Chapter Four. We are very shortly going to be continuing the study of power in an AC circuit. However, there is a couple of trig identities which I very quickly want to go over because they are key to developing formulas in those AC circuits. The first trig identity that I want to look at is the cosine of the sum of two angles x and y is equal to cosine x times cosine y minus sine x times sine y. And I'm going to start with a right angle triangle that you see here. And I'm going to designate the contained angle between the hypotenuse and the adjacent side as angle y. I am now going to place another right angle triangle right on top of the First triangles such that the high pot news is the same length as the adjacent side of the new triangle.

And I'm going to designate the contained angle between its high pot news and the adjacent side as angle x. I'm going to draw a perpendicular line from the peak of those two triangles to the adjacent side of the first triangle. And I'm going to drop another line from the peak of the first triangle, such that it meets the first line that I drew there at 90 degrees, in other words is perpendicular to the first line. I'm going to scale these triangles such that the hypotenuse of the second triangle is one. And it doesn't matter whether it's one foot one inch or one, whatever, it's just going to be one and everything else is going to be proportional to the hypothesis. Use of that triangle. In order to keep track of what we're talking about, I'm going to designate points on this dual triangle configuration as A, B, C, D, E and F. Looking at the second triangle, which I've shaded in yellow, the sine of the contained angle x is equal to the length of the line DC, all over one.

Now any number divided by one is the number itself, so we don't have to write DC over one we can just write DC. And in fact, DC then is equal to sine x, which I've written on the diagram there. Now if we take the cosine of the contained angle it is a C over one. And again, we can rewrite that equation such that A c is equal to the cosine of x, which I've written on the diagram as well. Now, because the line EC is parallel to the line a B, then the intersecting line of bipod news so the first triangle means that the angle y is reflected in the parallel angle formed by e C, A. I would like to look at the little triangle that I've shaded in yellow there. And if we look at the angle D, C, E, it is only a portion of the writing angle of the second triangle that we plotted.

And the angle there is 90 degrees minus the angle y. Switching to the other angle in that triangle now, C, D E, and we know that the angles of a triangle have to add to 180 degrees. Then that angle that I've shaded there that we're going to look at C, D, E is equal to 180 degrees minus the angles opposite, which are the angles, D, E, C, which happens to be 90 degrees and the C, which we've just looked at, which is anger. Goal, the C E is equal to 90 degrees minus y. So we can put those values in, and we have the angle C D is equal to 180 degrees minus the 90 degrees minus what we've what we've calculated in the first place of the CV to be 90 degrees minus y. If we take the brackets a wave in that angle equation, we have C, D is equal to 180 degrees minus 90 degrees minus 90 degrees plus y, leaving us with the angle of C D equal to y. I would now like to look at the triangle that I've shaded in yellow.

A, F d And if I was to take the cosine of the contained angle that is made up of the angle x plus y, cosine of x plus y is a f all over one. And as we have seen before any number divided by one is that number itself. So, the cosine of x plus y is a F and it is the link to the line a F. Now, a F is made up of a b the line a b minus the portion f b but FB is equal to the line E, C. So I can add that to the equation such that A f is equal to A B minus ec. And we're talking about lengths of lines here. I now want to concentrate on my original right angle triangle, which I've shaded in yellow. And if we were to take the cosine of the angle why it is a B, all over coasts x, the hypothesis we already figured out to be cosine x. I can rewrite that equation such that A B is equal to cosine x times cosine y would now like to go back to our little triangle I'd like to take the sine of the angle y, which is equal to EC all over sine x.

And I can rewrite that equation such that EC is equal to sine x times sine y. Now I've added color to a lot of things as I'm going along here, not for any specific reason that that color is significant only to separate it out so that we can follow the equations a little bit easier. Now, looking at our first equation that we had there for cosine y, a B is equal to cosine x times cosine y and I can substitute cosine x times cosine y for a B in the equation A f is equal to A B minus ec. And I could also substitute sine x times sine y for EC in that same equation. And I could substitute for a F in that equation with what I have developed down on the lower left hand side. And that is cosine of x plus y is equal to A f. So if we make those substitutions, you can see that cosine of x plus y is equal to cosine x times cosine y minus sine x times sine y, which is what we set out to prove in the first place.

This time, we're going to look at the trigger data into the four cosine of x minus y, which is equal to cos x times cos y plus sine x times sine y. Now there's a big long proof for this similar to the one we just went through. And I'm going to avoid that one. For now I'm going to go through a simpler version. And we already proved this equation where it's the sum of angles the cosine of the sum of the angles. And I'm going to look at two scenarios here, one with a triangle with y as the angle and he has the high pot news and oh is the opposite side and a as the adjacent side.

And I'm also going to look at a similar triangle but this time minus y angle with h as the apart news and a is the adjacent side, but this time, the opposite will have to be minus Oh, I'm going to rewrite that equation that we just wrote that we just proved actually. And I'm going to substitute minus y for the angle y. Now, in the term on the left hand side of the equation, it's cosine x plus the minus y angle is equal to cosine x minus y, because x plus the quantity minus y is just x minus y. So we can replace the quantity or the term on the left hand side of the equation with the actual quantity cosine x minus y. Now in looking at the top triangle that we have there with the positive y contained angle, cosine y is given by a over h, which is the Jason side, all over the high pot news.

In the case of the cosine minus y, it's given by the adjacent side, which is plus a, all over h. So it's the same thing. So cosine y is equal to coast minus y, which we probably already knew anyway, however, I just wanted to go through the exercise. So we can take the first term on the right hand side of the equation. And we can rewrite that as coasts x times coasts y. Now looking at sine y, of the of the equator of the triangle that's on the top there, sine y is opposite all over hypotenuse. Sine minus y is minus opposite all over hypothesis or minus old All over h. But that can be rewritten as minus taking it outside the whole quantity minus positive whole all over positive H. Which can be substituted or we can substitute for sine.

Why? Because sine y is nothing more than all over h. And we can rewrite that as minus sine y. So looking at the last term of that equation, sine x times sine minus y, we can substitute minus sine y for sinus sine minus y, and we are left with plus sine x times sine y because the two minus signs will change it to a plus where you can plug back in the equal sign In our equation, and lo and behold, we have the exact equation that we're looking to prove in the first place. So the trig identity coasts, x minus y is equal to cosine x times cosine y plus sine x times sine y. I'm going to go through one more trig identity. And this one's a short one, so bear with me. And in fact, I'm just going to use what we've already proven.

And that is the identities involving coasts x minus y and coasts, x plus y. And you can see them there in front of you. I'm going to subtract the second one from the first one, which one you're going to do equation addition and subtraction. You do it from with the left hand side and the right hand side independently, and I'm going to be left with this equation which is on the left hand side coasts x minus y minus one x plus y. And on the right hand side, the coasts, x times coasts, y disappears because it's one minus the other. And I'm going to be left with two sine x sine y.

And I'm just going to do an algebraic manipulation here. I'm going to move the two to the left hand side. And then just to make sure that we can follow what's happening next, I'm going to just switch about the equation which you can do. I'm going to put the sine x sine y on the left hand side, and I'm going to rewrite the right hand side so it looks more like what we're setting out to prove in the first place, which is sine x times sine y is equal to one half the quantity coasts x minus y minus cosine x plus y. A Yes, indeed, that's exactly what we set out. To prove in the first place This ends the chapter