 Hi, I'm Zor. Welcome to Nezor Education. Today we will continue solving relatively simple trigonometric problems. Still, it's not exactly like exercise on knowing the theory, but yeah, they do require certain ingenuity, but really not a lot. So just one little push and you will get to a solution. Now, this course is presented at Unizor.com. The name of this course is Mass Plus and Problems. There is a prerequisite course called Mass 14s on the same website. All the courses are totally free, no advertisement, no strings at age, even sign-in is optional. The name of this lecture is Trigonometry 06. In this course, Mass Plus and Problems also, you might find interesting on the same website, Physics 14s and Relativity for All. Okay, so solving problems is probably one of the most important purposes of all these courses. I just could not present a problem without the theory. So first you have to know about what Trigonometry is all about and then solve the problem. So that's why there is a course called Mass 14s which is basically for high school and maybe a little bit higher. And then I have Mass Plus and Problems. Okay, so let's just solve these few problems in Trigonometry. Okay, let's assume that angles alpha, beta and gamma all are angles in some triangle. Something like alpha, beta and gamma. Now, what I have to prove is that cosine of alpha plus cosine of beta plus cosine of gamma less than or equal to 3 seconds. Okay, well, I mean, obviously we know that Trigonometry functions sine and cosine are between minus 1 and plus 1. So definitely, if sum of 3 cosines will be less than 3. I would like to have it a little bit more strict. It's less than 3 half. If the angles are related by this condition that they all belong to one triangle. Well, obviously if it says they belong to one triangle it means that the sum of these angles is equal to pi which is 180 degrees. So we know the theorem from the geometry. So sum of these angles is always equal to 180 degrees or pi in radians. Okay, so this is the condition alpha plus beta plus gamma equals to pi and this is something which you have to prove that some of their cosines is less or equal to 3 seconds. Okay, so at this particular point you obviously need some ingenuity. Well, one simple thing is that alpha plus beta is equal to pi minus gamma. That might actually help to reduce the number of variables from 3 to 2. If instead of gamma you will put alpha plus beta minus pi or pi minus alpha minus beta and somehow open this cosine of the sum or difference between angles. I would like actually to do it slightly differently. I would like to cosine plus cosine somehow express in terms of alpha plus beta and then replace it with pi minus gamma and I will have only one variable gamma. Okay, how can I do it? Well, in this particular case and this is the piece which kind of requires some ingenuity. I would like to convert the sum of these two cosines into product of trigonometric functions. Now I do not remember quite frankly all these identities trigonometric. What I do remember is what is the sum, the angle, the sine of sum of angles and cosine of sum of angles. So like sine of alpha plus beta and cosine of alpha plus beta. That I do remember. And I also remember how to basically the way how to convert sum of cosines or sum of any two trigonometric functions into product. Here it is. Now I know that angle alpha can be represented as alpha plus beta plus alpha minus beta over 2, right? So it would be alpha over 2 plus alpha over 2 which is alpha beta over 2 minus beta over 2 which is 0. So this is correct, okay? And beta can be represented as alpha plus beta over 2 minus alpha minus beta over 2. Alpha will cancel out and beta will be minus and minus will be plus plus will be. From this I will do this cosine of alpha is equal to. And now I will use the formula which I do remember. The cosine of sum of two angles is cosine times cosine minus sine times sine. So that would be cosine alpha plus beta times cosine alpha minus beta minus sine of alpha plus beta minus sine of alpha minus beta. Now cosine beta is equal to, this is the cosine of difference of two angles. That would be plus here. So that would be the same cosine alpha plus beta over 2 cosine alpha minus beta over 2 plus sum. And if I will add them together, cosine plus cosine, this one. This thing would cancel out and I will have that instead of cosine plus cosine I can put 2 cosine alpha plus beta times cosine alpha minus beta. So this is this piece plus cosine gamma. So this is equivalent to whatever I have on the left side. Now immediately I know that this thing is less than 1 or equal to 1, right? Because it's a cosine. So I immediately can put that this is less than 2 cosine alpha plus beta over 2 without this plus cosine gamma. So I immediately establish that the expression on the left is less or equal to this. And as you see now everything is in terms of alpha plus beta or gamma and alpha plus beta can be expressed as gamma. So I will have equal to 2 cosine of alpha plus beta per column over 2 per column. So it's pi over 2 minus gamma over 2, right? Plus cosine gamma. Now here is something which I can express as sine of gamma over 2, right? Cosine of pi minus 2 is like a right triangle. Sum of alpha and beta is pi over 2, sine of this is equal to cosine of this. Plus cosine of gamma, since this is gamma over 2, I will probably have to express this as a function of gamma over 2. Now the same formula gamma is equal to gamma over 2 plus gamma over 2. So it's cosine cosine, which is cosine square minus sine of sine, which is sine square. But the cosine can be expressed as 1 minus sine square. Cosine square can be expressed as 1 minus sine square. So that would be what? 1 minus 2 sine square of gamma over 2. Okay, so I have established that this expression, sum of cosine, is less than equal to this. Now this is kind of a polynomial of sine of gamma. So I can put minus 2z square plus 2z plus 1, where z is sine of gamma over 2. Now this is some kind of a polynomial as a function of z. Well z is basically sine of some angle, which can be between 1 minus 1 and 1, can be anything. Now but what's important is, you see, this is minus, which means that the graph of this is a parabola which is directed with its horns down, and it has a maximum. That would be something like this. So question is, what is this maximum? If this maximum is not equal to, I mean less than or equal to 3 seconds, then I'm fine. And indeed, well, what I can definitely know that this is between this and this, I do not remember again the formula for the maximum, but I do remember the formula for roots. It's, if you have expression like ax square plus bx plus c, then roots are minus b over 2a plus minus some radical. So since I need the middle point, I have to basically have middle between left and right roots. So if I will add plus to minus, it will cancel out. I will have midpoint as minus b over 2a. So minus b, which is minus 2, divided by 2a, which is minus 2 times 2 minus 4. So it's 1 half. So midpoint is at 1 half. And what's the value of my polynomial at 1 half? This is 1 quarter minus 1 half plus 2 times 1 half plus 1 plus 1. So it's minus 1 half plus 1 plus 1. So it's 2 minus 1 half, which is 3 seconds, exactly. So this polynomial has a maximum of 3 seconds. No matter what's the value of sine of gamma over 2 is. And that proves the whole thing, the whole inequality. Okay, so that's problem number one. Now, obviously, I didn't say it in the very beginning, but I think I'm talking about this in every lecture related to problems. Try to solve these problems yourself. Immediately after I present the problem, pause the video if you watch it in the video and try to solve it yourself. Or, alternatively, you can go to Unisor.com, you can go to this course, Mass Plus and Problems. And if you will choose this lecture, you will have both video and textual presentation of the same thing. And in this case, the full solution is presented. So again, you read the condition of the problem and then solve it yourself. Don't look into the solution itself. Sometimes I present the hints, not a solution, just a hint. Sometimes don't at all. Sometimes it's a very easy problem, and I don't really put in writing the solution. But I will solve it here. Okay, let's assume you have a set of angles, which are all acute angles. Well, let's just take this out. So they're really acute angles. So they are basically sequenced in ascending order, these angles. And they all belong to the interval from 0 to pi over 2 to 90 degrees. Now, what I have to do is I have to prove the following inequality. Tangent of alpha 1 means less or equal to sin of alpha 1 plus sin of alpha 2 plus, etc., plus sin of alpha n divided by sum of cosines, alpha 1, alpha 2, cosine alpha n. And it's alpha n. So that's what I have to prove. Again, this is an easy problem, so it definitely makes sense to pause the video and solve it yourself. Now, here is how I approach it. Now, remember the graph of sin. Sin is something like this. This is pi over 2. Cosine is... So this is sin. And this is cosine. So sin is increasing. Cosine is decreasing. Okay. Now, if sin is increasing, then if angles are increasing, the sines also are increasing. So in the numerator, these values are basically in ascending order. And I can definitely say that n times sin of alpha 1 less than or equal to sin of alpha 1 plus sin of alpha 2 plus, etc., plus sin of alpha n and n times sin of alpha n. Since sin of alpha n is the biggest among them and sin of alpha 1 is the smallest of one of them, then n times the smallest is less than some and the sum is less than n times the maximum of them. Great. So that's the numerator. How about the denominator? Now, cosine is decreasing. So this is in decreasing order. So what I can say that this is the smallest and this is the biggest, which means n times cosine alpha n is less than sum of cosines. Cosine alpha n. And that's less than n times cosine alpha 1. So alpha n is the biggest but cosine of alpha n is the smallest because it's a decreasing function. And alpha 1 is the minimum of those alphas. So cosine of alpha 1 is the maximum of them. So that's why it's true. All right. Great. Now, what does it mean? What is this tangents of alpha 1? Tangent of alpha 1 is well, n times sine of alpha 1 divided by n times cosine of alpha 1, right? Because n and n is the cancelling out. Now, this is the smallest. This is the smaller than. And this is the biggest. So the numerator is the smallest, the smaller of this sum and denominator is bigger than this sum. So this is definitely less than sine of alpha 1 plus et cetera plus sine of alpha n divided by cosine of alpha 1 plus et cetera plus cosine of alpha n. So again, this is smaller than this. This is bigger than this. So we are decreasing numerator and increasing denominator. That's why the whole thing would be smaller than this. Analogously. Tangent of alpha n. I can put it as n sine of alpha n divided by n cosine of alpha n. Now, this is bigger than this and this is the smallest among them. That's why, again, that would be greater than sigma of sine divided by sigma of cosine of alpha i. i from 1 to n. i from 1 to n. So that's the proof. Just replaced with the smallest and the biggest and we got these two inequalities. That's easy, right? Now, the third one involves some calculations but it's straightforward. So I basically have to solve the equation sine of Ax plus sine of Bx equals 0. Where A and B are some real numbers. Alright, since it's 0 the best thing is obviously to convert it into product. If you convert it into product then equating each component to 0 will give you a solution, right? How to convert to product. Well, exactly the same thing. So A is equal to A plus B over 2 plus A minus B over 2. And B is A plus B over 2 minus A minus B over 2. So sine of Ax is equal to instead of Ax I have A plus B over 2 times Ax and A minus B. So it's a sine of sum that would be sine of the first A plus B over 2 times cosine second A minus B over 2x plus cosine sine plus cosine A plus B over 2x times sine A minus B over 2x. Now, sine of Bx is equal to Now, this is the difference between two angles. So it's a sine cosine minus cosine sine. So it would be sine A plus B over 2x times cosine A plus B minus in this case minus cosine times sine. And this thing will cancel out if I add them together. So sine of Ax plus sine Bx is equal to 2 sine A plus B over 2x times cosine A minus B over 2 times x. And this is supposed to be equal to 0, right? Now, that's easy. It's either this is equal to 0 or this is equal to 0. So both gives you different solutions. Now, when sine is equal to 0 when angle is equal to 180 degrees times any number, times pi times n. So A plus B over 2 times x is equal to pi times n where n is any integer number. It can be 0, it can be 1, it can be minus 1, etc. In these cases, this angle gives you sine of this angle gives you 0 from which x is equal to 2 pi n divided by A plus B. Okay. Now, obviously if A plus B not equal to 0 which means that in this particular case if A is equal to minus B it's like sine x plus sine minus x. Sine is an odd function so sine of minus x would be minus sine of x so it would be basically sine x minus sine x is equal to 0 which means any solution would fit. So if A plus B not equal to 0 then this is the solution. If A plus B equal to 0 so any x would be a solution. Now, another set of solutions would be when A minus B divided by 2 times x cosine is equal to pi over 2 plus pi n. We have to shift by pi over 2. Remember, sine has this graph so it's 0, it's pi, it's minus pi, so it's pi times any number. Cosine has this same thing but shifted by pi over 2. So it's pi over 2 plus the same pi times n from which x is equal to 2 times so it's pi plus 2 pi n divided by A minus B. Again, if A not equal to B if A is equal to B I will have sine of A x basically times 2 so this is 0 so it's sine of A x is equal to 0 and I will have basically all these solutions when sine is equal to 0 that's a different kind of thing but that's a trivial case so let's not consider A plus B or A equal to 0 or A minus B equal to 0 in which case the equation would be very kind of simple but for this and this not equal to 0 denominators this is a set of solutions and this is a set of solutions basically. But, yeah, basically that's it. Alright, so what I do recommend you to do right now is go to the Unisor that can't go to the course mass plus and problems lecture 3-dynamic 0.06 you will have all the problems there I think in a couple of cases I do have solutions in one case I don't have a solution presented in writing but read the condition of the problem try to solve it yourself if I have a hint use the hint if I don't have a hint don't use a hint but in any case it's very important for you to again do exactly the same as I did during the lecture do it for yourself just have a piece of paper and basically solve it yourself because right now you definitely are equipped to do this type of things and it's very important to repeat something like you will be next time you will be more comfortable with this Ok, that's it, thank you very much and good luck