This article contains tables of sines, cosines, tangents and cotangents. First we will provide a table of basic values trigonometric functions, that is, a table of sines, cosines, tangents and cotangents of angles 0, 30, 45, 60, 90, ..., 360 degrees ( 0, π/6, π/4, π/3, π/2, …, 2π radian). After this, we will give a table of sines and cosines, as well as a table of tangents and cotangents by V. M. Bradis, and show how to use these tables when finding the values of trigonometric functions.
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Table of sines, cosines, tangents and cotangents for angles of 0, 30, 45, 60, 90, ... degrees
References.
- Algebra: Textbook for 9th grade. avg. school/Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova; Ed. S. A. Telyakovsky. - M.: Education, 1990. - 272 pp.: ill. - ISBN 5-09-002727-7
- Bashmakov M. I. Algebra and the beginnings of analysis: Textbook. for 10-11 grades. avg. school - 3rd ed. - M.: Education, 1993. - 351 p.: ill. - ISBN 5-09-004617-4.
- Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
- Gusev V. A., Mordkovich A. G. Mathematics (a manual for those entering technical schools): Proc. allowance.- M.; Higher school, 1984.-351 p., ill.
- Bradis V. M. Four-digit math tables: For general education. textbook establishments. - 2nd ed. - M.: Bustard, 1999.- 96 p.: ill. ISBN 5-7107-2667-2
Let's construct an arbitrary triangle inscribed in a circle. Let's denote it as ABC.
To prove the entire theorem, since the dimensions of the triangle are chosen arbitrarily, it is enough to prove that the ratio of one arbitrary side to the angle opposite it is equal to 2R. Let it be 2R = a / sin α, that is, if we take from the drawing 2R = BC / sin A.
Let us calculate the diameter BD for the circumscribed circle. The resulting triangle BCD is right-angled because its hypotenuse lies on the diameter of the circumscribed circle (the property of angles inscribed in a circle).
Since the angles inscribed in a circle based on the same arc are equal, then the angle CDB is either equal to angle CAB (if points A and D lie on the same side of line BC), or equal to π - CAB (otherwise).
Let us turn to the properties of trigonometric functions. Since sin(π − α) = sin α, the indicated options for constructing a triangle will still lead to the same result.
Let's calculate the value 2R = a / sin α, according to the drawing 2R = BC / sin A. To do this, replace sin A with the ratio of the corresponding sides right triangle.
2R = BC / sin A
2R = BC / (BC / DB)
2R = DB
And, since DB was constructed as the diameter of a circle, then the equality is satisfied.
Repeating the same reasoning for the other two sides of the triangle, we get: 
The sine theorem has been proven.
Theorem of sines
Note. This is part of a lesson with geometry problems (section theorem of sines). If you need to solve a geometry problem that is not here, write about it in the forum. In tasks, instead of the "square root" symbol, the sqrt() function is used, in which sqrt is the symbol square root, and the radical expression is indicated in brackets.
Theorem of sines:
The sides of a triangle are proportional to the sines of the opposite angles, or, in an expanded formulation:
a / sin α = b / sin β = c / sin γ = 2R
where R is the radius of the circumscribed circle
For the theory - the formulation and proof of the theorem, see in detail in the chapter "Theorem of Sines" .
Task
In triangle XYZ, angle X=30, angle Z=15. The perpendicular YQ to ZY divides the side XZ into parts XQ and QZ. Find XY if QZ = 1.5 m
Solution.
The height formed two right triangles XYQ and ZYQ.
To solve the problem, we will use the theorem of sines.
QZ / sin(QYZ) = QY / sin(QZY)
QZY = 15 degrees, Accordingly, QYZ = 180 - 90 - 15 = 75
Since the length of the altitude of the triangle is now known, we will find XY using the same theorem of sines.
QY / sin(30) = XY / sin(90)
Let's take into account table values some trigonometric functions:
- sine of 30 degrees is equal to sin(30) = 1 / 2
- the sine of 90 degrees is equal to sin(90) = 1
QY = XY sin (30)
3/2 (√3 - 1) / (√3 + 1) = 1/2 XY
XY = 3 (√3 - 1) / (√3 + 1) ≈ 0.8 m
Answer: 0.8 m or 3 (√3 - 1) / (√3 + 1)
Theorem of sines (part 2)
Note. This is part of a lesson with geometry problems (section theorem of sines). If you need to solve a geometry problem that is not here, write about it in the forum .
See the theory in detail in the chapter "Theorem of Sines" .
Task
Side AB of triangle ABC is 16 cm. Angle A is 30 degrees. Angle B is 105 degrees. Calculate the length of side BC. 
Solution.
According to the law of sines, the sides of a triangle are proportional to the sines of the opposite angles:
a / sin α = b / sin β = c / sin γ
Thus
BC / sin α = AB / sin γ
We find the size of angle C based on the fact that the sum of the angles of a triangle is equal to 180 degrees.
C = 180 - 30 -105 = 45 degrees.
Where:
BC / sin 30° = 16 / sin 45°
BC = 16 sin 30° / sin 45°
Referring to the table of trigonometric functions, we find:
BC = (16 * 1 / 2) / √2/2 = 16 / √2 ≈ 11.3 cm
Answer: 16 / √2
Task.
IN triangle ABC angle A = α, angle C = β, ВС = 7cm, ВН - height of the triangle.
Find AN 
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What is the cosine theorem? Imagine this... Pythagorean theorem for an arbitrary triangle.
The cosine theorem: formulation.
The cosine theorem states: square any side of the triangle equal to the sum the squares of the other two sides of the triangle minus twice the product of these sides and the cosine of the angle between them.
And now I’ll explain why this is so and what the Pythagorean theorem has to do with it.
After all, what does the Pythagorean theorem say?
What happens if, say, it’s spicy?
What if I'm stupid?
Now we’ll find out, or rather, we’ll first formulate it and then prove it.
So, for every (acute-angled, obtuse-angled and even rectangular!) triangle, the following is true: cosine theorem.
Cosine theorem:
What is and?
can be expressed from a triangle (rectangular!).
And here it is (from again).
Let's substitute:
We reveal:
We use what we have and... that's it!
2 Case: let.
So, that is, stupid.
And now, attention, the difference!
This is from, which is now outside, and
We remember that
(read the topic if you completely forgot why this is so).
So, that’s it! The difference is over!
As it was, that is:
Well, there is one last case left.
3 Case: let.
So, . But then the cosine theorem simply turns into the Pythagorean theorem:
In what problems is the cosine theorem useful?
Well, for example, if you have given two sides of a triangle and the angle between them, then you right away can you find a third party.
Or if you all three sides are given, then you will find it right away cosine any angle according to the formula
And even if you given two sides and an angle NOT between them, then the third side can also be found by solving quadratic equation. True, in this case, sometimes you get two answers and you need to figure out which one to choose, or leave both.
Try to use it and don’t be afraid - the cosine theorem is almost as easy to use as the Pythagorean theorem.
THEOREM OF COSINES. BRIEFLY ABOUT THE MAIN THINGS
Cosine theorem: The square of a side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of these sides and the cosine of the angle between them:
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successful completion Unified State Exam, for admission to college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who received good education, earn much more than those who did not receive it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because there is much more open before them more possibilities and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
You won't be asked for theory during the exam.
You will need solve problems against time.
And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.
It's like in sports - you need to repeat it many times to win for sure.
Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!
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Find problems and solve them!
Graduates who are preparing to take the Unified State Exam in mathematics and want to get enough high scores, must definitely master the principle of solving problems using the theorem of sines and cosines. Many years of practice show that similar tasks from the section “Geometry on a Plane” are mandatory part certification testing programs. Therefore, if one of your weak points are problems on the theorem of cosines and sines, we recommend that you definitely repeat the basic theory on this topic.
Prepare for the exam with the Shkolkovo educational portal
Exercising before passing the Unified State Exam, many graduates face the problem of finding basic theory required to solve practical problems on the application of the theorem of sines and cosines.
The textbook is not always at hand in right moment. And find necessary formulas sometimes it can be quite problematic even on the Internet.
Preparing for the certification test with educational portal“Shkolkovo” will be of the highest quality and efficiency. To make problems on the theorem of sines and cosines easy, we recommend brushing up on the entire theory on this topic. Our experts prepared this material based on extensive experience and presented it in in an understandable form. You can find it in the “Theoretical Information” section.
Knowledge of basic theorems and definitions is half the success when passing the certification test. Appropriate exercises allow you to hone your skills in solving examples. To find them, just go to the “Catalog” section on the Shkolkovo educational website. There is a large list of tasks various levels complexity, which is constantly supplemented and updated.
Students can complete problems on the theorems of sines and cosines, similar to those found in the Unified State Examination in mathematics, online, while in Moscow or any other Russian city.
If necessary, any exercise, for example, can be saved in the “Favorites” section. This will allow you to return to it in the future to once again analyze the algorithm for finding the correct answer and discuss it with a teacher at school or a tutor.
When studying triangles, the question of calculating the relationship between their sides and angles involuntarily arises. Geometry and Sines provides the most complete answer to solve this problem. In the abundance of various mathematical expressions and formulas, laws, theorems and rules, there are those that are distinguished by their extraordinary harmony, conciseness and simplicity of presentation of the meaning contained in them. The sine theorem is a shining example similar mathematical formulation. If in the verbal interpretation there is also a certain obstacle in understanding this mathematical rule, then when looking at mathematical formula everything immediately falls into place.
The first information about this theorem was discovered in the form of a proof in the framework of the mathematical work of Nasir ad-Din At-Tusi, dating back to the thirteenth century.
Moving closer to considering the ratio of sides and angles in any triangle, it is worth noting that the theorem of sines allows us to solve the mass mathematical problems, while this law geometry finds application in various types practical activities person.
The sine theorem itself states that any triangle is characterized by the proportionality of its sides to the sines of opposite angles. There is also a second part of this theorem, according to which the ratio of any side of a triangle to the sine opposite corner equal to that described around the triangle in question.
In formula form, this expression looks like
a/sinA = b/sinB = c/sinC = 2R
The sine theorem has a proof, which is offered in a wide variety of versions in various textbooks.
As an example, consider one of the proofs that explains the first part of the theorem. To do this, we set ourselves the goal of proving the correctness of the expression asinC= csinA.
IN arbitrary triangle ABC will construct the height BH. In one of the construction options, H will lie on the segment AC, and in the other outside it, depending on the size of the angles at the vertices of the triangles. In the first case, the height can be expressed in terms of the angles and sides of the triangle, as BH = a sinC and BH = c sinA, which is the required proof.
In the case when point H is outside the segment AC, we can obtain the following solutions:
VN = a sinC and VN = c sin(180-A)= c sinA;
or VN = a sin(180-C) = a sinC and VN = c sinA.
As you can see, regardless of the construction options, we arrive at the desired result.
The proof of the second part of the theorem will require us to draw a circle around the triangle. Using one of the altitudes of the triangle, for example B, we construct the diameter of the circle. We connect the resulting point on circle D to one of the altitudes of the triangle, let it be point A of the triangle.
If we consider the resulting triangles ABD and ABC, we will notice that the angles C and D are equal (they rest on the same arc). And given that angle A is equal to ninety degrees, then sin D = c/2R, or sin C = c/2R, which is what needed to be proven.
The sine theorem is the starting point for solving a wide range of various tasks. Its particular attractiveness lies in its practical application; as a consequence of the theorem, we get the opportunity to connect with each other the values of the sides of a triangle, opposite angles and the radius (diameter) of the circle described around the triangle. Simplicity and accessibility of the formula describing this mathematical expression, made it possible to widely use this theorem to solve problems using various mechanical calculating devices, tables, etc.), but even the advent of powerful computing devices in human service did not reduce the relevance of this theorem.
This theorem is not only included in the compulsory geometry course high school, but is also further used in some areas of practical activity.
