 Hi, I'm Zor. Welcome to Ineasier Education. Today is the fifth lecture in series of trigonometrical problems presented as part of the course called Mass Plus and Problems on website Unisor.com. On the same website there is a prerequisite course called Mass for Teens, which is basically a theory of mathematics as basically on the level of high school maybe just a little bit more than that. And so the Mass for Teens is a theoretical course. Mass Plus and Problems is the course where I present different problems. Problems in algebra, geometry, trigonometry, etc. So this is trigonometric series and as I said lecture number five. Also on the same website Unisor.com we can find physics for teens and relativity for all courses. They are all presented on the website and the site is totally free. There are no advertisements, no strings at age, even login is optional. It's really needed only if you're studying under somebody's supervision. So we need to establish the connection between the supervisor and the student. Okay, so let's go back to problems. Problem number one. What's given is that there are three acute angles of a bacon gamma, which are, if you summarize them together, will give you pi over two radians, 90 degrees. My problem is I would like to evaluate an expression. Tangent alpha times tangent beta plus tangent beta times tangent gamma plus tangent gamma times tangent alpha. So my angles are not given. The only thing which is given is that their sum is equal to pi over two. Nevertheless, it's sufficient to evaluate this whole expression to a concrete number. It would be number one actually. Okay, so as usually I suggest you after I present the problem, you can stop the video, pause it and think about the problem yourself. The most important part of problem solving is to solve them yourself. Now, even if you don't really come up with a solution, just think about the problem, how to approach it. That's very important. And I will present you my solution, which is probably one of few solutions which can be presented. Okay, so what should I do? First of all, from this I can derive very simple thing that alpha plus beta is equal to pi over two minus gamma, right? So, if I will present tangent gamma as tangent by definition is sine divided by cosine. At the same time, sine of some angle, if we are talking about the cubed angles and they are cubed angles, remember from the right triangle if this is angle X and this is angle Y, then the cosine of X is equal to sine of Y and sine of X is equal to cosine of Y. I hope you remember this and the explanation is simple because for a right triangle, sine of X is this category divided by hypotenuse and that's exactly the definition of the cosine of Y and vice versa. So, and also X plus Y is equal to pi over two, so X is equal to pi over two minus Y. Now, using this very simple property, I can say that sine of gamma is equal to cosine of Y. Pi over two minus gamma and cosine of gamma is equal to sine of pi over two minus gamma, which is equal to pi over two minus gamma we can express as alpha plus beta. So, this is cosine of alpha plus beta divided by sine of alpha plus beta. Okay, so using this, I can substitute it here and my expression which contains three angles using this condition will have only two variables, alpha and beta. Okay, now before doing that, I actually open up what is the cosine and sine of some angles. This is the cosine alpha times cosine beta. That's one of the few formulas which I actually remember. Everything else can be derived from them. So, I remember what the cosine of some of angles and sine of two angles is. Obviously, it's presented and proven in the theoretical course math for teens on the same website as a prerequisite. So, it's cosine cosine minus sine sine divided by sine is sine cosine plus cosine sine. I will simplify it even more. I will divide both numerator and denominator by product of cosines. So, if I will divide it, I will have one. Now, sine times sine divided by cosine cosine will be tangent times tangent, right? So, it will be tangent alpha times tangent beta. Here, sine cosine divided by cosine sine, it will be just tangent of alpha. And this, if I divide this, it will be sine over sine of beta plus, which is tangent beta. And this is my final formula. So, tangent of gamma can be presented as this expression. All I have to do is substitute it into my original expression, which I need. And what will I have? Tangent alpha times tangent beta plus. Now, tangent gamma and tangent gamma, I will just factor out. So, it will be tangent gamma times tangent alpha plus tangent beta. Tangent gamma times tangent alpha and tangent gamma times tangent beta. That's why. And now, I will substitute instead of tangent gamma this expression. So, what will be? My tangent plus tangent will cancel with this one. And remaining will be one minus tangent times tangent. And this is tangent times tangent. So, it will be one. Very simple. So, that's the answer. Okay, next. Okay, I have, again, two acute angles. Acute means they are in the first quadrant from zero to pi over two. And I know that sine square of x is equal to one-fifths. And sine of square of y is equal to one-tenths. What I have to do, I have to evaluate x plus y. Well, I mean, in theory, you can say that, okay, sine of x is equal to square root of one over five. Which means x is equal to arc sine of square root of one over five. Well, that's an answer. So, you can say that this is arc sine of one over square root of five. Plus y, which is similarly arc sine of this. Yes, you can do that. But that's not what I'm asking. I'm asking about something much more simpler than this. So, apparently, this plus this gives you a much simpler answer if you will be able to come up with this answer. So, again, I suggest you to pause the video and think about this problem. And I will present my solution. So, that's not the way how it's supposed to be done. What we will do is differently. Look, cosine square of x is equal to what? I mean, cosine square plus sine square is always one for any angle, right? So, this is four-fifths. And cosine square of y is equal to nine-tenths. So, from this, we have sine x is equal to one over square root of five. Cosine x is equal to two over square root of five, right? Square root of four is two. Now, sine of y is equal to one over square root of ten. And cosine of y is equal to three over square root of ten. Square root of nine is three. So, we have sine and cosine. And now, we can do something like sine of x plus y, which is equal to what? Sine x times cosine y plus cosine x times sine y. And since I know the values of all these, I can substitute it. And I will get a nice expression for sine of x plus y. And then, I will get arc sine of it, right? Okay, so sine x times cosine of y. Sine x times cosine of y. That's three over square root of five, square root of ten. Plus cosine of x, which is this, times sine. So, it's two divided by, again, square root of five, square root of ten equals two. Well, it's the same denominator. So, I can just add numerator. So, it's five over square root of five, square root of ten. Now, five over square root of five is square root of five. And this is square root of two times square root of five, right? Ten, which is what? One over square root of two. Now, one over square root of two, or square root of two over two. This is a sine of what? And that's something which, again, I do remember. It's actually of 45 degrees. So, if you have a triangle, right triangle with the same categories, which is 45 degrees, 45 degrees. If this is one, this is one, this is square root of two, right? It's a Guerin theorem. So, one over square root of two would be sine or cosine of 45 degrees of pi over four. So, the answer is pi over four, radians, or 45 degrees, okay? That's the second problem. And the third and the last one of this lecture would be, okay, I have to simplify the expression cosine of arc sine of, I have to simplify it somehow. Well, what do we know about arc sine? Arc sine, by definition, is such an angle, let's say alpha is equal to arc sine of x. Such an angle, sine of which is equal to x. That's the definition of arc sine. Now, if I will define it just like that, there are many different angles, sine of which is equal to x, because sine is a periodic function. The period is two pi. So, if I find one particular angle, then this angle plus two pi will also be angle sine of which would be equal to the same x. So, for this purpose, if you remember, I'll just repeat a little bit some theoretical stuff. Arc sine is not function which is defined like everywhere. Arc sine is a function which defined only with a domain. Well, domain is obvious. It's from minus one to one, because sine is always, x cannot be greater than one or less than minus one, because that's the sine of it. Now, the angle alpha, I mean, sorry, that's a codomane. It's a codomane. That's one second. No, that's the main. Now, the value which is angles themselves, angles are from minus pi over two to pi over two. Why? Because if you will take a look at the graph of sine, the only period, the only interval where the function is outside of the periodicity is this one, from minus pi over two, this is zero, to pi over two. So, only this period, only this interval function does not have a period. It's monotonically increasing, and since it's monotonically, then for every value, this is one and this is minus one. From every value of x, from minus one to one, we can find the corresponding value of this is x, this is alpha. So, alpha is in this particular range, and x is in this particular range. So, the function is defined on domain from minus one to minus two to plus one, with the codomane with various of this function between minus pi over two and pi over two. Now, this is the graph of sine of x, sine of alpha, sorry, sine of alpha. X is equal to sine of alpha. The graph of this function will be, well, this is the bisector, so it will be correspondingly from minus one to one, but the function would be like this. So, this is pi over two and this is minus pi over two. So, this is arc sine of x. So, again, this is sine of alpha, x is equal to sine of alpha, and this is alpha equals to arc sine of x. So, these two graphs are symmetrical relatively to bisector. So, all these issues have been covered in the theoretical course, which is prerequisite for this one. On the same website, it's called mass patins, so I suggest you to repeat it if you don't remember. I just put some refresher because the problem is actually simple if you will kind of understand all these issues. All right, so I know that by definition sine of arc sine is equal to x. Now, but we are interested in the cosine. Now, the cosine of alpha, alpha is arc sine, which is substituted, is equal to square root of one minus sine square of alpha, right? Because cosine square plus sine square are equal to one. Now, in theory, there are two different sines, plus or minus, but I have said in the very beginning that we have certain restrictions on domain and co-domain, etc. And let's just think about this particular issue. My x is supposed to be from minus one to one, and my angle is supposed to be from minus pi over two to pi over two. Now, if the angle this alpha is between minus pi over two and plus pi over two, let's think about the cosine. Now, I'll put it in the same graph. The cosine actually has maximum at zero, and then it goes this way. That's the cosine. So, from minus pi over two to pi over two is always positive, which means I can always put only the plus sine. And that actually solves the whole problem, because sine of alpha, we know what it is, it's x. So, the whole thing is square root of one minus x square. So, I can say that cosine of arc sine of x is equal to square root of one minus x square, always for any x which belongs to a domain from minus one to plus one. And again, why is it only plus? Because the arc sine for this particular x, by definition, is an angle from minus pi over two to pi over two, and cosine is positive in this particular area. Okay, that concludes my third and last problem. So, what I suggest you to do is open the website Unisor.com, go to course called Mass Plus and Problems among all the courses. Now, if you will click on this course, you will see the trigonometry as one of the options. You go there and the trigonometry zero five is this lecture. All the problems are presented on the textual part of this particular page, and you will be able just to basically again get into these problems and try to solve them yourself. Sometimes I put some hints, maybe sometimes solutions in writing, but the conditions of the problem always there, obviously. Solutions are not always there. So, read the problem itself and try to come up with your own solution. Maybe it will be the same as presented here, maybe it will be something different, because any problem can be probably solved in more than one way. And just spend some time, what's important is to spend some time thinking about the problem. If you think about the problem, how to approach it, etc., that's the best way how you prepare yourself to real life problems. In real life problems you will not see something like this, but if your mind is trained to seek the solution using whatever different ways you can, that's the way how you prepare yourself. And the more mass problems you can solve, the better you will be prepared to real life problems. Okay, on this note, let me say thank you and good luck.