}\right)$

• A. $\frac12$
• B. $2$
• C. $\frac{\sqrt3}{2}$
• D. $\frac{2\sqrt3}{2}$

Answer: (B)

Here’s what students ask on this topic:

What is the unit circle and why is it important in trigonometry?

The unit circle is a circle with a radius of 1, centered at the origin (0, 0) on a coordinate plane. It is crucial in trigonometry because it provides a geometric framework for understanding the relationships between angles and trigonometric functions. Each angle on the unit circle corresponds to a specific point (x, y), where x represents the cosine of the angle and y represents the sine. The unit circle also helps in defining trigonometric functions for all angles, including those greater than 90° or negative angles. Additionally, it simplifies calculations involving radians, reference angles, and periodicity of trig functions. The equation of the unit circle is:

$x^2+y^2=1$

.

How do you find the sine, cosine, and tangent of an angle using the unit circle?

To find the sine, cosine, and tangent of an angle using the unit circle, follow these steps:

1. Locate the angle on the unit circle. If the angle is in degrees or radians, find its corresponding point (x, y) on the circle.

2. The cosine of the angle is the x-coordinate of the point, and the sine is the y-coordinate.

3. To find the tangent, divide the sine (y) by the cosine (x):

$\tan \left(\theta \right)=\frac{y}{x}$

.

For example, at 45° (or π/4 radians), the point is (√2/2, √2/2). Thus, cos(45°) = √2/2, sin(45°) = √2/2, and tan(45°) = 1.

What are reference angles, and how are they used in the unit circle?

A reference angle is the acute angle formed between the terminal side of a given angle and the x-axis. It is always positive and lies between 0° and 90° (or 0 and π/2 radians). Reference angles are used to simplify the calculation of trigonometric values for angles outside the first quadrant by relating them to their corresponding first-quadrant angles.

To find a reference angle:

- For angles in Quadrant II, subtract the angle from 180° (or π).

- For angles in Quadrant III, subtract 180° (or π) from the angle.

- For angles in Quadrant IV, subtract the angle from 360° (or 2π).

For example, the reference angle for 150° is 30° because 180° - 150° = 30°. The trigonometric values of 150° are the same as those of 30°, but with appropriate signs based on the quadrant.

How do you determine the signs of trigonometric functions in different quadrants?

The signs of trigonometric functions depend on the quadrant in which the angle lies. Use the mnemonic "All Students Take Calculus" to remember:

- Quadrant I: All trigonometric functions (sine, cosine, tangent) are positive.

- Quadrant II: Only sine is positive; cosine and tangent are negative.

- Quadrant III: Only tangent is positive; sine and cosine are negative.

- Quadrant IV: Only cosine is positive; sine and tangent are negative.

This rule is based on the signs of the x and y coordinates of points on the unit circle. For example, in Quadrant II, x is negative, and y is positive, so cosine (x) is negative, sine (y) is positive, and tangent (y/x) is negative.

What are reciprocal trigonometric functions, and how are they calculated on the unit circle?

Reciprocal trigonometric functions are derived from the basic trig functions (sine, cosine, tangent) and are defined as follows:

- Cosecant (csc): Reciprocal of sine, csc(θ) = 1/sin(θ) = 1/y.

- Secant (sec): Reciprocal of cosine, sec(θ) = 1/cos(θ) = 1/x.

- Cotangent (cot): Reciprocal of tangent, cot(θ) = 1/tan(θ) = x/y.

To calculate these on the unit circle, locate the angle and its corresponding point (x, y). Use the x and y values to compute the reciprocal functions. For example, at 30° (or π/6 radians), the point is (√3/2, 1/2). Thus, csc(30°) = 2, sec(30°) = 2/√3, and cot(30°) = √3.