Chapter I. Solving Right Triangles
§3 (37). Basic relationships and problems
Trigonometry deals with problems in which it is necessary to calculate certain elements of a triangle from a sufficient number of numerical values of its given elements. These problems are usually called problems on solution triangle.
Let ABC be right triangle, C - right angle, A And b- legs opposite acute angles A and B, With- hypotenuse (Fig. 3);
then we have:
Cosine acute angle is the ratio of the adjacent leg to the hypotenuse:
cos A = b/ c, cos V = a/ c (1)
The sine of an acute angle is the ratio of the opposite side to the hypotenuse:
sin A = a/ c, sin B = b/ c (2)
The tangent of an acute angle is the ratio of the opposite side to the adjacent side:
tan A = a/ b, tan B = b/ a (3)
The cotangent of an acute angle is the ratio of the adjacent side to the opposite:
ctg A = b/ a, ctg B = a/ b (4)
The sum of acute angles is 90°.
Basic problems on right triangles.
Task I. Given the hypotenuse and one of the acute angles, calculate the other elements.
Solution. Let them be given With and A. Angle B = 90° - A is also known; the legs are found from formulas (1) and (2).
a = c sinA, b = c cos A.
Problem II . Given a leg and one of the acute angles, calculate the other elements.
Solution. Let them be given A and A. Angle B = 90° - A is known; from formulas (3) and (2) we find:
b = a tan B (= a ctg A), With = a/sinA
Task III. Given a leg and a hypotenuse, calculate the remaining elements.
Solution. Let them be given A And With(and A< с ). From equalities (2) we find angle A:
sin A = a/ c and A = arc sin a/ c ,
and finally the leg b:
b = With cos A (= With sin B).
Task IV. Given sides a and b, find the other elements.
Solution. From equalities (3) we find an acute angle, for example A:
tg A = a/ b, A = arc tg a/ b ,
angle B = 90° - A,
hypotenuse: c = a/ sin A (= b/sinB; = a/ cos B)
Below is an example of solving a right triangle using logarithmic tables*.
* Calculation of elements of right triangles using natural tables is known from the VIII grade geometry course.
When calculating using logarithmic tables, you should write out the appropriate formulas, take logarithms, substitute numerical data, and use the tables to find the required logarithms of known elements (or their trigonometric functions), calculate the logarithms of the required elements (or their trigonometric functions) and use the tables to find the required elements.
Example. Legs are given A= 166.1 and hypotenuse With= 187.3; calculate acute angles, other side and area.
Solution. We have:
sin A = a/ c; log sin A = log a-lg c; 
A ≈ 62°30", B ≈ 90° - 62°30" ≈ 27°30".
Calculating the leg b:
b = a tan B ; lg b= log b+ log tan B ; 
The area of a triangle can be calculated using the formula
S = 1/2 ab = 0,5 a 2 tg V; 
To control, let’s calculate angle A on a slide rule:
A = arc sin a/ c= arc sin 166 / 187 ≈ 62°.
Note. Leg b can be calculated using the Pythagorean theorem, using tables of squares and square roots(Tables III and IV):
b= √187,3 2 - 166,1 2 = √35080 - 27590 ≈ 86,54.
Discrepancy from the previously obtained value b= 86.48 is explained by the errors of the tables, which give approximate values of the functions. The result of 86.54 is more accurate.
Unified State Exam for 4? Won't you burst with happiness?
The question, as they say, is interesting... It is possible, it is possible to pass with a 4! And at the same time not to burst... The main condition is to exercise regularly. Here is the basic preparation for the Unified State Exam in mathematics. With all the secrets and mysteries of the Unified State Exam, which you will not read about in textbooks... Study this section, solve more tasks from various sources - and everything will work out! It is assumed that the basic section "A C is enough for you!" does not cause you any difficulties. But if suddenly... Follow the links, don’t be lazy!
And we will start with a great and terrible topic.
Trigonometry
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
This topic causes a lot of problems for students. It is considered one of the most severe. What are sine and cosine? What are tangent and cotangent? What's happened number circle? As soon as you ask these harmless questions, the person turns pale and tries to divert the conversation... But in vain. These are simple concepts. And this topic is no more difficult than others. You just need to clearly understand the answers to these very questions from the very beginning. This is very important. If you understand, you will like trigonometry. So,
What are sine and cosine? What are tangent and cotangent?
Let's start with ancient times. Don’t worry, we’ll go through all 20 centuries of trigonometry in about 15 minutes. And, without noticing it, we’ll repeat a piece of geometry from 8th grade.
Let's draw a right triangle with sides a, b, c and angle X. Here it is.
Let me remind you that the sides that form a right angle are called legs. a and c– legs. There are two of them. The remaining side is called the hypotenuse. With– hypotenuse.
Triangle and triangle, just think! What to do with it? But the ancient people knew what to do! Let's repeat their actions. Let's measure the side V. In the figure, the cells are specially drawn, as in Unified State Exam assignments It happens. Side V equal to four cells. OK. Let's measure the side A. Three cells.
Now let's divide the length of the side A per side length V. Or, as they also say, let’s take the attitude A To V. a/v= 3/4.
On the contrary, you can divide V on A. We get 4/3. Can V divide by With. Hypotenuse With It’s impossible to count by cells, but it is equal to 5. We get high quality= 4/5. In short, you can divide the lengths of the sides by each other and get some numbers.
So what? What is the point of this interesting activity? None yet. A pointless exercise, to put it bluntly.)
Now let's do this. Let's enlarge the triangle. Let's extend the sides in and with, but so that the triangle remains rectangular. Corner X, of course, does not change. To see this, hover your mouse over the picture, or touch it (if you have a tablet). Parties a, b and c will turn into m, n, k, and, of course, the lengths of the sides will change.
But their relationship is not!
Attitude a/v was: a/v= 3/4, became m/n= 6/8 = 3/4. The relationships of other relevant parties are also won't change . You can change the lengths of the sides in a right triangle as you like, increase, decrease, without changing the angle x – the relationship between the relevant parties will not change . You can check it, or you can take the ancient people’s word for it.
But this is already very important! The ratios of the sides in a right triangle do not depend in any way on the lengths of the sides (at the same angle). This is so important that the relationship between the parties has earned its own special name. Your names, so to speak.) Meet me.
What is the sine of angle x ? This is the ratio of the opposite side to the hypotenuse:
sinx = a/c
What is the cosine of the angle x ? This is the ratio of the adjacent leg to the hypotenuse:
Withosx= high quality
What is tangent x ? This is the ratio of the opposite side to the adjacent side:
tgx =a/v
What is the cotangent of angle x ? This is the ratio of the adjacent side to the opposite:
ctgx = v/a
It's very simple. Sine, cosine, tangent and cotangent are some numbers. Dimensionless. Just numbers. Each angle has its own.
Why am I repeating everything so boringly? Then what is this need to remember. It's important to remember. Memorization can be made easier. Is the phrase “Let’s start from afar…” familiar? So start from afar.
Sinus angle is a ratio distant from the leg angle to the hypotenuse. Cosine– the ratio of the neighbor to the hypotenuse.
Tangent angle is a ratio distant from the leg angle to the near one. Cotangent- vice versa.
It's easier, right?
Well, if you remember that in tangent and cotangent there are only legs, and in sine and cosine the hypotenuse appears, then everything will become quite simple.
This whole glorious family - sine, cosine, tangent and cotangent is also called trigonometric functions.
Now a question for consideration.
Why do we say sine, cosine, tangent and cotangent corner? We are talking about the relationship between the parties, like... What does it have to do with it? corner?
Let's look at the second picture. Exactly the same as the first one.
Hover your mouse over the picture. I changed the angle X. Increased it from x to x. All relationships have changed! Attitude a/v was 3/4, and the corresponding ratio t/v became 6/4.
And all other relationships became different!
Therefore, the ratios of the sides do not depend in any way on their lengths (at one angle x), but depend sharply on this very angle! And only from him. Therefore, the terms sine, cosine, tangent and cotangent refer to corner. The angle here is the main one.
It must be clearly understood that the angle is inextricably linked with its trigonometric functions. Each angle has its own sine and cosine. And almost everyone has their own tangent and cotangent. This is important. It is believed that if we are given an angle, then its sine, cosine, tangent and cotangent we know ! And vice versa. Given a sine, or any other trigonometric function, it means we know the angle.
There are special tables where for each angle its trigonometric functions are described. They are called Bradis tables. They were compiled a very long time ago. When there were no calculators or computers yet...
Of course, it is impossible to remember the trigonometric functions of all angles. You are required to know them only for a few angles, more on this later. But the spell I know an angle, which means I know its trigonometric functions” - always works!
So we repeated a piece of geometry from 8th grade. Do we need it for the Unified State Exam? Necessary. Here is a typical problem from the Unified State Exam. To solve this problem, 8th grade is enough. Given picture:
All. There is no more data. We need to find the length of the side of the aircraft.
The cells do not help much, the triangle is somehow incorrectly positioned.... On purpose, I guess... From the information there is the length of the hypotenuse. 8 cells. For some reason, the angle was given.
This is where you need to immediately remember about trigonometry. There is an angle, which means we know all its trigonometric functions. Which of the four functions should we use? Let's see, what do we know? We know the hypotenuse and the angle, but we need to find adjacent catheter to this corner! It’s clear, the cosine needs to be put into action! Here we go. We simply write, by the definition of cosine (the ratio adjacent leg to hypotenuse):
cosC = BC/8
Our angle C is 60 degrees, its cosine is 1/2. You need to know this, without any tables! So:
1/2 = BC/8
Elementary linear equation. Unknown – Sun. For those who have forgotten how to solve equations, follow the link, the rest solve:
BC = 4
When ancient people realized that each angle has its own set of trigonometric functions, they had a reasonable question. Are sine, cosine, tangent and cotangent somehow related to each other? So that knowing one angle function, you can find the others? Without calculating the angle itself?
They were so restless...)
Relationship between trigonometric functions of one angle.
Of course, the sine, cosine, tangent and cotangent of the same angle are related to each other. Any connection between expressions is given in mathematics by formulas. In trigonometry there are a colossal number of formulas. But here we will look at the most basic ones. These formulas are called: basic trigonometric identities. Here they are:
You need to know these formulas thoroughly. Without them, there is generally nothing to do in trigonometry. Three more auxiliary identities follow from these basic identities:
I warn you right away that the last three formulas quickly fall out of your memory. For some reason.) You can, of course, derive these formulas from the first three. But, in difficult times... You understand.)
In standard problems, like the ones below, there is a way to avoid these forgettable formulas. AND dramatically reduce errors due to forgetfulness, and in calculations too. This practice is in Section 555, lesson "Relationships between trigonometric functions of the same angle."
In what tasks and how are the basic trigonometric identities used? The most popular task is to find some angle function if another is given. In the Unified State Examination such a task is present from year to year.) For example:
Find the value of sinx if x is an acute angle and cosx=0.8.
The task is almost elementary. We are looking for a formula that contains sine and cosine. Here is the formula:
sin 2 x + cos 2 x = 1
We substitute here a known value, namely 0.8 instead of cosine:
sin 2 x + 0.8 2 = 1
Well, we count as usual:
sin 2 x + 0.64 = 1
sin 2 x = 1 - 0.64
That's practically all. We have calculated the square of the sine, all that remains is to extract the square root and the answer is ready! The root of 0.36 is 0.6.
The task is almost elementary. But the word “almost” is there for a reason... The fact is that the answer sinx= - 0.6 is also suitable... (-0.6) 2 will also be 0.36.
There are two different answers. And you need one. The second one is incorrect. How to be!? Yes, as usual.) Read the assignment carefully. For some reason it says:... if x is an acute angle... And in tasks, every word has a meaning, yes... This phrase is additional information for the solution.
An acute angle is an angle less than 90°. And at such corners All trigonometric functions - sine, cosine, and tangent with cotangent - positive. Those. We simply discard the negative answer here. We have the right.
Actually, eighth graders don’t need such subtleties. They only work with right triangles, where the corners can only be acute. And they don’t know, happy ones, that there are both negative angles and angles of 1000°... And all these terrible angles have their own trigonometric functions, both plus and minus...
But for high school students, without taking into account the sign - no way. Much knowledge multiplies sorrows, yes...) And for the correct solution, additional information is necessarily present in the task (if it is necessary). For example, it can be given by the following entry:
Or some other way. You will see in the examples below.) To solve such examples you need to know which quarter does it fall into? specified angle x and what is the sign of the desired trigonometric function in this quadrant.
These basics of trigonometry are discussed in the lessons on what a trigonometric circle is, the measurement of angles on this circle, the radian measure of an angle. Sometimes you need to know the table of sines, cosines of tangents and cotangents.
So, let's note the most important thing:
1. Remember the definitions of sine, cosine, tangent and cotangent. It will be very useful.
2. We clearly understand: sine, cosine, tangent and cotangent are tightly connected with angles. We know one thing, which means we know another.
3. We clearly understand: sine, cosine, tangent and cotangent of one angle are related to each other by basic trigonometric identities. We know one function, which means we can (if we have the necessary additional information) calculate all the others.
Now let’s decide, as usual. First, tasks in the scope of 8th grade. But high school students can do it too...)
1. Calculate the value of tgA if ctgA = 0.4.
2. β is an angle in a right triangle. Find the value of tanβ if sinβ = 12/13.
3. Determine the sine of the acute angle x if tgх = 4/3.
4. Find the meaning of the expression:
6sin 2 5° - 3 + 6cos 2 5°
5. Find the meaning of the expression:
(1-cosx)(1+cosx), if sinx = 0.3
Answers (separated by semicolons, in disarray):
0,09; 3; 0,8; 2,4; 2,5
Did it work? Great! Eighth graders can already go get their A's.)
Didn't everything work out? Tasks 2 and 3 are somehow not very good...? No problem! There is one beautiful technique for such tasks. Everything can be solved practically without formulas at all! Well, therefore, without errors. This technique is described in the lesson: “Relationships between trigonometric functions of one angle” in Section 555. All other tasks are also dealt with there.
These were problems Unified State Exam type, but in a stripped down version. Unified State Exam - light). And now almost the same tasks, but in a full-fledged format. For knowledge-burdened high school students.)
6. Find the value of tanβ if sinβ = 12/13, and
7. Determine sinх if tgх = 4/3, and x belongs to the interval (- 540°; - 450°).
8. Find the value of the expression sinβ cosβ if ctgβ = 1.
Answers (in disarray):
0,8; 0,5; -2,4.
Here in problem 6 the angle is not specified very clearly... But in problem 8 it is not specified at all! This is on purpose). Additional information not only taken from the task, but also from the head.) But if you decide, one correct task is guaranteed!
What if you haven't decided? Hmm... Well, Section 555 will help here. There the solutions to all these tasks are described in detail, it is difficult not to understand.
This lesson provides a very limited understanding of trigonometric functions. Within 8th grade. And the elders still have questions...
For example, if the angle X(look at the second picture on this page) - make it stupid!? The triangle will completely fall apart! So what should we do? There will be no leg, no hypotenuse... The sine has disappeared...
If ancient people had not found a way out of this situation, we would not have cell phones, TV, or electricity now. Yes, yes! Theoretical basis all these things without trigonometric functions are zero without a stick. But the ancient people did not disappoint. How they got out is in the next lesson.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
Knowing one of the legs in a right triangle, you can find the second leg and hypotenuse using trigonometric ratios - sine and tangent of a known angle. Since the ratio of the leg opposite the angle to the hypotenuse is equal to the sine of this angle, therefore, to find the hypotenuse, you need to divide the leg by the sine of the angle. a/c=sinα c=a/sinα
The second leg can be found from the tangent of a known angle, as the ratio of the known leg to the tangent. a/b=tanα b=a/tanα
To calculate the unknown angle in a right triangle, you need to subtract the angle α from 90 degrees. β=90°-α
The perimeter and area of a right triangle can be expressed in terms of the leg and the angle opposite it by substituting the previously obtained expressions for the second leg and the hypotenuse into the formulas. P=a+b+c=a+a/tanα +a/sinα =a tanα sinα+a sinα+a tanα S=ab/2=a^2/( 2 tanα)
You can also calculate the height through trigonometric ratios, but in the internal right triangle with side a, which it forms. To do this, you need to multiply side a, as the hypotenuse of such a triangle, by the sine of angle β or cosine α, since according to trigonometric identities they are equivalent. (Fig. 79.2) h=a cosα
The median of the hypotenuse is equal to half the hypotenuse or the known leg a divided by two sines α. To find the medians of the legs, we reduce the formulas to the corresponding form for the known sides and angles. (Fig.79.3) m_с=c/2=a/(2 sinα) m_b=√(2a^2+2c^2-b^2)/2=√(2a^2+2a^2+2b^ 2-b^2)/2=√(4a^2+b^2)/2=√(4a^2+a^2/tan^2α)/2=(a√(4 tan^2 α+1))/(2 tanα) m_a=√(2c^2+2b^2-a^2)/2=√(2a^2+2b^2+2b^2-a^2)/ 2=√(4b^2+a^2)/2=√(4b^2+c^2-b^2)/2=√(3 a^2/tan^2α +a^2/sin ^2α)/2=√((3a^2 sin^2α+a^2 tan^2α)/(tan^2α sin^2α))/2=(a√( 3 sin^2α+tan^2α))/(2 tanα sinα)
Since the bisector of a right angle in a triangle is the product of two sides and the root of two, divided by the sum of these sides, then replacing one of the legs with the ratio of the known leg to the tangent, we obtain the following expression. Similarly, by substituting the ratio into the second and third formulas, you can calculate the bisectors of the angles α and β. (Fig.79.4) l_с=(a a/tanα √2)/(a+a/tanα)=(a^2 √2)/(a tanα+a)=(a√2)/ (tanα+1) l_a=√(bc(a+b+c)(b+c-a))/(b+c)=√(bc((b+c)^2-a^2))/ (b+c)=√(bc(b^2+2bc+c^2-a^2))/(b+c)=√(bc(b^2+2bc+b^2))/(b +c)=√(bc(2b^2+2bc))/(b+c)=(b√(2c(b+c)))/(b+c)=(a/tanα √(2c (a/tanα +c)))/(a/tanα +c)=(a√(2c(a/tanα +c)))/(a+c tanα) l_b=√ (ac(a+b+c)(a+c-b))/(a+c)=(a√(2c(a+c)))/(a+c)=(a√(2c(a+a /sinα)))/(a+a/sinα)=(a sinα √(2c(a+a/sinα)))/(a sinα+a)
The middle line runs parallel to one of the sides of the triangle, while forming another similar right-angled triangle with the same angles, in which all sides are half the size of the original one. Based on this, the middle lines can be found using the following formulas, knowing only the leg and the angle opposite it. (Fig.79.7) M_a=a/2 M_b=b/2=a/(2 tanα) M_c=c/2=a/(2 sinα)
The radius of the inscribed circle is equal to the difference between the legs and the hypotenuse divided by two, and to find the radius of the inscribed circle, you need to divide the hypotenuse by two. We replace the second leg and hypotenuse with the ratios of leg a to sine and tangent, respectively. (Fig. 79.5, 79.6) r=(a+b-c)/2=(a+a/tanα -a/sinα)/2=(a tanα sinα+a sinα-a tanα)/(2 tanα sinα) R=c/2=a/2sinα
What is sine, cosine, tangent, cotangent of an angle will help you understand a right triangle.
What are the sides of a right triangle called? That's right, hypotenuse and legs: the hypotenuse is the side that lies opposite the right angle (in our example this is the side \(AC\)); legs are the two remaining sides \(AB\) and \(BC\) (those adjacent to right angle), and, if we consider the legs relative to the angle \(BC\), then the leg \(AB\) is the adjacent leg, and the leg \(BC\) is the opposite. So, now let’s answer the question: what are sine, cosine, tangent and cotangent of an angle?
Sine of angle– this is the ratio of the opposite (distant) leg to the hypotenuse.
In our triangle:
\[ \sin \beta =\dfrac(BC)(AC) \]
Cosine of angle– this is the ratio of the adjacent (close) leg to the hypotenuse.
In our triangle:
\[ \cos \beta =\dfrac(AB)(AC) \]
Tangent of the angle– this is the ratio of the opposite (distant) side to the adjacent (close).
In our triangle:
\[ tg\beta =\dfrac(BC)(AB) \]
Cotangent of angle– this is the ratio of the adjacent (close) leg to the opposite (far).
In our triangle:
\[ ctg\beta =\dfrac(AB)(BC) \]
These definitions are necessary remember! To make it easier to remember which leg to divide into what, you need to clearly understand that in tangent And cotangent only the legs sit, and the hypotenuse appears only in sinus And cosine. And then you can come up with a chain of associations. For example, this one:
Cosine→touch→touch→adjacent;
Cotangent→touch→touch→adjacent.
First of all, you need to remember that sine, cosine, tangent and cotangent as the ratios of the sides of a triangle do not depend on the lengths of these sides (at the same angle). Don't believe me? Then make sure by looking at the picture:
Consider, for example, the cosine of the angle \(\beta \) . By definition, from a triangle \(ABC\) : \(\cos \beta =\dfrac(AB)(AC)=\dfrac(4)(6)=\dfrac(2)(3) \), but we can calculate the cosine of the angle \(\beta \) from the triangle \(AHI \) : \(\cos \beta =\dfrac(AH)(AI)=\dfrac(6)(9)=\dfrac(2)(3) \). You see, the lengths of the sides are different, but the value of the cosine of one angle is the same. Thus, the values of sine, cosine, tangent and cotangent depend solely on the magnitude of the angle.
If you understand the definitions, then go ahead and consolidate them!
For the triangle \(ABC \) shown in the figure below, we find \(\sin \ \alpha ,\ \cos \ \alpha ,\ tg\ \alpha ,\ ctg\ \alpha \).
\(\begin(array)(l)\sin \ \alpha =\dfrac(4)(5)=0.8\\\cos \ \alpha =\dfrac(3)(5)=0.6\\ tg\ \alpha =\dfrac(4)(3)\\ctg\ \alpha =\dfrac(3)(4)=0.75\end(array) \)
Well, did you get it? Then try it yourself: calculate the same for the angle \(\beta \) .
Answers: \(\sin \ \beta =0.6;\ \cos \ \beta =0.8;\ tg\ \beta =0.75;\ ctg\ \beta =\dfrac(4)(3) \).
Unit (trigonometric) circle
Understanding the concepts of degrees and radians, we considered a circle with a radius equal to \(1\) . Such a circle is called single. It will be very useful when studying trigonometry. Therefore, let's look at it in a little more detail.
As you can see, this circle is constructed in the Cartesian coordinate system. The radius of the circle is equal to one, while the center of the circle lies at the origin of coordinates, the initial position of the radius vector is fixed along the positive direction of the \(x\) axis (in our example, this is the radius \(AB\)).
Each point on the circle corresponds to two numbers: the coordinate along the \(x\) axis and the coordinate along the \(y\) axis. What are these coordinate numbers? And in general, what do they have to do with the topic at hand? To do this, we need to remember about the considered right triangle. In the figure above, you can see two whole right triangles. Consider the triangle \(ACG\) . It is rectangular because \(CG\) is perpendicular to the \(x\) axis.
What is \(\cos \ \alpha \) from triangle \(ACG \)? That's right \(\cos \ \alpha =\dfrac(AG)(AC) \). In addition, we know that \(AC\) is the radius unit circle, which means \(AC=1\) . Let's substitute this value into our formula for cosine. Here's what happens:
\(\cos \ \alpha =\dfrac(AG)(AC)=\dfrac(AG)(1)=AG \).
What is \(\sin \ \alpha \) from the triangle \(ACG \) equal to? Well of course \(\sin \alpha =\dfrac(CG)(AC)\)! Substitute the value of the radius \(AC\) into this formula and get:
\(\sin \alpha =\dfrac(CG)(AC)=\dfrac(CG)(1)=CG \)
So, can you tell what coordinates the point \(C\) belonging to the circle has? Well, no way? What if you realize that \(\cos \ \alpha \) and \(\sin \alpha \) are just numbers? What coordinate does \(\cos \alpha \) correspond to? Well, of course, the coordinate \(x\)! And what coordinate does \(\sin \alpha \) correspond to? That's right, coordinate \(y\)! So the point \(C(x;y)=C(\cos \alpha ;\sin \alpha) \).
What then are \(tg \alpha \) and \(ctg \alpha \) equal to? That’s right, let’s use the corresponding definitions of tangent and cotangent and get that \(tg \alpha =\dfrac(\sin \alpha )(\cos \alpha )=\dfrac(y)(x) \), A \(ctg \alpha =\dfrac(\cos \alpha )(\sin \alpha )=\dfrac(x)(y) \).
What if the angle is larger? For example, like in this picture:
What has changed in in this example? Let's figure it out. To do this, let's turn again to a right triangle. Consider a right triangle \(((A)_(1))((C)_(1))G \) : angle (as adjacent to angle \(\beta \) ). What is the value of sine, cosine, tangent and cotangent for an angle \(((C)_(1))((A)_(1))G=180()^\circ -\beta \ \)? That's right, we adhere to the corresponding definitions of trigonometric functions:
\(\begin(array)(l)\sin \angle ((C)_(1))((A)_(1))G=\dfrac(((C)_(1))G)(( (A)_(1))((C)_(1)))=\dfrac(((C)_(1))G)(1)=((C)_(1))G=y; \\\cos \angle ((C)_(1))((A)_(1))G=\dfrac(((A)_(1))G)(((A)_(1)) ((C)_(1)))=\dfrac(((A)_(1))G)(1)=((A)_(1))G=x;\\tg\angle ((C )_(1))((A)_(1))G=\dfrac(((C)_(1))G)(((A)_(1))G)=\dfrac(y)( x);\\ctg\angle ((C)_(1))((A)_(1))G=\dfrac(((A)_(1))G)(((C)_(1 ))G)=\dfrac(x)(y)\end(array) \)
Well, as you can see, the value of the sine of the angle still corresponds to the coordinate \(y\) ; the value of the cosine of the angle – coordinate \(x\) ; and the values of tangent and cotangent to the corresponding ratios. Thus, these relations apply to any rotation of the radius vector.
It has already been mentioned that the initial position of the radius vector is along the positive direction of the \(x\) axis. So far we have rotated this vector counterclockwise, but what happens if we rotate it clockwise? Nothing extraordinary, you will also get an angle of a certain value, but only it will be negative. Thus, when rotating the radius vector counterclockwise, we get positive angles, and when rotating clockwise – negative.
So, we know that the whole revolution of the radius vector around the circle is \(360()^\circ \) or \(2\pi \) . Is it possible to rotate the radius vector by \(390()^\circ \) or by \(-1140()^\circ \)? Well, of course you can! In the first case, \(390()^\circ =360()^\circ +30()^\circ \), thus, the radius vector will make one full revolution and stop at the position \(30()^\circ \) or \(\dfrac(\pi )(6) \) .
In the second case, \(-1140()^\circ =-360()^\circ \cdot 3-60()^\circ \), that is, the radius vector will make three full turns and stop at the position \(-60()^\circ \) or \(-\dfrac(\pi )(3) \) .
Thus, from the above examples we can conclude that angles that differ by \(360()^\circ \cdot m \) or \(2\pi \cdot m \) (where \(m \) is any integer ), correspond to the same position of the radius vector.
The figure below shows the angle \(\beta =-60()^\circ \) . The same image corresponds to the corner \(-420()^\circ ,-780()^\circ ,\ 300()^\circ ,660()^\circ \) etc. This list can be continued indefinitely. All these angles can be written by the general formula \(\beta +360()^\circ \cdot m\) or \(\beta +2\pi \cdot m \) (where \(m \) is any integer)
\(\begin(array)(l)-420()^\circ =-60+360\cdot (-1);\\-780()^\circ =-60+360\cdot (-2); \\300()^\circ =-60+360\cdot 1;\\660()^\circ =-60+360\cdot 2.\end(array) \)
Now, knowing the definitions of the basic trigonometric functions and using the unit circle, try to answer what the values are:
\(\begin(array)(l)\sin \ 90()^\circ =?\\\cos \ 90()^\circ =?\\\text(tg)\ 90()^\circ =? \\\text(ctg)\ 90()^\circ =?\\\sin \ 180()^\circ =\sin \ \pi =?\\\cos \ 180()^\circ =\cos \ \pi =?\\\text(tg)\ 180()^\circ =\text(tg)\ \pi =?\\\text(ctg)\ 180()^\circ =\text(ctg)\ \pi =?\\\sin \ 270()^\circ =?\\\cos \ 270()^\circ =?\\\text(tg)\ 270()^\circ =?\\\text (ctg)\ 270()^\circ =?\\\sin \ 360()^\circ =?\\\cos \ 360()^\circ =?\\\text(tg)\ 360()^ \circ =?\\\text(ctg)\ 360()^\circ =?\\\sin \ 450()^\circ =?\\\cos \ 450()^\circ =?\\\text (tg)\ 450()^\circ =?\\\text(ctg)\ 450()^\circ =?\end(array) \)
Here's a unit circle to help you:
Having difficulties? Then let's figure it out. So we know that:
\(\begin(array)(l)\sin \alpha =y;\\cos\alpha =x;\\tg\alpha =\dfrac(y)(x);\\ctg\alpha =\dfrac(x )(y).\end(array)\)
From here, we determine the coordinates of the points corresponding to certain angle measures. Well, let's start in order: the corner in \(90()^\circ =\dfrac(\pi )(2) \) corresponds to a point with coordinates \(\left(0;1 \right) \) , therefore:
\(\sin 90()^\circ =y=1 \) ;
\(\cos 90()^\circ =x=0 \) ;
\(\text(tg)\ 90()^\circ =\dfrac(y)(x)=\dfrac(1)(0)\Rightarrow \text(tg)\ 90()^\circ \)- does not exist;
\(\text(ctg)\ 90()^\circ =\dfrac(x)(y)=\dfrac(0)(1)=0 \).
Further, adhering to the same logic, we find out that the corners in \(180()^\circ ,\ 270()^\circ ,\ 360()^\circ ,\ 450()^\circ (=360()^\circ +90()^\circ)\ \ ) correspond to points with coordinates \(\left(-1;0 \right),\text( )\left(0;-1 \right),\text( )\left(1;0 \right),\text( )\left(0 ;1 \right) \), respectively. Knowing this, it is easy to determine the values of trigonometric functions at the corresponding points. Try it yourself first, and then check the answers.
Answers:
\(\displaystyle \sin \180()^\circ =\sin \ \pi =0\)
\(\displaystyle \cos \180()^\circ =\cos \ \pi =-1\)
\(\text(tg)\ 180()^\circ =\text(tg)\ \pi =\dfrac(0)(-1)=0 \)
\(\text(ctg)\ 180()^\circ =\text(ctg)\ \pi =\dfrac(-1)(0)\Rightarrow \text(ctg)\ \pi \)- does not exist
\(\sin \270()^\circ =-1\)
\(\cos \ 270()^\circ =0 \)
\(\text(tg)\ 270()^\circ =\dfrac(-1)(0)\Rightarrow \text(tg)\ 270()^\circ \)- does not exist
\(\text(ctg)\ 270()^\circ =\dfrac(0)(-1)=0 \)
\(\sin \360()^\circ =0\)
\(\cos \360()^\circ =1\)
\(\text(tg)\ 360()^\circ =\dfrac(0)(1)=0 \)
\(\text(ctg)\ 360()^\circ =\dfrac(1)(0)\Rightarrow \text(ctg)\ 2\pi \)- does not exist
\(\sin \ 450()^\circ =\sin \ \left(360()^\circ +90()^\circ \right)=\sin \ 90()^\circ =1 \)
\(\cos \ 450()^\circ =\cos \ \left(360()^\circ +90()^\circ \right)=\cos \ 90()^\circ =0 \)
\(\text(tg)\ 450()^\circ =\text(tg)\ \left(360()^\circ +90()^\circ \right)=\text(tg)\ 90() ^\circ =\dfrac(1)(0)\Rightarrow \text(tg)\ 450()^\circ \)- does not exist
\(\text(ctg)\ 450()^\circ =\text(ctg)\left(360()^\circ +90()^\circ \right)=\text(ctg)\ 90()^ \circ =\dfrac(0)(1)=0 \).
Thus, we can make the following table:
There is no need to remember all these values. It is enough to remember the correspondence between the coordinates of points on the unit circle and the values of trigonometric functions:
\(\left. \begin(array)(l)\sin \alpha =y;\\cos \alpha =x;\\tg \alpha =\dfrac(y)(x);\\ctg \alpha =\ dfrac(x)(y).\end(array) \right\)\ \text(You must remember or be able to output it!! \) !}
But the values of the trigonometric functions of angles in and \(30()^\circ =\dfrac(\pi )(6),\ 45()^\circ =\dfrac(\pi )(4)\) given in the table below, you must remember:
Don’t be scared, now we’ll show you one example of a fairly simple memorization of the corresponding values:
To use this method, it is vital to remember the sine values for all three measures of angle ( \(30()^\circ =\dfrac(\pi )(6),\ 45()^\circ =\dfrac(\pi )(4),\ 60()^\circ =\dfrac(\pi )(3)\)), as well as the value of the tangent of the angle in \(30()^\circ \) . Knowing these \(4\) values, it is quite simple to restore the entire table - the cosine values are transferred in accordance with the arrows, that is:
\(\begin(array)(l)\sin 30()^\circ =\cos \ 60()^\circ =\dfrac(1)(2)\ \ \\\sin 45()^\circ = \cos \ 45()^\circ =\dfrac(\sqrt(2))(2)\\\sin 60()^\circ =\cos \ 30()^\circ =\dfrac(\sqrt(3 ))(2)\ \end(array) \)
\(\text(tg)\ 30()^\circ \ =\dfrac(1)(\sqrt(3)) \), knowing this, you can restore the values for \(\text(tg)\ 45()^\circ , \text(tg)\ 60()^\circ \). The numerator "\(1 \)" will correspond to \(\text(tg)\ 45()^\circ \ \) and the denominator "\(\sqrt(\text(3)) \)" will correspond to \(\text (tg)\ 60()^\circ \ \) . Cotangent values are transferred in accordance with the arrows indicated in the figure. If you understand this and remember the diagram with the arrows, then it will be enough to remember only \(4\) values from the table.
Coordinates of a point on a circle
Is it possible to find a point (its coordinates) on a circle, knowing the coordinates of the center of the circle, its radius and angle of rotation? Well, of course you can! Let's derive a general formula for finding the coordinates of a point. For example, here is a circle in front of us:
We are given that point \(K(((x)_(0));((y)_(0)))=K(3;2) \)- center of the circle. The radius of the circle is \(1.5\) . It is necessary to find the coordinates of the point \(P\) obtained by rotating the point \(O\) by \(\delta \) degrees.
As can be seen from the figure, the coordinate \(x\) of the point \(P\) corresponds to the length of the segment \(TP=UQ=UK+KQ\) . The length of the segment \(UK\) corresponds to the coordinate \(x\) of the center of the circle, that is, it is equal to \(3\) . The length of the segment \(KQ\) can be expressed using the definition of cosine:
\(\cos \ \delta =\dfrac(KQ)(KP)=\dfrac(KQ)(r)\Rightarrow KQ=r\cdot \cos \ \delta \).
Then we have that for the point \(P\) the coordinate \(x=((x)_(0))+r\cdot \cos \ \delta =3+1.5\cdot \cos \ \delta \).
Using the same logic, we find the value of the y coordinate for the point \(P\) . Thus,
\(y=((y)_(0))+r\cdot \sin \ \delta =2+1.5\cdot \sin \delta \).
So, in general view coordinates of points are determined by the formulas:
\(\begin(array)(l)x=((x)_(0))+r\cdot \cos \ \delta \\y=((y)_(0))+r\cdot \sin \ \delta \end(array) \), Where
\(((x)_(0)),((y)_(0)) \) - coordinates of the center of the circle,
\(r\) - radius of the circle,
\(\delta \) - rotation angle of the vector radius.
As you can see, for the unit circle we are considering, these formulas are significantly reduced, since the coordinates of the center are equal to zero and the radius is equal to one:
\(\begin(array)(l)x=((x)_(0))+r\cdot \cos \ \delta =0+1\cdot \cos \ \delta =\cos \ \delta \\y =((y)_(0))+r\cdot \sin \ \delta =0+1\cdot \sin \ \delta =\sin \ \delta \end(array) \)
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The ratio of the opposite side to the hypotenuse is called sinus of an acute angle right triangle.
\sin \alpha = \frac(a)(c)
Cosine of an acute angle of a right triangle
The ratio of the adjacent leg to the hypotenuse is called cosine of an acute angle right triangle.
\cos \alpha = \frac(b)(c)
Tangent of an acute angle of a right triangle
The ratio of the opposite side to the adjacent side is called tangent of an acute angle right triangle.
tg \alpha = \frac(a)(b)
Cotangent of an acute angle of a right triangle
The ratio of the adjacent side to the opposite side is called cotangent of an acute angle right triangle.
ctg \alpha = \frac(b)(a)
Sine of an arbitrary angle
The ordinate of a point on the unit circle to which the angle \alpha corresponds is called sine of an arbitrary angle rotation \alpha .
\sin \alpha=y
Cosine of an arbitrary angle
The abscissa of a point on the unit circle to which the angle \alpha corresponds is called cosine of an arbitrary angle rotation \alpha .
\cos \alpha=x
Tangent of an arbitrary angle
The ratio of the sine of an arbitrary rotation angle \alpha to its cosine is called tangent of an arbitrary angle rotation \alpha .
tan \alpha = y_(A)
tg \alpha = \frac(\sin \alpha)(\cos \alpha)
Cotangent of an arbitrary angle
The ratio of the cosine of an arbitrary rotation angle \alpha to its sine is called cotangent of an arbitrary angle rotation \alpha .
ctg\alpha =x_(A)
ctg \alpha = \frac(\cos \alpha)(\sin \alpha)
An example of finding an arbitrary angle
If \alpha is some angle AOM, where M is a point on the unit circle, then
\sin \alpha=y_(M) , \cos \alpha=x_(M) , tg \alpha=\frac(y_(M))(x_(M)), ctg \alpha=\frac(x_(M))(y_(M)).
For example, if \angle AOM = -\frac(\pi)(4), then: the ordinate of point M is equal to -\frac(\sqrt(2))(2), abscissa is equal \frac(\sqrt(2))(2) and therefore
\sin \left (-\frac(\pi)(4) \right)=-\frac(\sqrt(2))(2);
\cos \left (\frac(\pi)(4) \right)=\frac(\sqrt(2))(2);
tg;
ctg \left (-\frac(\pi)(4) \right)=-1.
Table of values of sines of cosines of tangents of cotangents
The values of the main frequently occurring angles are given in the table:
| 0^(\circ) (0) | 30^(\circ)\left(\frac(\pi)(6)\right) | 45^(\circ)\left(\frac(\pi)(4)\right) | 60^(\circ)\left(\frac(\pi)(3)\right) | 90^(\circ)\left(\frac(\pi)(2)\right) | 180^(\circ)\left(\pi\right) | 270^(\circ)\left(\frac(3\pi)(2)\right) | 360^(\circ)\left(2\pi\right) | |
| \sin\alpha | 0 | \frac12 | \frac(\sqrt 2)(2) | \frac(\sqrt 3)(2) | 1 | 0 | −1 | 0 |
| \cos\alpha | 1 | \frac(\sqrt 3)(2) | \frac(\sqrt 2)(2) | \frac12 | 0 | −1 | 0 | 1 |
| tg\alpha | 0 | \frac(\sqrt 3)(3) | 1 | \sqrt3 | — | 0 | — | 0 |
| ctg\alpha | — | \sqrt3 | 1 | \frac(\sqrt 3)(3) | 0 | — | 0 | — |
