Trigonometric Functions on the Unit Circle - Video Tutorials & Practice Problems

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 0.10. 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 not 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 going to 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 3/5 4 5th on that unit. Now, if I trace my xmiy 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 ask you to find the sign of 53 degrees and the cosine of 53 degrees, using this triangle, you would actually find that the sign 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 trick functions as going in alphabetical order. I know that C comes before S in the alphabet. So cosine comes before S here, my cosine is my X value and my sign 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 sign 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 s and our cosine correspond to our Y 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 quadrant 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 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 217 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, OK, X Y CS telling me which is my cosine and which is my sign. So looking at this sign of 217 degrees, I know that that's my Y value. So the sign of 217 degrees is negative three fits. 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 cosign here is our X value, remember negative four fits 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 217 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 3/5 and divide it by a negative 4/5. Now whenever we divide fractions, we're just multiplying by the reciprocal. So this would be negative 3/5 times negative five fourth, the reciprocal here. Now those fives are going to cancel and the negatives will cancel as well leaving me with an answer of 3/4. 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 270 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, that triangle if it makes you happy and then we can go ahead and identify our trade 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 sign value is that Y route 3/2. 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 root 3/1, 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. $\sin\theta=\frac{5}{13},\cos\theta=\frac{-12}{13},\tan\theta=\frac{5}{12}$

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. $\sin\theta=\frac{\sqrt2}{2},\cos\theta=\frac{\sqrt2}{2},\tan\theta=\frac{\sqrt2}{1}$