Formulation: Square side of triangle equal to the sum the squares of its two other sides minus twice the product of these sides and the cosine of the angle between them.
For arbitrary triangle ABC and his sides a,b and c (opposite to the corresponding vertices) this equality can be written for the other two sides:
The cosine theorem is used to solve triangles in two main situations:
1) When two sides and the angle between them are given, and you need to find the last side: 
2) When all three sides of a triangle are given, and you need to find its angles: 
Sometimes a math tutor recommends using the cosine theorem in a problem with two given sides and an angle that does not lie between them. In this case a) you will have to decide quadratic equation and select the length of the real side from the resulting roots. b) this situation is not typical for problems with the Unified State Exam in mathematics, since it does not always uniquely define a triangle. If the angle does not lie between the sides, then using a compass and a ruler you can construct two different triangles with such elements.
The cosine theorem is sometimes called the extended Pythagorean theorem or a generalization of the Pythagorean theorem, because at an angle of 90 degrees, the above equalities yield . Like any generalization, it is much more universal and effective than a particular case and applies to more real situations (as opposed to artificial problems GIA and Unified State Examination in mathematics, designed for the 8th grade program).
All the proofs I know involve vectors and coordinates. In Atanasyan’s textbook, it is carried out through the coordinates of points, and in Pogorelov’s textbook, the concept of “scalar product of vectors” is used. Let's carry out the proof according to Atanasyan. It seems to me that it is most suitable for a math tutor to work with, since it has less dependence on neighboring topics. 
Let's prove equality for the side A and angle A. To do this, we introduce a coordinate system as shown in the figure (the Ox axis is directed along the side AC). Point B will then receive coordinates B (cCosA;cSinA). This is the only fact that is difficult for a weak or average student, which a mathematics tutor working from Atanasyan’s textbook should separately consider. It is often complicated due to the fact that it is not supported by a sufficient number of tasks in the program and is not used after studying the cosine theorem. In the case of a given arrangement of points (when it is acute), a math tutor just needs to refer to the definition of cosine and sine acute angle V right triangle x with dotted sides.
The further proof is based on algebraic and trigonometric calculations. To them you need to add knowledge of the formula distance between two points.
We apply the abbreviated multiplication formula to the square of the sum:
We put it out of brackets: . We use the basic trigonometric identity and we get
and in the end
A math tutor can show an inquisitive student a rare proof of the cosine theorem. 
We will spend in triangle ABC height BH and write AB=AN+HB or c=bCosA+aCosB. If angle B is obtuse, then AB = AN-NV and taking into account the fact that the cosines adjacent corners are opposite, we again obtain the equality c=bCosA+aCosB. Therefore, it does not depend on the type of triangle. Let's write similar formulas for a and b:
a=cCosB+bCosC and b=aCosC+cCosA. Multiplying them by a and b respectively and subtracting from their sum the equality c=bCosA+aCosB we get the equality
The torema of cosines allows us to explain a property of the diagonals of a parallelogram that is very useful in practice: 
The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of its sides. In order to verify this, it is enough to write down the cosine theorem for each diagonal and add the resulting equalities.
Examples of problems in which one way or another you can (or need) to use the cosine theorem:
1) In a triangle with sides 2,3 and 4, find the length of the median drawn to the longer side.
2) In the same triangle, find the length of the bisector drawn to the longer side.
3) In triangle ABC, the segment connecting the midpoints AB and BC is equal to 3 dm, side AB is equal to 7 dm, angle C is equal to . Find the sun.
4) Center of a circle inscribed in a rectangular triangle ABC with a right angle C is at a distance from vertices A and B. Place the legs of the triangle.
Full preparation for the Unified State Exam in mathematics is impossible without solving problems on the cosine theorem. IN version of the Unified State Exam it can be found either in room B4 or in C4. Gradually I will transfer to the page interesting tasks C4 from my didactic base and from trial exams. Tutors, do not forget that in the GIA, as in the Unified State Exam, the cosine theorem can appear in both the first and second parts of the variant.
Kolpakov Alexander Nikolaevich,
math tutor in Moscow. Preparation for the Unified State Exam
Not all schoolchildren, and especially adults, know that the cosine theorem is directly related to the Pythagorean theorem. More precisely, the latter is a special case of the former. This point, as well as two ways to prove the cosine theorem, will help you become more knowledgeable person. In addition, practice in expressing quantities from initial expressions develops well logical thinking. The long formula of the theorem being studied will definitely force you to work hard and improve.
Starting a conversation: introducing notation
This theorem is formulated and proven for an arbitrary triangle. Therefore, it can always be used, in any situation, if two sides are given, and in some cases three, and an angle, and not necessarily between them. Whatever the type of triangle, the theorem will always work.
And now about the designation of quantities in all expressions. It is better to agree right away, so as not to have to explain several times later. The following table has been compiled for this purpose.
Formulation and mathematical notation
So, the cosine theorem is formulated in the following way:
The square of a side of any triangle is equal to the sum of the squares of its two other sides minus twice the product of these same sides and the cosine of the angle lying between them.
Of course, it is long, but if you understand its essence, it will be easy to remember. You can even imagine drawing a triangle. It's always easier to remember visually.
The formula of this theorem will look like this:
A little long, but everything is logical. If you look a little more closely, you can see that the letters are repeated, which means it’s not difficult to remember.
Common proof of the theorem
Since it is true for all triangles, you can choose any of the types for reasoning. Let it be a figure with all sharp angles. Consider an arbitrary acute triangle, whose angle C is greater than angle B. From the vertex with this large angle you need to lower a perpendicular to the opposite side. The height drawn will divide the triangle into two rectangular ones. This will be required for proof.
The side will be divided into two segments: x, y. They need to be expressed through known quantities. The part that ends up in a triangle with a hypotenuse equal to b will be expressed through the notation:
x = b * cos A.
The other will be equal to this difference:
y = c - in * cos A.
Now you need to write down the Pythagorean theorem for the two resulting right triangles, taking the height as the unknown value. These formulas will look like this:
n 2 = in 2 - (in * cos A) 2,
n 2 = a 2 - (c - b * cos A) 2.
In these equalities there are identical expressions left. This means that their right sides will also be equal. It's easy to write it down. Now you need to open the brackets:
in 2 - in 2 * (cos A) 2 = a 2 - c 2 + 2 c * in * cos A - in 2 * (cos A) 2.
If you perform a transfer and cast here similar terms, then it will work out starting formula, which is written after the formulation, that is, the cosine theorem. The proof is complete.
Proof of the theorem using vectors
It is much shorter than the previous one. And if you know the properties of vectors, then the cosine theorem for a triangle will be proven simply.
If sides a, b, c are designated by the vectors BC, AC and AB, respectively, then the equality holds:
BC = AC - AB.
Now you need to do some steps. The first of these is squaring both sides of the equality:
BC 2 = AC 2 + AB 2 - 2 AC * AB.
Then the equality needs to be rewritten in scalar form, taking into account that the product of vectors is equal to the cosine of the angle between them and their scalar values:
BC 2 = AC 2 + AB 2 - 2 AC * AB * cos A.
All that remains is to return to the old notation, and again we get the cosine theorem:
a 2 = b 2 + c 2 - 2 * b * c * cos A.
Formulas for other sides and all angles
To find the side, you need to take the square root of the cosine theorem. The formula for the squares of one of the other sides will look like this:
c 2 = a 2 + b 2 - 2 * a * b * cos C.
To write the expression for the square of a side V, you need to replace in the previous equality With on V, and vice versa, and put angle B under the cosine.
From the basic formula of the theorem, we can express the value of the cosine of angle A:
cos A = (in 2 + c 2 - a 2) / (2 in * c).
Formulas for other angles are derived similarly. This good practice, so you can try to write them yourself.
Naturally, there is no need to memorize these formulas. It is enough to understand the theorem and be able to derive these expressions from its main notation.
The original formula of the theorem makes it possible to find the side if the angle does not lie between two known ones. For example, you need to find V, when the values are given: a, c, A. Or unknown With, but there are meanings a, b, A.
In this situation, you need to transfer all terms of the formula to left side. You get the following equality:
с 2 - 2 * в * с * cos А + в 2 - а 2 = 0.
Let's rewrite it in a slightly different form:
c 2 - (2 * in * cos A) * c + (in 2 - a 2) = 0.
You can easily see the quadratic equation. There is an unknown quantity in it - With, and all the rest are given. Therefore, it is sufficient to solve it using a discriminant. This way the unknown side will be found.
The formula for the second side is obtained similarly:
in 2 - (2 * c * cos A) * in + (c 2 - a 2) = 0.
From other expressions, such formulas are also easy to obtain independently.
How can you find out the type of angle without calculating the cosine?
If you look closely at the angle cosine formula derived earlier, you will notice the following:
- the denominator of the fraction is always positive number, because it contains the product of sides that cannot be negative;
- the value of the angle will depend on the sign of the numerator.
Angle A will be:
- acute in a situation where the numerator is greater than zero;
- stupid if this expression is negative;
- direct when it is equal to zero.
By the way, last situation converts the cosine theorem into the Pythagorean theorem. Because for an angle of 90º its cosine is equal to zero, and the last term disappears.
First task
Condition
The obtuse angle of some arbitrary triangle is 120º. About the sides by which it is limited, it is known that one of them is 8 cm larger than the other. The length of the third side is known, it is 28 cm. It is required to find the perimeter of the triangle.
Solution
First you need to mark one of the sides with the letter “x”. In this case, the other one will be equal to (x + 8). Since there are expressions for all three sides, we can use the formula provided by the cosine theorem:
28 2 = (x + 8) 2 + x 2 - 2 * (x + 8) * x * cos 120º.
In the tables for cosines you need to find the value corresponding to 120 degrees. This will be the number 0.5 with a minus sign. Now you need to open the brackets, following all the rules, and bring similar terms:
784 = x 2 + 16x + 64 + x 2 - 2x * (-0.5) * (x + 8);
784 = 2x 2 + 16x + 64 + x 2 + 8x;
3x 2 + 24x - 720 = 0.
This quadratic equation is solved by finding the discriminant, which will be equal to:
D = 24 2 - 4 * 3 * (- 720) = 9216.
Since its value is greater than zero, the equation has two root answers.
x 1 = ((-24) + √(9216)) / (2 * 3) = 12;
x 2 = ((-24) - √(9216)) / (2 * 3) = -20.
The last root cannot be the answer to the problem, because the side must be positive.
Trigonometry is widely used not only in the section of algebra - the beginning of analysis, but also in geometry. In this regard, it is reasonable to assume the existence of theorems and their proofs related to trigonometric functions. Indeed, the theorems of cosines and sines derive very interesting, and most importantly useful, relationships between the sides and angles of triangles.
Using this formula, you can derive any of the sides of the triangle:
The proof of the statement is derived based on the Pythagorean theorem: the square of the hypotenuse is equal to the sum of the squares of the legs.
Consider an arbitrary triangle ABC. From vertex C we lower the height h to the base of the figure, at in this case Its length is absolutely not important. Now, if we consider an arbitrary triangle ACB, then we can express the coordinates of point C in terms of trigonometric cos functions and sin.
Let's remember the definition of cosine and write down the ratio of the sides of the triangle ACD: cos α = AD/AC | multiply both sides of the equality by AC; AD = AC * cos α.
We take the length AC as b and obtain an expression for the first coordinate of point C:
x = b * cosα. Similarly, we find the value of the ordinate C: y = b * sin α. Next, we apply the Pythagorean theorem and express h alternately for the triangle ACD and DCB:
It is obvious that both expressions (1) and (2) are equal to each other. Let’s equate the right-hand sides and present similar ones:
On practice this formula allows you to find the length unknown side triangle by given angles. The cosine theorem has three consequences: for direct, acute and obtuse angle triangle.
Let us replace the value of cos α with the usual variable x, then for the acute angle of triangle ABC we obtain:
If the angle turns out to be right, then 2bx will disappear from the expression, since cos 90° = 0. Graphically, the second consequence can be represented as follows:
In the case of an obtuse angle, the “-” sign in front double argument in the formula will change to “+”:
As can be seen from the explanation, there is nothing complicated in the relationships. The cosine theorem is nothing more than a translation of the Pythagorean theorem into trigonometric quantities.
Practical application of the theorem
Exercise 1. Given a triangle ABC, whose side BC = a = 4 cm, AC = b = 5 cm, and cos α = ½. You need to find the length of side AB.
To make the calculation correctly, you need to determine the angle α. To do this, you should refer to the table of values for trigonometric functions, according to which the arc cosine is equal to 1/2 for an angle of 60°. Based on this, we use the formula of the first corollary of the theorem:
Task 2. For triangle ABC, all sides are known: AB =4√2,BC=5,AC=7. You need to find all the angles of the figure.
In this case, you cannot do without a drawing of the conditions of the problem.
Since the angle values remain unknown, you should use full formula for an acute angle.
By analogy, it is not difficult to create formulas and calculate the values of other angles:
The sum of the three angles of the triangle should be 180°: 53 + 82 + 45 = 180, therefore, the solution has been found.
Theorem of sines
The theorem states that all sides of an arbitrary triangle are proportional to the sines of the opposite angles. The relations are written in the form of triple equality:
The classical proof of the statement is carried out using the example of a figure inscribed in a circle.
To verify the veracity of the statement using the example of triangle ABC in the figure, it is necessary to confirm the fact that 2R = BC / sin A. Then prove that the other sides are also related to sines opposite corners, like 2R or D circles.
To do this, draw the diameter of the circle from vertex B. From the property of angles inscribed in a circle, ∠GCB is a straight line, and ∠CGB is either equal to ∠CAB or (π - ∠CAB). In the case of sine, the latter circumstance is not significant, since sin (π –α) = sin α. Based on the above conclusions, it can be stated that:
sin ∠CGB = BC/ BG or sin A = BC/2R,
If we consider other angles of the figure, we obtain an extended formula for the theorem of sines:
Typical tasks for practicing the sine theorem boil down to finding an unknown side or angle of a triangle.
As can be seen from the examples, the solution similar tasks does not cause difficulties and consists in carrying out mathematical calculations.
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When solving geometry problems from the Unified State Exam and the Unified State Exam in mathematics, quite often the need arises, knowing the two sides of a triangle and the angle between them, to find the third side. Or, knowing all the sides of a triangle, find its angles. To solve these problems you will need the value of the cosine theorem for a triangle. In this article, a mathematics and physics tutor talks about how this theorem is formulated, proven and applied in practice when solving problems.
Formulation of the cosine theorem for a triangle
The cosine theorem for a triangle relates the two sides of a triangle and the angle between them with the side opposite that angle. For example, let us denote by the letters , and the lengths of the sides of the triangle ABC, lying respectively opposite the angles A, B And C.
Then the cosine theorem for this triangle can be written as:
In the figure, for the convenience of further discussion, the angle WITH indicated by angle. In words, this can be formulated as follows: “The square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of these sides by the cosine of the angle between them.”
It is clear that if you were expressing the other side of the triangle, for example, side, then in the formula you would need to take the cosine of the angle A, that is, lying opposite the desired side in the triangle, and on the right in the equation the sides and would be in their places. The expression for the square of the side is obtained similarly:
Proof of the cosine theorem for a triangle
The proof of the cosine theorem for a triangle is usually carried out as follows. They split the original triangle into two right-angled triangles with a height, and then play with the sides of the resulting triangles and the Pythagorean theorem. As a result, after long tedious transformations I get desired result. I personally don't like this approach. And not only because of the cumbersome calculations, but also because in this case we have to separately consider the case when the triangle is obtuse. There are too many difficulties.
I propose to prove this theorem using the concept " dot product vectors." I consciously take this risk for myself, knowing that many schoolchildren prefer to avoid this topic, believing that it is somehow murky and it is better not to deal with it. But the reluctance to tinker separately with obtuse triangle It still overwhelms me. Moreover, the resulting proof turns out to be surprisingly simple and memorable. Now you will see this.
Let's replace the sides of our triangle with the following vectors:
Using the cosine theorem for a triangle ABC. The square of a side is equal to the sum of the squares of the sides minus twice the product of these sides by the cosine of the angle between them:
Since , the result is:
Means, . It is clear that we do not take a negative solution, because the length of the segment is a positive number.
The required angle is indicated in the figure. Let us rewrite the cosine theorem for a triangle ABC. Since we have retained all the notation, the formula expressing the cosine theorem for this triangle will remain the same:
Let us now substitute into this formula all the quantities that are given. As a result, we get the following expression:
After all the calculations and transformations we get the following simple expression:
What should be the size of the acute angle so that its cosine is equal? We look at the table, which can be found in, and we get the answer: .
This is how geometry problems are solved using the cosine theorem for a triangle. If you are going to take the OGE or the Unified State Exam in mathematics, then you definitely need to master this material. Relevant problems will almost certainly be on the exam. Practice solving them yourself. Complete the following tasks:
- In a triangle ABC side AB equal to 4 cm, side B.C. equal to 6 cm, angle B equal to 30°. Find the side A.C..
- In a triangle ABC side AB equal to 10, side B.C. equal to 8, side A.C. is equal to 9. Find the cosine of the angle A.
Write your answers and solutions in the comments. Good luck to you!
Material prepared by Sergey Valerievich