 In this video, we will prove that the classic geometric problem of trisect in the angle is impossible to solve that is given any angle. So this is the statement of the problem given any angle. It is not always possible to construct with a straight edge and compass alone, an angle whose measure is one third of the original angle. Okay. Now, in order to analyze this, we have to translate what it means to construct an angle back into constructable numbers, because a constructable number is the length of a segment. Not an angle measure. So how can we correlate those things together? So be aware that if you have some angle constructed like so, without the loss of generality, we can assume this angle is in standard position so that its initial side is on the positive x-axis and at the terminal side, well, it terminates wherever the heck it wants to, right? And we can associate the vertex of the angle with the origin of our plane and then constructing a circle, the unit circle centered at the origin, we then have our unit circle like so. We will then get this point of intersection, which we know from fundamental trigonometry, this point of intersection will be cosine theta comma sine theta, like so. If we take the perpendicular line dropped from this point onto the x-axis, so this does form a right angle, then this point right here would have as its coordinates cosine theta comma zero. And so that means that this distance right here is cosine of theta. So if the angle theta is constructable, that means the distance cosine of theta is constructable and this process is reversible. So we can construct the angle theta if and only if we can construct the distance cosine of theta. So we're going to argue that cosine of theta is not a constructable number for all theta. Now, in particular, I want you to think about the following trigonometric identity. Cosine of three theta, this is relevant because we want to trisect angles right here. Cosine of three theta follows the trigonometric identity, cosine of three theta is equal to four cosine cubed of theta minus three cosine theta. Basically, you're going to use the fact that this is cosine of two theta plus theta. You use the classic angle sum identity plus the double angle identity for cosine. You can simplify that to the following equation. I'll leave it as an exercise to the viewer to verify that. And so then if you plug in the angle 20 degrees into this formula, well, then on the left hand side, you're going to get three times 20, which is 60 degrees. This is one of those special angles we memorized in trigonometry class. Cosine of 60 degrees is equal to one half. For simplicity, say that alpha is the cosine of theta here, cosine of 20 degrees. And so then plugging this into the above equation, we then get that equation becomes four alpha cubed minus three alpha is equal to one half. If you times both sides by two, you're going to get equal eight. If you times both sides by two, you'll get eight alpha cubed minus six alpha is equal to one. Move the one to the other side. You then get that eight alpha cubed minus six alpha minus one is equal to zero. Therefore, alpha is a root to the polynomial eight x cubed minus six x minus one. This is a degree three polynomial. If it's irreducible, that happens if and only if it has no roots, because if it's reducible has a linear factor and if it has a linear factor, it has a root. By the rational roots theorem, the only rational, only possible rational roots of this polynomial are the divisors of one divided by the divisors of eight. So you get plus or minus one plus minus one half plus or minus one fourth plus or minus one eighth. I will leave it as a as a as a exercise to the viewer here to verify that none of these eight numbers are roots of this polynomial. Therefore, this polynomial is in fact irreducible because it has no rational roots. Therefore, this polynomial must then be the minimal polynomial for the cosine of 20 degrees. Since this is a degree three polynomial, if we take q adjoin alpha and look at its degree over q, we end up with three. But a property of constructible numbers is if you take q adjoin a constructible number alpha and you look at its field extension over q as a degree, this always looks like a power of two. Three is not a power of two. Therefore, this cannot be a constructible extension. Therefore, the cosine of 20 degrees is not a constructible number. Therefore, a 20 degree angle is not constructible, and that shows you that a 60 degree angle is that impossible to trisect using a compass and straight edge alone.