 As we've seen earlier, the unit circle provides geometric representations for both sine of theta and cosine of theta. That is to say cosine is just the x-coordinate of points on the unit circle and sine is just the y-coordinate of points on the unit circle. Do the other trigonometric functions have similar representations, geometric representations on the unit circle? The answer is yes, and I want to explain those in this video. So consider this diagram you see on the screen right here. So this in yellow is going to be our unit circle, and we have some angle theta in play right here. And just for the sake of argument, allow it to terminate in the first quadrant, although similar diagrams could also be constructed for the second, third, and fourth quadrants. So if we take the point on the unit circle for which the terminal side intersects the unit circle, you're going to get this point P equals x, y. And to make it a little bit easier, I'm going to zoom into this picture and focus mostly just what's happening in the first quadrant. Again, just to make it a little bit easier for the viewer to see here. So we have this point P whose coordinates are x and y for where x coincides with cosine and y coincides with sine right here. So associated to this point P is the right triangle given right here. Let's call the origin O that's going to be the center of the circle. It's also one of the vertices of this right triangle. You're going to have vertex A, which is where the right angle coincides. And then this point P right here. So there's the circle A or excuse me, the triangle AOP, which is a right triangle, which you have a one side whose length is cosine theta. One side is sine theta. And as it's a it's a triangle in the unit circle, the hypotenuse here is just a radius. It's one. So you get this triangle here AOP. And so I'm going to kind of keep this track on the side over here, kind of pulling it out of the picture to kind of avoid the clutter. You have angle theta, you have OAP, which again, the hypotenuse is one. The opposite side is sine and then the adjacent side is cosine. This we know from our previous unit circle discussion. All right. So what I want to do next is consider the following triangle you see right here in the diagram. This would be the triangle BOR. All right. So where does that come from? Well, it'll have as one of its vertices again, the center of the circle, the origin O, take one of the sides of the triangle to be a radius of the of the unit circle. This is the radius that coincides with the positive X axis. And then if we take the line perpendicular to this point. So B is going to be the point one zero. This is the point on the positive X axis, which is on the radius of the circle right there. Take the line that's perpendicular to the X axis coming out of B. Eventually that point will intersect the terminal side of this angle theta, call it point of intersection R. In which case then we are going to get this much larger triangle. But if we label all the appropriate parts, you get theta, you have O, you have the point B with coincides to the right angle. And then you have this other point R right here. What do we know about this triangle? Well, we do know that the side length OB, I guess you'd call BO if you prefer, that's going to be one because it's a radius of the unit circle. We also know that these two triangles, the triangle AOP and the triangle BOR, these triangles are similar to each other. Notice how they're both right triangles. So they both have a 90 degree angle and they both overlap at the angle theta. So these angles right here, the angle P and the angle R are going to be compliments to theta and thus their congruence to each other. So you get these two triangles are similar to each other. So because they're similar triangles, we can set up ratios to compare things. So if we take the distance BR and we put it over OB like so. So notice that BR is this side right here. It's unknown. This will coincide with PA and then the side OB, which is the adjacent side. This will coincide with the side OA like so. So because the triangles are similar, we get the following. But what do we know about this? OB is a radius of the unit circle. So it's equal to one. So we're going to get BR over one. OA is this side right here. It's the adjacent side, which we already know is cosine. This could be the X coordinate for the unit circle. That's just going to be cosine of theta. And then PA is just the opposite side here of the unit circle triangle. And so that's going to be a sign right here. So simplifying this, we're going to get that BR is equal to sine over cosine. This is equal to tangent theta. And so following this similar triangle argument, we get that this distance BR is equal to tangent. Let me clean up the screen a little bit here and label will be just discovered. So this side right here is tangent of theta. And so you already see that labeled on the diagram right here. The line segment right here is of length tangent. Let me give you a little bit explanation of where the name tangent comes from. When it comes to a circle, if you have a line that intersects the circle in just one point, that is the line just kisses the circle right there, just intersects at one point, we call this a tangent line. And so you'll notice that the line determined by the points B and R, even though we didn't draw the whole line if we continued it on, this is a tangent line. It touches the circle only at the point B. And so the trigonometric ratio tangent got its name because tangent, the tangent of theta represents the length of a segment on a tangent line to the circle. So the idea of a tangent line actually existed prior to trigonometry as we know it today. And the name tangent came about because it's measuring the length of a tangent line to the circle. Okay, continuing with this similarity right here, I want to show that the hypotenuse of this triangle is going to be secant. Let's again do a similar argument, a similar triangle type argument. Take the line segment OR and divide it by OB. Alright, OR is the hypotenuse, so it'll coincide with OP. And then OB is the adjacent side, so it'll coincide with OA. So by similar triangles, we get that right there. But we know some of these things. OB is still length one because it's the radius of the unit circle. So we get OR is equal to OP. Well, OP is also the radius of the unit circle, so it's a 1, and OA is going to be cosine. So we get 1 over cosine, which of course is equal to secant theta, like so. So I'm going to erase this stuff that's on the screen right now and then summarize what we discovered. The line segment OR is equal to secant of theta. And so this distance right here, the distance from O to R, is equal to secant of the angle theta. Now let me give you a little bit of an explanation of where secant came from. I'm going to have to zoom out a little bit to do this correctly. A tangent line was a line that intersects the circle at exactly one point, but a secant line is a line that intersects the circle at two distinct points right here. Notice the line determined by O and R. It has two points of intersection with the circle. One of those is P, and one of those is on the other side of the circle. We didn't give it a name, but this line OR is an example of a secant line intersects the circle at two different points. And therefore the trigonometric ratio secant of theta got the name secant because it's measuring the line segment of an associated secant line of the unit circle. So tangent was named after a tangent line, secant was named after a secant line. The other two, cotangent and cosecant, they'll get similar considerations, but it comes from taking complements here. So we're going to have to set up a different similar triangle here, but we can do as much. We're going to show that the triangle AOP is similar to the triangle. Make sure I get this one correct here. We're going to get the triangle QOC, which if you haven't noticed, that's this triangle right here on our diagram, QOC. So let's consider what's going on here. So we have a line segment right here. So C is giving me the point 01. So this is one radius that the radius of the unit circle that coincides with the positive Y axis. The point of intersection between the positive Y axis and the unit circle will be the point 01, call that point C. Let's take the perpendicular that comes out, the perpendicular line that comes out of this point C that's perpendicular to the Y axis, that is. This line will eventually intersect the terminal side of our angle theta, call that point of intersection Q. And so we get this line segment right here, the CQ line segment. And then these three points CQ in the center of the circle, the origin form a triangle. And I claim this triangle is similar to the original triangle AOP. And the way we're going to see that is actually using the alternate interior angle theorem. This line segment right here, this line CQ is parallel to the line segment OA. And the reason we know that is that these are both horizontal lines. They're both perpendicular to the Y axis. So these are parallel lines. And therefore the terminal side of our angle transverses these two parallel lines. So by the alternate interior angle theorem, this angle Q right here is in fact congruent to angle theta down here. And so smoothing things around, you're going to see, I'm going to do this in a different color to emphasize it here. If you take this triangle and rotate it like this, you're going to get angle Q, which has, excuse me, vertex Q, which has angle theta. You have C, which is the right angle, and then O is over here. So again, if you rotate this triangle over here, that's what our diagram looks like. What do we know about this triangle? We know that the side CO is the radius of the unit circle. So that distance is one. We want to show that the other sides are the complementary trigonometric functions, cotangent and cosecant. So we have the similarity now established. All right. So what are the comparisons we want to do? So take the segment CQ and divide it by OC. So you can kind of see what we're always doing here. We're taking an unknown side of the triangle and we're always dividing it by the one side we do know, which is length one. It's a radius of the unit circle. So we're always going to take an unknown side like CQ and divide it by the known side OC like so. And then we're going to compare it, right? CQ, this is the adjacent side. So it's going to compare it to OA. But OA, getting ahead of ourselves, is cosine of theta. Then OC, OC is the adjacent side, excuse me, the opposite side. It coincides with PA, Pennsylvania. But we know that's equal to sine. And if you put that together, cosine over sine, this is the same thing as cotangent. And so, but also OC is just one. So we get that CQ is equal to cotangent and that's this distance right here. So erasing these for a moment, we get that QC is the same thing as cotangent of theta, which you'll notice that the line QC is also a tangent line to the circle. Although you could say it's complementary to the tangent one we had before. This was a vertical line. This is a horizontal line. So that's where it gets to then cotangent. It's complementary to the tangent, but it also is measuring the distance of a tangent line associated to the unit circle here. Now for the last one, we need to know what this side is right here. So we would take OQ, right? Divided by OC, because we know OC is one, because it's a radius of unit circle. OQ is the hypotenuse. So it's going to coincide with OP, which is equal to one. And then OC, like we saw before, it coincides with the opposite side of PA, which was equal to sine theta. And so we see this equals to cosecant theta, just like we expected it to. So we get cosecant theta right here. And so moving these things down, we can summarize the six trigonometric ratios. Oh, can I get the whole picture in there? If I zoom out, I can do that. So we get that the six trigonometric ratios represent sides of triangles associated to various parts of this diagram, involving secant lines and tangent lines and triangles. So this discussion gives us geometric interpretations of tangent, cotangent, secant, and cosecant, representing that they're distances related to associated tangent lines and secant lines. And that explains the name tangent and secant. Cotangent is the complement of tangent and cosecant is the complement of secant. So those all make sense now, given this geometric interpretation. Well, cosine is clearly the complement of sine. So it begs the question, where did the name sine come from? The geometric interpretation, we never talked about anything called a sine, other than this meaning of sine. Well, it's kind of a funny little history. It turns out that the English word sine is actually a transliteration of the Latin word sinus. Which sinus would be translated as a curved line, something that concave curves, something like that. And when you think of the graph of a sine wave, I guess that kind of makes sense. A sine wave is kind of curvy, but it turns out this Latin word is actually mistranslation of an Arabic word, which I'll apologize if not pronouncing this correctly, jiba, which jiba, and this translation miss up is due to Robert of Chester. But I'll give Robert some credit here. Jiba is not actually an Arabic word. This is actually a transliteration of a Sanskrit word of jaw. Again, I don't claim that's the correct pronunciation there. For which jaw in Sanskrit, that is to the Indian mathematicians, this would represent a half cord, which a cord is kind of like a secant line. A secant line is a line that intersects a circle and two points. A cord is a line segment that intersects a circle and two different points. And so if you think of the line sine right here, sine is half of a cord, the other half being down here. And then cosine is also half of a cord in a complementary manner here. And so if you go back to the original Sanskrit, sine has that same name. It's a half cord. But because of some various translation complications, we get this artificial term sine in the English language. Kind of funny, huh?