 In a previous video we talked about 3060 90 triangle which allowed us to compute sine and cosine for 30 and 60 degrees in a Separate video. We were able to talk about an esosceles right triangle That is a 45 45 90 degree triangle which allowed us to Compute sine and cosine of 45 degrees it turns out there are two other very Special very important angles we to talk about that is a zero degree angle and a 90 degree angle How do you compute sine and cosine of those things right there? Well, if we try to think about in terms of triangles we might get a picture that looks something like the following So we have a right triangle right? And again, it should be a right triangle so we'll mark it up right here It's a right triangle Can we make a right triangle for which one of the angles is zero degrees? That seems kind of like a curious thing for a moment But just let's just go with the thought experiment just for the moment here What would have to be true if that was the case you had a zero degree angle well The total sum of all the angles has to be 180 degrees and the right angle is 90 degrees and we have zero degrees So it's compliment would have to be 90 degrees So we have a triangle which has two right angles. That's kind of bizarre. This isn't legitimately a triangle here This would be a degenerate triangle, but just again follow the thought experiment for a little bit longer here Let's say that the The length on the bottom here is just one make a standard unit here Well, if this is a zero-degree angle, there's really no space between these things This would be just collapsed on top of each other, which means this distance here would be zero In which case then if we drew this more properly, right? You have this point right here We have point a point B and point C like so well in that case then really if we drew this to scale You'd have line segment a B right here But then because of a zero-degree angle you have to draw the line segment again a to C And they'd overlap each other because again, there's no distance between B and C right there So the hypotenuse is legitimately the same distance again one and that does match up with the Pythagorean equation Right notice that one squared plus zero squared is in fact equal to one squared because both the left-hand side on the right-hand side equals one So okay, I guess I can live with that a little bit now if this is if this were a triangle Then cosine and sine can be computed in the following right way, right with respect to a zero-degree angle Coasts a sign. Let's do that one first sign would be opposite over hypotenuse You get zero over one which then gives you zero so sign of zero degrees ought to be zero by this suggestive Diagram right here on the other hand if we want to compute cosine of zero degrees We should take the adjacent side divided by the hypotenuse So one over one so we think that cosine of zero degrees should be one And then tangent is gonna be then the ratio of these two things Sign divided by cosine would be zero over one in which case we then get zero so tangent of zero ought to be zero What if we change our perspective now? What if we look at angle C? Right, and we consider with respect to this 90-degree angle Well sign should then be opposite over hypotenuse one over one that is sign of 90 degrees Equals one and then cosine would be adjacent over hypotenuse zero over one So we see that cosine in this situation is now zero. What about tangent though? Well tangent should equal sign of 90 degrees divided by cosine of 90 degrees and by our observations we end up with In this case one over zero when you divide it by zero and that's not a real number right there This is undefined so tangent of 90 degrees ought to be undefined if we accept this sort of fake Zero-degree triangle business right here Well, the good news is if you didn't feel comfortable with this thought experiment with triangles our original definition of the trigonometric ratio It isn't based upon triangles. It's it's based upon points, right? So we really can be thinking of it as we have the x-axis right here We have the y-axis right here and then generally speaking We just take a point in space right there a point in the plane and we then form the angle Between the origin and this point that is we take the ray that emanates from the origin to this point and then the angle in question Form between the positive x-axis. This is our angle theta And so if this point has coordinates x comma y and the distance between the origin and this point is r then we can define the The trigonometric ratio sine cosine tangent seeking cosecant and cotangent using these numbers x y and r So the trigonometric ratios are defined from points. It's just given this angle Generally speaking we can form a right triangle for which angle theta is one of the interior angles to the triangle There of course are some exceptions to this, right? I mean if you think about this for example in the second quadrant Right, you have an angle over here, and then your angle theta Would be something like this that you can't make a right triangle with that so While the right triangle definition makes sense when your angles are in the first quadrant that is their acute They're between 0 and 90 degrees these extremal values that is the angles 0 degrees and 90 degrees kind of lead to this funky sounding Degenerate triangle and then when you start getting the larger quadrants Things start to not make any sense with your right triangles anymore. That's not exactly true This is what we're leading up to with our idea of a reference angle turns out that given any any angle We can always form a right triangle with the exception of the Quadrantal angles that is 0 degrees 90 degrees 180 etc But again with respect to points we can make all we can make sense out of all these things So what happens when the point is in this case 1 0? All right, so let's move our line right here. Let's put our point to be 1 0 1 0 right there So then the ray emanating from the origin going through 1 0 be that line right there And then the distance between the origin and the point would it's be 1 and so using r equals 1 x equals 1 and Y equals 0 then we get that sign is going to equal y over r which is 0 over 1 which is 0 We get that cosine should equal x over r which is 1 over 1 which equals 1 and then we get that tangent should equal Y over x which equals 0 over 1 which gives a 0 right there Similarly, we could do the same game for a 90-degree angle using the point 0 1 in that case 0 1 would be this point right here Take the ray emanating from the origin through that point like so Then the angle form between the positive x-axis and this ray would be 90 degrees In which case then again r is going to be 1 once again x is this time 0 and y equals 1 then we can do all the ratios all over again sign is Y over r so we get 1 over 1 which is 1 cosine is x over r Which equals 0 over 1 which is 0 and then tangent again as we saw would be y over x Which gives you 1 over 0 which is undefined does not exist So you don't need to have a right triangle to define sign cosine and tangent or the other three trigonometric ratios as well It's this idea of having a terminal point that determines the trigonometric ratios But we can connect it to this idea of triangles if we sort of consider these degenerate triangle situations like we have right here a flat triangle So if you take the trigonometry of zero degrees and 90 degrees which we've discussed in this video and we combined it with the trigonometry of 30 60 and 45-degree angles that we that we considered in previous videos we get our so-called special angles These are angles of extreme importance and the typical trigonometric student will do themselves a huge favor by memorizing this table So if we need to memorize what sign of zero degrees 30 degrees 45 degrees 60 degrees 90 degrees What's cosine of those things and then what's tangent of those things? So I wrote the sine column or the sine row excuse me right here out But I wrote it out an unsimplified format We saw on the previous slide that sign of zero degrees is zero well I'm gonna write that as unsimplified here. I'm gonna get the square root of zero over two What why that notice the square root of square root of excuse me zero over two This is the same thing as zero over two, which is the same thing as zero So it's right and then we're gonna think of sine of 30 degrees as the square root of one over two Notice the square root of one over two is the same thing as one half, which is what we saw in a previous video The square root of two over two This is exactly what we saw when we talked about sine of 45 degrees Coast our sine of six degrees. We also saw was root three over two. No worries there, but then Sine of 90 degrees. I'm gonna write this on the table as a square root of four over two Why is that we'll notice square root of four over two square root of four is two two over two is equal to one Oh sine of 90 degrees is equal to one. We saw that's the moment ago So all of these values are correct But this unsimplified form leads to a very simple mnemonic device Notice what we have here the square root of zero over two square root of one over two the square root of two over two Square root of three over two square root of four over two So there's sort of like a sequence where it looks like the square root of n over two Whereas n ranges from zero to four and you get these five special angles zero 30 40 60 and 90 degree angles So that can help you memorize these these these special angles because again They are critical that you know all each and every one of these Cosine does the same pattern but backwards because cosine is the Complements the co-function of sine so it goes to the same sequence backwards sine goes from zero all the way up to one Cosine start it starts at one and goes all the way down to zero. So it does it backwards. So you here you have Zero one half or two over two three over three and then one right there. What do you do with tangent? Well, honestly, I don't memorize the ones for tangent I just have to think about in tangents to sign over cosine So if you take zero over one you get zero if you take one half divided by the square root of three You're gonna end up with one over the square root of three for which if you rationalize that you get root three over three At 45 degrees notice that sine and cosine are actually the same value So the ratios can equal one tangents one at 45 degrees And then at 60 degrees you get the square root of three Why is that you take the square root of three over two divided by one over two that simplifies just to be a square of three and then lastly Tangent nine degrees is undefined for the reasons we saw before because it's one divided by zero And so you get that tangent nine degrees is D&E I highly recommend that all the viewers of this video memorize this table as it'll make it much easier for you in the future As you're studying and computing these trigonometric ratios