 Hi, I'm Zor. Welcome to Unisor Education. I would like to offer you a few problems, very trivial problems, very simple problems, which are basically related only to the definition of trigonometric function. So basically, all you have to know to solve these problems is the definitions. I do recommend you to try to solve these problems just yourself, go to Unisor.com and in notes for this particular lecture, you will find all these problems. There are nine of them. I apologize for noise. Anyway, so I will consider these problems right now, and begin they are only based on the definitions of trigonometric functions. All right, so let's just do it one by one. First and probably the most famous trigonometric problem is this. Now, this is an identity or equality which is true for any animal. Now, how can we prove that this is true? Well, let's just consider the definitions. So you remember that for any point, let me actually take the point, not in the first quadrant but something like here. It has coordinates x and y. So by definition, sign of this angle phi is its originate, and cosine is subsystem. This is y and this is x. Now, no matter where this point is, first quadrant, first, second, first, third, whatever, it's always, you can always consider this right triangle and considering that the hypotenuse is equal to one, this constitutes the Pythagorean theory. So basically, this fundamental identity between trigonometric functions, sine and cosine, sine squared plus cosine squared of any angle is equal to one. It's a direct consequence from the definition of the trigonometric functions sine and cosine and the Pythagorean theory. Now, obviously, if we go back to more narrow definition of the sine and cosine, which I have introduced in the beginning for right triangle, and that's obviously true only for acute angles. So if you have this angle phi, this is a, b, c, and the sine of phi is a divided by c and the cosine of phi is b divided by c. Now, this is obvious identity because again, you convert it into this, c squared of the common denominator, and again, you have a squared plus b squared in the numerator, which is sum of the squares of the category and the c squared, which is the hypotenuse square, and that's again the Pythagorean theorem because a squared plus b squared is equal to c squared, so it's equal to one. So that's proven. Again, this is for a narrow definition of sine and cosine for acute angles, but the general definition works exactly the same way. We just have to consider different right triangles in any case. It doesn't matter the origin of this negative or if this is negative since everything is squared, the sine is not important, so that's all for number one. And this is the only meaningful identity which I will be talking about. Everything else is really trivial and follows straight for the definition. Okay, tangent of phi times cotangent of phi is equal to one. Well, first of all, can I say that this is always true? Absolutely not, because tangent and cotangent have their domain as functions of an angle phi, have their domain not all the different angles, but there are certain exceptions. Now, let's go back to definition. Tangent by definition is sine over cosine, which means that wherever the cosine is equal to zero, tangent is not defined. And these are phi is equal to phi over two plus phi n of these capital letters. So it's phi over two minus phi over two, three phi over two, et cetera, n can be any integer number, positive, negative, whatever. So for these angles, cosine is not defined. Now, the cotangent is cosine over sine. And in this case, I should say not equal, phi not equal to phi n, which is zero, pi, two pi, minus pi, minus two pi, minus three pi, et cetera. That's where the sine is equal to zero. Now, everywhere else, this is obviously true, because the tangent is sine over cosine, cotangent is cosine over sine. Now, the denominators are not equal to zero. And what do we have as a result? Since it's not equal to zero, it's all reduced to one. That goes straight from the definition. And the only thing I would say you have to be accurate is that this particular identity is not always true. I mean, in some cases, tangent or cotangent are undefined, primarily these cases. So for all phi not equal to pi n, n's not equal to pi over two plus pi n. That's true. Just in case, I'm not sure, just as a reminder, again, this is our unit circle. Now, this is our point x, y. We are talking about sine of phi equals two y and cosine of phi is equal to x. Now, I was talking about tangent and cotangent, defined wherever cotangent cosine not equal to zero and for cotangent cosine and for cotangent sine not equal to zero. So, none of these, originate or abstissive should be equal to zero. And where are these angles? Geometrically, this point where your originate is equal to zero. This point is where your axis is equal to zero. This point is where again, originate is equal to zero and this point is where axis is equal to zero. So these four points which are pi over two plus pi over two plus pi over two plus pi over two. So these points are where either tangent or cotangent are not defined. So you have to exclude all these points. And instead of writing something like this, which I have written just a second ago and phi not equal to pi n, instead of this, I can put phi not equal to pi over two n. Right? Because this is pi n, which means zero, pi, two pi, et cetera, which is this and this and this and this and this. Now, this means pi over two plus pi n, which is this and this and this and this. So this constitute this pair and this constitute this pair and this constitute this pair of points. But if I'm combining them, if I'm saying that pi should not be equal to neither these points nor these points, I can combine them in this particular format so that actually my ability is there. So always think about this geometrical meaning of trigonometric functions. So this last one, the tangent times cotangent not is equal to one is always true for pi not equal to pi over two times whatever n. What did you do? Next, one plus tangent square phi equals to second square of phi. Okay, let's remember tangent is sine over cosine. Second is one over cosine. So, first of all, we have to think about where this function is defined. Now, in both cases, I have denominator cosine. So that's where the cosine not equal to zero, which is phi not equal to pi over two plus pi n, right? These two points, pi over two, 90 degree plus pi n, plus another pi n, plus another pi n, or minus, that's information. So in those cases, my cosine is not equal to zero and that's why this is defined. But this is defined, I'll just substitute. What is it? One plus sine over cosine phi square should be equal to one over cosine phi squared. Now, this is exactly the same as this for these angles. So instead of proving this, I can prove this. They are completely equivalent because they just substitute it for tangent and secant their definitions. Now, how about this? Well, we can always square separately. So it would be one plus sine square phi, the cosine square phi equals to one cosine square phi. Now, you'll notice that instead of writing cosine f square, many times we are using cosine square phi. It's just notation. It's exactly the same thing and it's traditionally to use this notation instead of this notation. Now, this we can use the common denominator which is cosine square, cosine square phi divided by cosine square phi plus sine square phi divided by cosine square phi. And this is equal to cosine square phi. Why? Because this is equal to one, cosine square plus sine square is always equal to one for any angle phi. So that's why we have this. So this is an obvious equality which is exactly equivalent to these. So the proof, if you want to do it like relatively rigorously, from here, we came to here through these transformations but everything is reversible. So from this particular equality which is kind of obvious, you can always go back to the original one which is the proof. And again, don't forget that the whole thing is very, very important to remember there is always the domain which we have to take into consideration. This is not always true, it's always true only within this domain where tangent and secant are defined. Next, next is absolutely symmetrical one but it's about cotangent. One plus cotangent cotangent square phi is equal to cosecant square phi. All right, cotangent cotangent by definition is cosine divided by sine. So sine should not be equal to zero which is phi not equal to pi n. Now, the cosecant definition is one over sine, same thing. So this is the common domain for left and right. So let's consider that phi is not equal to pi n. Now, both sides of this identity are defined. So we can just go straight from the definition. One plus cosine square phi divided by sine square phi is equal to one over sine square phi. I can already squared separate numerator and denominator. And this is also obvious because we can go to the common denominator which is sine square and I will have sine square phi over sine square phi. That's my one and plus cosine square phi. And this is obviously one. So it's equal to one over sine square phi and everything obviously is reversible. So from a true statement, we go back to the one that you want to prove. That's simple. And basically all of these examples are like that. First, you have to define the domain where left and right parts are defined. And secondly, very simple manipulations which are primarily based on the definition of the trigonometric functions are sufficient to prove. Next is tangent phi plus cotangent phi equals secant phi plus cosecant. Back to the roots, back to the definition. Tangent is sine over cosine. Cotangent is cosine over sine. Secant is one over cosine. And cosecant is one over sine. Right? So what is the domain of left and the right? Well, in both cases, I have cosine in the numerator and sine in the numerator and also cosine and sine. So that's basically the same situation which we were considering when we were multiplying tangent by cotangent because sine and cosine in the numerator is denominator which we have proven that this is the formula. So the angle should not be equal to pi over two times n which is 90 degree 180 degree 270, zero, 360, minus 90, minus 270, et cetera. So for all these angles, let's just substitute all these into this. What happens? All right, tangent is sine phi divided by cosine phi plus cosine phi divided by sine phi. That's the cotangent. And I made a mistake. Here, that's the multiplication. And that's supposed to be equal to what? One over cosine phi times sine. True or not? True or false? Well, first of all, let's just say that this is exactly equivalent to this. I just wrote the definition of all these functions. So proving this is the same as proving this. Now, how about this? Well, we do know that within this area, sine and cosine are not equal to zero. So I can manipulate the way how I want it. Well, right now obviously I have to use the common denominator. In this case, it's sine times cosine, right? So sine over cosine, if the denominator is sine times cosine, would be what? Sine squared phi divided by sine phi times cosine phi, right? Sine over cosine, but I have to multiply it by sine, both numerator and denominator. So sine and another sine would be sine squared. Now this, common denominator again, sine times cosine, but in this case I have to multiply by cosine. And I will get cosine squared. That's what I will get from the left. Now, on the right, I still have the same thing. But this is one, as has been proven in the problem number one, right? So we have the equality for all these angles. This is true and this is true. And that's why this is true. Nothing but analyzing the domain of left and right parts of the equality and basically using the definition of the functions. All of these problems are like that. Next, tangent phi divided by secant phi should be equal to sine of phi, well. Now, here is an interesting fact. Is secant equal to zero? No, so this, you can always divide one by one. One by another. However, tangent is not always defined and secant also is not always defined. So it's not the division per set which restricts us but the definition of the functions themselves. On the right, however, I don't have any restrictions. Phi can be any. Any angle can have the sine, but not any angle can have a tangent and not any angle has a secant. So what's the right approach to this? Well, obviously right approach is to take the smallest domain where both have sense. Now, this has sense everywhere, but this has sense only if tangent is defined which is when cosine not equal to zero, right? Because tangent is sine over cosine. Now, secant is also one over cosine. So again, it's the same thing. So everywhere where cosine is not equal to zero, left part exists, which is what? Phi not equal to pi over two plus Pm. So that's the domain where we have to really consider the left part and the right part obviously as well. So if pi is equal to let's say pi over two, well, all we can say that the right part exists but the left part is not. That's why there is no equation. So we are only considering these angles not equal to pi over two plus Pm. Geometrically, it's this. This one and this one are excluded. Every other angle is fine. Now, if this is the domain which we're talking about, then let's just use sine over cosine. That's my tangent divided by one over cosine pi. That's the left, which is actually equal to what? Sine over cosine, that's this one, divided by one and multiplied by cosine of pi. And we can reduce it and we get the sine, what we want. And again, I reduced it here, noting that cosine is not equal to zero, so I can do it safely. It's very important when you're given something like this to understand that this is not like 100% true statement. You always have to think about what's the definition, what's the domain actually of the function on the left, on the right. And if these domains are different or overlap or something, you have to really use the smallest one where both sides have sense and are defined. Now I have kind of symmetrical problem. You have a cotangent pi divided by cosecant pi and it should be equal to cosine. Fine, right? Let's think about it. Cotangent is cosine over sine, so sine should not be equal to zero, that's the condition. Now cosecant is one over sine, which is also restricted to these angles. And where is sine is equal to zero? That's where. It's zero, that's where the sine is equal to zero, sine is ordinate, and pi. And zero and pi, minus pi, and 360 degrees is two pi, et cetera, et cetera. So all the multiples of pi are excluded and all other angles are fine. And now we just have to go by definition. Instead of a cotangent, we put cosine pi over sine. Instead of cosecant, we put one over sine, which is equal to cosine phi divided by sine, divided by one, and multiplied by sine. Sine is not equal to pi, sine of pi is not equal to zero, so we can reduce it and we get the cosine. Two more. I hope it's not very boring. I think what's really important is to understand all these exceptions in the domain where left part and right part are defined. Everything else is really straightforward. Okay, now, sine to the fourth minus cosine to the fourth equals two sine square minus one. All right. It's only sines and cosines, which means everything is fine and any angle is good and the functions are defined. So now this is, well, I would like to remind you, I hope you remember this. If you don't, just check it out. A square minus AB plus AB, so AB is reduced and minus B square, so that's the same thing. Now, using this, I will substitute A equals two sine square and B equals cosine square. So A square is sine to the fourth, right? And B square is cosine to the fourth. So I can say that this difference is equal to A minus B times A plus B, which is sine square phi minus cosine square phi times sine square phi plus cosine square phi. That's the left part. This is equal to one, as we know. So I have only sine square phi minus cosine square phi. That's the left part. Now, because of this, sine square phi is equal to one minus cosine square phi, right? Or if you wish, cosine square phi is equal to one minus sine square phi, right? That's the direct consequence from the sine square plus cosine square is equal to one. So I can express sine through the cosine or cosine through the sine. So I will do this. Sine square phi minus, instead of cosine square phi, I will take this, which obviously is sine square phi minus one plus sine square phi, which is exactly what this is. So I transform this left part of this identity using the representation as a product of two terms. One is equal to one. So I just basically got rid of it. And the other, I was using the main trigonometric identity. I should really say sine square plus cosine square equals one is a main trigonometric identity. That's how probably everybody remembers it. So I just use the expression of cosine square through the cosine square and I got the right part. So that's simple, right? So what's the creative part in this particular case? This representation of difference of force power to the product of these two members. That's the only creative part. Everything else is just straightforward. Okay. And the last problem, sine phi minus cosine phi plus one times cosine phi equals sine phi plus cosine phi plus one, one plus sine phi. Well, this is kind of a trivial exercise as far as I understand. Well, let's just open all the differences. What do we have? Sine times cosine minus cosine square plus cosine. That's the left part. The right part is sine plus cosine plus one plus sine square phi plus sine phi cosine phi plus sine phi. Well, let's just think about it. This can be reduced, right? And this can be reduced. So I have cosine square with a minus sign. I think I mixed the sign somewhere. It should be something else because the sign should reduce. I think it's minus one. It's minus one. Am I right? Now I would have minus one here and minus here. Okay, now it works fine. Because now these signs are reduced. And what do I have left? Minus cosine square phi equals minus one plus sine square phi, right? Which is a true statement because I can add one to both sides. I will get this. And add cosine square to both parts. I will have this, which is a true identity, the main identity of trigonometry. So from this, I came to this through equivalent transformations which obviously are all reversible. There are nothing here which is irreversible. I didn't divide by anything. I didn't multiply by anything. So everything is just pluses and minuses. And again, the main trigonometric identity. Well, that's it. That's all the problems I have. I do recommend you to go through them again. Just go to Unisor.com notes for this lecture and just do it yourself. It's really very, very simple. And the purpose was primarily to bring your attention to two things. Number one, the definitions of trigonometric functions. The tangent is sine over cosine, cotangent is cosine over sine, secant, cosecant, et cetera. So number one is to bring your attention to definition. Number two, to bring your attention to domains of expressions of trigonometric expressions. If trigonometric expressions contain functions which are not defined everywhere like tangent, cotangent, secant, cosecant, that's the restriction. You have to really understand that wherever these functions are not defined, the whole identity doesn't make any sense. Also, if your identity involves some operations like division, obviously wherever the denominator equals to zero makes more restrictions onto the values of whatever the arguments are. So again, the definitions and the domains, these are very, very important when you're talking about all these equalities. That's it for today. Thank you very much and good luck.