Transformation formulas for trigonometric functions. Basic trigonometry formulas

There are many formulas in trigonometry.

It is very difficult to memorize them mechanically, almost impossible. In the classroom, many schoolchildren and students use printouts on the endpapers of textbooks and notebooks, posters on the walls, cribs, and finally. What about the exam?

However, if you take a closer look at these formulas, you will find that they are all interconnected and have a certain symmetry. Let's analyze them in terms of definitions and properties. trigonometric functions to determine the minimum that is really worth memorizing.

Group I. Basic identities

sin 2 α + cos 2 α = 1;

tgα = ____ sinα cosα; ctgα = ____ cosα sinα ;

tgα · ctgα = 1;

1 + tg 2 α = _____ 1 cos 2 α; 1 + ctg 2 α = _____ 1 sin 2 α.

This group contains the simplest and most popular formulas. Most of the students know them. But if there are still difficulties, then in order to remember the first three formulas, mentally imagine right triangle with a hypotenuse equal to one. Then his legs will be equal, respectively, sinα by definition of sine (the ratio of the opposite leg to the hypotenuse) and cosα by the definition of cosine (the ratio of the adjacent leg to the hypotenuse).

The first formula is the Pythagorean theorem for such a triangle - the sum of the squares of the legs is equal to the square of the hypotenuse (1 2 = 1), the second and third are the definitions of the tangent (the ratio of the opposite leg to the adjacent one) and the cotangent (the ratio of the adjacent leg to the opposite one).
The product of the tangent and the cotangent is 1 because the cotangent written as a fraction (formula three) is an inverted tangent (formula two). The latter consideration, by the way, makes it possible to exclude from the number of formulas that must be memorized, all subsequent long formulas with a cotangent. If in any difficult task You will encounter ctgα, just replace it with a fraction ___ 1 tgα and use the formulas for the tangent.

The last two formulas do not need to be memorized pre-symbolically. They are less common. And if required, you can always re-print them on a draft. To do this, it is enough to substitute instead of the tangent or contangent of their definitions through a fraction (formulas second and third, respectively) and reduce the expression to common denominator... But it is important to remember that such formulas that connect the squares of the tangent and cosine, and the squares of the cotangent and sine exist. Otherwise, you may not guess what transformations are needed to solve a particular problem.

Group II. Addition formulas

sin (α + β) = sinα · cosβ + cosα · sinβ;

sin (α - β) = sinα · cosβ - cosα · sinβ;

cos (α + β) = cosα · cosβ - sinα · sinβ;

cos (α - β) = cosα · cosβ + sinα · sinβ;

tg (α + β) = tgα + tgβ _________ 1 - tgα · tgβ;

tg (α - β) =

Recall the odd / even parity properties of trigonometric functions:

sin (−α) = - sin (α); cos (−α) = cos (α); tg (−α) = - tg (α).

Of all trigonometric functions, only the cosine is an even function and does not change its sign when the argument (angle) sign changes, the rest of the functions are odd. The oddness of the function, in fact, means that the minus sign can be introduced and removed outside the function sign. Therefore, if you come across a trigonometric expression with the difference of two angles, you can always understand it as the sum of positive and negative angles.

For example, sin ( x- 30º) = sin ( x+ (−30º)).
Next, we use the formula for the sum of two angles and deal with the signs:
sin ( x+ (−30º)) = sin x· Cos (−30º) + cos x Sin (−30º) =
= sin x· Cos30º - cos x· Sin30º.

Thus, all formulas containing the difference in angles can simply be skipped during the first memorization. Then it is worth learning how to restore them to general view first on a draft, and then mentally.

For example, tan (α - β) = tan (α + (−β)) = tgα + tg (−β) ___________ 1 - tgα tg (−β) = tgα - tgβ _________ 1 + tgα · tgβ.

This will help in the future to quickly guess what transformations need to be applied to solve a particular task from trigonometry.

Sh group. Multiple Argument Formulas

sin2α = 2 sinα cosα;

cos2α = cos 2 α - sin 2 α;

tg2α = 2tgα _______ 1 - tg 2 α;

sin3α = 3sinα - 4sin 3 α;

cos3α = 4cos 3 α - 3cosα.

The need to use formulas for the sine and cosine of a double angle arises very often, for the tangent, too, quite often. These formulas should be known by heart. Moreover, there are no difficulties in memorizing them. First, the formulas are short. Secondly, they are easy to control according to the formulas of the previous group, based on the fact that 2α = α + α.
For example:
sin (α + β) = sinα · cosβ + cosα · sinβ;
sin (α + α) = sinα · cosα + cosα · sinα;
sin2α = 2sinα cosα.

However, if you quickly learned these formulas, and not the previous ones, then you can do the opposite: you can remember the formula for the sum of two angles using the corresponding formula for a double angle.

For example, if you need a formula for the cosine of the sum of two angles:
1) remember the formula for the cosine of a double angle: cos2 x= cos 2 x- sin 2 x;
2) we paint it long: cos ( x + x) = cos x Cos x- sin x Sin x;
3) replace one NS by α, the second by β: cos (α + β) = cosα cosβ - sinα sinβ.

Practice in the same way to restore the formulas for the sine of the sum and the tangent of the sum. In critical cases, such as, for example, the USE, check the accuracy of the restored formulas using the known first quarter: 0º, 30º, 45º, 60º, 90º.

Checking the previous formula (obtained by replacing in line 3):
let be α = 60 °, β = 30 °, α + β = 90 °,
then cos (α + β) = cos90 ° = 0, cosα = cos60 ° = 1/2, cosβ = cos30 ° = √3 _ / 2, sinα = sin60 ° = √3 _ / 2, sinβ = sin30 ° = 1/2;
substitute the values ​​into the formula: 0 = (1/2) ( √3_ /2) − (√3_ / 2) (1/2);
0 ≡ 0, no errors were found.

Formulas for a triple angle, in my opinion, do not need to be specially "crammed". They are quite rare in examinations like the exam. They are easily deduced from the formulas that were above, since sin3α = sin (2α + α). And for those students who, for some reason, still need to learn these formulas by heart, I advise you to pay attention to their certain "symmetry" and memorize not the formulas themselves, but the mnemonic rules. For example, the order in which the numbers are located in the two formulas "33433433", etc.

IV group. Sum / difference - into product

sinα + sinβ = 2 sin α + β ____ 2 Cos α - β ____ 2 ;

sinα - sinβ = 2 sin α - β ____ 2 Cos α + β ____ 2 ;

cosα + cosβ = 2cos α + β ____ 2 Cos α - β ____ 2 ;

cosα - cosβ = −2 sin α - β ____ 2 Sin α + β ____ 2 ;

tgα + tgβ = sin (α + β) ________ cosα cosβ ;

tgα - tgβ = sin (α - β) ________ cosα cosβ .

Using the odd properties of the sine and tangent functions: sin (−α) = - sin (α); tg (−α) = - tg (α),
it is possible to reduce the formulas for the differences of two functions to formulas for their sums. For example,

sin90º - sin30º = sin90º + sin (−30º) = 2 · sin 90º + (−30º) __________ 2 Cos 90º - (−30º) __________ 2 =

2 · sin30º · cos60º = 2 · (1/2) · (1/2) = 1/2.

Thus, the formulas for the difference of sines and tangents do not have to be learned by heart right away.
The situation with the sum and difference of cosines is more complicated. These formulas are not interchangeable. But again, using the parity of the cosine, you can remember the following rules.

The sum cosα + cosβ cannot change its sign for any change in the sign of the angles, therefore the product must also consist of even functions, i.e. two cosines.

The sign of the difference cosα - cosβ depends on the values ​​of the functions themselves, which means that the sign of the product should depend on the ratio of the angles, therefore the product should consist of odd functions, i.e. two sinuses.

And yet this group of formulas is not the easiest to memorize. This is the case when it is better to cram less, but check more. In order to avoid mistakes in the formula on the responsible exam, be sure to first write it down on a draft and check it in two ways. First, by substitutions β = α and β = −α, then by the known values ​​of the functions for prime angles. For this, it is best to take 90º and 30º, as it was done in the example above, because the half-sum and half-difference of these values ​​again give simple angles, and you can easily see how equality becomes an identity for the correct option. Or, on the contrary, it is not executed if you made a mistake.

Example checking the formula cosα - cosβ = 2 sin α - β ____ 2 Sin α + β ____ 2 for the difference of cosines with a mistake !

1) Let β = α, then cosα - cosα = 2 sin α - α _____ 2 Sin α + α _____ 2= 2sin0 sinα = 0 sinα = 0. cosα - cosα ≡ 0.

2) Let β = - α, then cosα - cos (- α) = 2 sin α - (−α) _______ 2 Sin α + (−α) _______ 2= 2sinα sin0 = 0 sinα = 0. cosα - cos (- α) = cosα - cosα ≡ 0.

These checks showed that the functions in the formula were used correctly, but due to the fact that the identity turned out to be of the form 0 ≡ 0, an error with a sign or a coefficient could be missed. We do the third check.

3) Let α = 90º, β = 30º, then cos90º - cos30º = 2 · sin 90º - 30º ________ 2 Sin 90º + 30º ________ 2= 2sin30º · sin60º = 2 · (1/2) · (√3 _ /2) = √3_ /2.

cos90 - cos30 = 0 - √3 _ /2 = −√3_ /2 ≠ √3_ /2.

The error was really in the sign and only in the sign before the work.

Group V. Product - in sum / difference

sinα · sinβ = 1 _ 2 (Cos (α - β) - cos (α + β));

cosα cosβ = 1 _ 2 (Cos (α - β) + cos (α + β));

sinα cosβ = 1 _ 2 (Sin (α - β) + sin (α + β)).

The very name of the fifth group of formulas suggests that these formulas are the reverse of the previous group. It is clear that in this case it is easier to restore the formula on a draft than to learn it again, increasing the risk of creating a "mess in your head." The only thing that makes sense to focus on for a faster recovery of the formula is the following equalities (check them):

α = α + β ____ 2 + α - β ____ 2; β = α + β ____ 2 − α - β ____ 2.

Consider example: need to convert the product sin5 x Cos3 x into the sum of two trigonometric functions.
Since the product includes both sine and cosine, we take from the previous group the formula for the sum of sines, which we have already learned, and write it down on a draft.

sinα + sinβ = 2 sin α + β ____ 2 Cos α - β ____ 2

Let 5 x = α + β ____ 2 and 3 x = α - β ____ 2, then α = α + β ____ 2 + α - β ____ 2 = 5x + 3x = 8x, β = α + β ____ 2 − α - β ____ 2 = 5x − 3x = 2x.

We replace in the formula on the draft the values ​​of the angles, expressed in terms of the variables α and β, by the values ​​of the angles, expressed in terms of the variable x.
We get sin8 x+ sin2 x= 2 sin5 x Cos3 x

We divide both parts of equality by 2 and write it down on the clean copy from right to left sin5 x Cos3 x = 1 _ 2 (sin8 x+ sin2 x). The answer is ready.

As an exercise: Explain why in the textbook there are only 3 formulas for converting the sum / difference to the product of 6, and the inverse (to convert the product to the sum or difference)?

VI group. Degree reduction formulas

cos 2 α = 1 + cos2α _________ 2;

sin 2 α = 1 - cos2α _________ 2;

cos 3 α = 3cosα + cos3α ____________ 4;

sin 3 α = 3sinα - sin3α ____________ 4.

The first two formulas of this group are very much needed. They are often used when solving trigonometric equations, including level unified exam, as well as when calculating integrals containing integrands of trigonometric type.

It may be easier to remember them in the next "one-story" form.
2cos 2 α = 1 + cos2α;
2 sin 2 α = 1 - cos2α,
and you can always divide by 2 in your head or on a draft.

The need to use the following two formulas (with cubes of functions) in exams is much less common. In a different setting, you will always have time to use the draft. In this case, the following options are possible:
1) If you remember the last two formulas of the III group, then use them to express sin 3 α and cos 3 α by simple transformations.
2) If in the last two formulas of this group you notice elements of symmetry that contribute to their memorization, then write down the "sketches" of the formulas on the draft and check them by the values ​​of the main angles.
3) If, in addition to the fact that such formulas for lowering the degree exist, you do not know anything about them, then solve the problem in stages, based on the fact that sin 3 α = sin 2 α · sinα and other learned formulas. Degree reduction formulas for a square and a formula for converting a product to a sum will be required.

VII group. Half argument

sin α _ 2 = ± √ 1 - cosα ________ 2; _____

cos α _ 2 = ± √ 1 + cosα ________ 2; _____

tg α _ 2 = ± √ 1 - cosα ________ 1 + cosα. _____

I see no point in memorizing this group of formulas in the form in which they are presented in textbooks and reference books. If you understand that α is half of 2α, then this is enough to quickly deduce the desired formula half argument, based on the first two formulas for decreasing the degree.

This also applies to the tangent of the half angle, the formula for which is obtained by dividing the sine expression by the corresponding cosine expression.

Do not forget only when retrieving square root put a sign ± .

VIII group. Universal substitution

sinα = 2tg (α / 2) _________ 1 + tan 2 (α / 2);

cosα = 1 - tan 2 (α / 2) __________ 1 + tan 2 (α / 2);

tgα = 2tg (α / 2) _________ 1 - tg 2 (α / 2).

These formulas can be extremely useful for solving trigonometric problems of all kinds. They allow implementing the principle of "one argument - one function", which allows you to make variable changes that reduce complex trigonometric expressions to algebraic ones. It is not without reason that this substitution is called universal.
We must learn the first two formulas. The third can be obtained by dividing the first two by each other according to the definition of the tangent tgα = sinα ___ cosα

IX group. Casting formulas.

To understand this group of trigonometric formulas, pass

X group. Values ​​for major angles.

The values ​​of the trigonometric functions for the main angles of the first quarter are given

So we do output: Trigonometry formulas need to know. The bigger, the better. But what to spend their time and efforts on - memorizing formulas or recovering them in the process of solving problems, everyone must decide on their own.

An example of a task for using trigonometry formulas

Solve the equation sin5 x Cos3 x- sin8 x Cos6 x = 0.

We have two different sin functions() and cos () and four! different arguments 5 x, 3x, 8x and 6 x... Without preliminary transformations, it will not work to reduce to the simplest types of trigonometric equations. Therefore, we first try to replace the products with the sums or differences of functions.
We do this in the same way as in the example above (see section).

sin (5 x + 3x) + sin (5 x − 3x) = 2 sin5 x Cos3 x
sin8 x+ sin2 x= 2 sin5 x Cos3 x

sin (8 x + 6x) + sin (8 x − 6x) = 2 sin8 x Cos6 x
sin14 x+ sin2 x= 2 sin8 x Cos6 x

Expressing the products from these equalities, we substitute them into the equation. We get:

(sin8 x+ sin2 x) / 2 - (sin14 x+ sin2 x)/2 = 0.

We multiply both sides of the equation by 2, open the brackets and give similar terms

Sin8 x+ sin2 x- sin14 x- sin2 x = 0;
sin8 x- sin14 x = 0.

The equation has become much simpler, but solve it like this sin8 x= sin14 x, therefore 8 x = 14x+ T, where T is the period, is incorrect, since we do not know the meaning of this period. Therefore, we will use the fact that there is 0 on the right side of the equality, with which it is easy to compare the factors in any expression.
To expand sin8 x- sin14 x by factors, you need to go from the difference to the product. To do this, you can use the formula for the difference of sines, or again the formula for the sum of sines and the oddness of the sine function (see the example in the section).

sin8 x- sin14 x= sin8 x+ sin (−14 x) = 2 sin 8x + (−14x) __________ 2 Cos 8x − (−14x) __________ 2 = sin (−3 x) Cos11 x= −sin3 x Cos11 x.

So the equation sin8 x- sin14 x= 0 is equivalent to the equation sin3 x Cos11 x= 0, which, in turn, is equivalent to the set of two simplest equations sin3 x= 0 and cos11 x= 0. Solving the latter, we get two series of answers
x 1 = π n/3, nϵZ
x 2 = π / 22 + π k/11, kϵZ

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Executing for all argument values ​​(from the common scope).

Universal substitution formulas.

With these formulas, it is easy to any expression that contains various trigonometric functions of one argument turns into a rational expression of one function tg (α / 2):

Formulas for converting sums to products and products to sums.

Previously, these formulas were used to simplify calculations. Calculated using logarithmic tables, and later - slide rule, since logarithms are best suited for multiplying numbers. That is why each initial expression was reduced to a form that would be convenient for taking the logarithm, that is, to products, for example:

2 sin α sin b = cos (α - b) - cos (α + b);

2 cos α cos b = cos (α - b) + cos (α + b);

2 sin α cos b = sin (α - b) + sin (α + b).

where is the angle for which, in particular,

Formulas for the tangent and cotangent functions are easily derived from the above.

Degree reduction formulas.

sin 2 α = (1 - cos 2α) / 2;	cos 2 α = (1 + cos 2α) / 2;
sin 3α = (3 sinα - sin 3α )/4;	cos 3 a = (3 cosα + cos 3α )/4.

Using these formulas trigonometric equations are easily reduced to equations with more low degrees... Reduction formulas are derived in the same way for more high degrees sin and cos.

Expression of trigonometric functions in terms of one of them of the same argument.

The sign in front of the root depends on the quarter of the angle α .

To solve some problems, a table of trigonometric identities will be useful, which will make it much easier to perform transformations of functions:

Simplest trigonometric identities

The quotient of dividing the sine of the angle alpha by the cosine of the same angle is equal to the tangent of this angle (Formula 1). See also the proof of the correctness of the transformation of the simplest trigonometric identities.
The quotient of dividing the cosine of the alpha angle by the sine of the same angle is equal to the cotangent of the same angle (Formula 2)
Angle secant is equal to one divided by the cosine of the same angle (Formula 3)
The sum of the squares of the sine and cosine of the same angle is equal to one (Formula 4). see also the proof of the sum of squares of cosine and sine.
The sum of the unit and the tangent of an angle is equal to the ratio of the unit to the square of the cosine of this angle (Formula 5)
The unit plus the cotangent of the angle is equal to the quotient of dividing one by the sine square of this angle (Formula 6)
The product of the tangent and the cotangent of the same angle is equal to one (Formula 7).

Convert negative angles of trigonometric functions (even and odd)

In order to get rid of the negative value of the degree measure of the angle when calculating the sine, cosine or tangent, you can use the following trigonometric transformations (identities) based on the principles of evenness or oddness of trigonometric functions.

As seen, cosine and the secant is even function, sine, tangent and cotangent are odd functions.

The sine of a negative angle is equal to the negative sine of that same positive angle (minus sine alpha).
The cosine "minus alpha" will give the same value as the cosine of the angle alpha.
The tangent minus alpha is equal to the minus tangent alpha.

Double angle reduction formulas (sine, cosine, tangent and cotangent of double angle)

If you need to divide an angle in half, or vice versa, go from a double angle to a single angle, you can use the following trigonometric identities:

Double angle conversion (sine of double angle, cosine of double angle and tangent of double angle) to single occurs according to the following rules:

Double angle sine equal to twice the product of sine and cosine of a single angle

Cosine of a double angle is equal to the difference between the square of the cosine of a single angle and the square of the sine of this angle

Cosine of a double angle equal to twice the square of the cosine of a single angle minus one

Cosine of a double angle equal to one minus double sine square of a single angle

Double angle tangent is equal to a fraction, the numerator of which is the double tangent of a single angle, and the denominator is equal to one minus the tangent of the square of a single angle.

Double angle cotangent is equal to a fraction, the numerator of which is the square of the cotangent of a single angle minus one, and the denominator is equal to twice the cotangent of a single angle

Universal trigonometric substitution formulas

The conversion formulas below can be useful when you need to divide the argument of a trigonometric function (sin α, cos α, tan α) by two and reduce the expression to half the angle. From the value of α we obtain α / 2.

These formulas are called universal trigonometric substitution formulas... Their value lies in the fact that the trigonometric expression with their help is reduced to the expression of the tangent of half an angle, regardless of which trigonometric functions ( sin cos tg ctg) were in the expression originally. After that, the equation with the tangent of half the angle is much easier to solve.

Trigonometric transformations of half an angle

The following are formulas for the trigonometric conversion of half an angle to an integer value.
The value of the argument of the trigonometric function α / 2 is reduced to the value of the argument of the trigonometric function α.

Trigonometric formulas for adding angles

cos (α - β) = cos α cos β + sin α sin β

sin (α + β) = sin α cos β + sin β cos α

sin (α - β) = sin α cos β - sin β cos α
cos (α + β) = cos α cos β - sin α sin β

Tangent and cotangent of the sum of angles alpha and beta can be converted according to the following trigonometric function conversion rules:

Tangent of the sum of angles is equal to a fraction, the numerator of which is the sum of the tangent of the first and the tangent of the second angle, and the denominator is one minus the product of the tangent of the first angle and the tangent of the second angle.

Angle difference tangent is equal to the fraction, the numerator of which is equal to the difference between the tangent of the reduced angle and the tangent of the subtracted angle, and the denominator is equal to one plus the product of the tangents of these angles.

Cotangent of the sum of angles is equal to a fraction, the numerator of which is equal to the product of the cotangents of these angles plus one, and the denominator is equal to the difference between the cotangent of the second angle and the cotangent of the first angle.

Angle difference cotangent is equal to the fraction, the numerator of which is the product of the cotangents of these angles minus one, and the denominator is equal to the sum the cotangents of these angles.

These trigonometric identities are convenient to use when you need to calculate, for example, the tangent of 105 degrees (tg 105). If you represent it as tg (45 + 60), then you can use the given identical transformations of the tangent of the sum of the angles, and then simply substitute the tabular values ​​of the tangent 45 and the tangent 60 degrees.

Sum or Difference Conversion Formulas for Trigonometric Functions

Expressions representing the sum of the form sin α + sin β can be transformed using the following formulas:

Triple angle formulas - convert sin3α cos3α tg3α to sinα cosα tgα

Sometimes it is necessary to convert the triple value of the angle so that the angle α becomes the argument of the trigonometric function instead of 3α.
In this case, you can use the triple angle transformation formulas (identities):

Transformation Formulas for the Product of Trigonometric Functions

If it becomes necessary to transform the product of sines of different angles of cosines of different angles, or even the product of sine and cosine, then you can use the following trigonometric identities:


In this case, the product of sine, cosine or tangent functions of different angles will be converted to the sum or difference.

Trigonometric function reduction formulas

You need to use the cast table as follows. In the line, select the function that interests us. The column contains the corner. For example, the sine of the angle (α + 90) at the intersection of the first row and the first column, we find out that sin (α + 90) = cos α.

V identical transformations trigonometric expressions the following algebraic techniques can be used: addition and subtraction of the same terms; taking the common factor out of the parentheses; multiplication and division by the same amount; application of abbreviated multiplication formulas; selection of a full square; factorization of a square trinomial; introduction of new variables in order to simplify transformations.

When converting trigonometric expressions that contain fractions, you can use the properties of proportion, reduction of fractions, or converting fractions to a common denominator. In addition, you can use the selection of the integer part of the fraction, multiplying the numerator and denominator of the fraction by the same value, as well as, if possible, take into account the homogeneity of the numerator or denominator. If necessary, you can represent a fraction as the sum or difference of several simpler fractions.

In addition, when applying all the necessary methods for converting trigonometric expressions, it is necessary to constantly take into account the permissible values expressions to be converted.

Let's look at a few examples.

Example 1.

Calculate А = (sin (2x - π) cos (3π - x) + sin (2x - 9π / 2) cos (x + π / 2)) 2 + (cos (x - π / 2) cos ( 2x - 7π / 2) +
+ sin (3π / 2 - x) sin (2x -5π / 2)) 2

Solution.

It follows from the reduction formulas:

sin (2x - π) = -sin 2x; cos (3π - x) = -cos x;

sin (2x - 9π / 2) = -cos 2x; cos (x + π / 2) = -sin x;

cos (x - π / 2) = sin x; cos (2x - 7π / 2) = -sin 2x;

sin (3π / 2 - x) = -cos x; sin (2x - 5π / 2) = -cos 2x.

Whence, by virtue of the formulas for the addition of arguments and the basic trigonometric identity, we obtain

A = (sin 2x cos x + cos 2x sin x) 2 + (-sin x sin 2x + cos x cos 2x) 2 = sin 2 (2x + x) + cos 2 (x + 2x) =
= sin 2 3x + cos 2 3x = 1

Answer: 1.

Example 2.

Convert the expression М = cos α + cos (α + β) cos γ + cos β - sin (α + β) sin γ + cos γ into a product.

Solution.

From the formulas for the addition of arguments and the formulas for the transformation of the sum of trigonometric functions into a product after the corresponding grouping, we have

М = (cos (α + β) cos γ - sin (α + β) sin γ) + cos α + (cos β + cos γ) =

2cos ((β + γ) / 2) cos ((β - γ) / 2) + (cos α + cos (α + β + γ)) =

2cos ((β + γ) / 2) cos ((β - γ) / 2) + 2cos (α + (β + γ) / 2) cos ((β + γ) / 2)) =

2cos ((β + γ) / 2) (cos ((β - γ) / 2) + cos (α + (β + γ) / 2)) =

2cos ((β + γ) / 2) 2cos ((β - γ) / 2 + α + (β + γ) / 2) / 2) cos ((β - γ) / 2) - (α + ( β + γ) / 2) / 2) =

4cos ((β + γ) / 2) cos ((α + β) / 2) cos ((α + γ) / 2).

Answer: М = 4cos ((α + β) / 2) cos ((α + γ) / 2) cos ((β + γ) / 2).

Example 3.

Show that the expression A = cos 2 (x + π / 6) - cos (x + π / 6) cos (x - π / 6) + cos 2 (x - π / 6) takes one and the same meaning. Find this value.

Solution.

Here are two ways to solve this problem. Applying the first method, by selecting a complete square and using the corresponding basic trigonometric formulas, we get

А = (cos (x + π / 6) - cos (x - π / 6)) 2 + cos (x - π / 6) cos (x - π / 6) =

4sin 2 x sin 2 π / 6 + 1/2 (cos 2x + cos π / 3) =

Sin 2 x + 1/2 cos 2x + 1/4 = 1/2 (1 - cos 2x) + 1/2 cos 2x + 1/4 = 3/4.

Solving the problem in the second way, consider A as a function of x from R and calculate its derivative. After transformations, we get

A´ = -2cos (x + π / 6) sin (x + π / 6) + (sin (x + π / 6) cos (x - π / 6) + cos (x + π / 6) sin (x + π / 6)) - 2cos (x - π / 6) sin (x - π / 6) =

Sin 2 (x + π / 6) + sin ((x + π / 6) + (x - π / 6)) - sin 2 (x - π / 6) =

Sin 2x - (sin (2x + π / 3) + sin (2x - π / 3)) =

Sin 2x - 2sin 2x cos π / 3 = sin 2x - sin 2x ≡ 0.

Hence, by virtue of the criterion for the constancy of a function differentiable on an interval, we conclude that

A (x) ≡ (0) = cos 2 π / 6 - cos 2 π / 6 + cos 2 π / 6 = (√3 / 2) 2 = 3/4, x € R.

Answer: A = 3/4 for x € R.

The main methods of proving trigonometric identities are:

a) reducing the left side of the identity to right way appropriate transformations;
b) reduction of the right side of the identity to the left;
v) reduction of the right and left parts of the identity to the same kind;
G) reduction to zero of the difference between the left and right sides of the identity being proved.

Example 4.

Check that cos 3x = -4cos x cos (x + π / 3) cos (x + 2π / 3).

Solution.

Transforming the right-hand side of this identity according to the corresponding trigonometric formulas, we have

4cos x cos (x + π / 3) cos (x + 2π / 3) =

2cos x (cos ((x + π / 3) + (x + 2π / 3)) + cos ((x + π / 3) - (x + 2π / 3))) =

2cos x (cos (2x + π) + cos π / 3) =

2cos x cos 2x - cos x = (cos 3x + cos x) - cos x = cos 3x.

The right side of the identity has been reduced to the left.

Example 5.

Prove that sin 2 α + sin 2 β + sin 2 γ - 2cos α cos β cos γ = 2 if α, β, γ are interior angles of some triangle.

Solution.

Taking into account that α, β, γ are interior angles of some triangle, we obtain that

α + β + γ = π and, therefore, γ = π - α - β.

sin 2 α + sin 2 β + sin 2 γ - 2 cos α cos β cos γ =

Sin 2 α + sin 2 β + sin 2 (π - α - β) - 2cos α cos β cos (π - α - β) =

Sin 2 α + sin 2 β + sin 2 (α + β) + (cos (α + β) + cos (α - β) (cos (α + β) =

Sin 2 α + sin 2 β + (sin 2 (α + β) + cos 2 (α + β)) + cos (α - β) (cos (α + β) =

1/2 · (1 - cos 2α) + ½ · (1 - cos 2β) + 1 + 1/2 · (cos 2α + cos 2β) = 2.

The original equality is proved.

Example 6.

To prove that for one of the angles α, β, γ of the triangle to be equal to 60 °, it is necessary and sufficient that sin 3α + sin 3β + sin 3γ = 0.

Solution.

The condition of this problem presupposes proof of both necessity and sufficiency.

First, let us prove need.

It can be shown that

sin 3α + sin 3β + sin 3γ = -4cos (3α / 2) cos (3β / 2) cos (3γ / 2).

Hence, taking into account that cos (3/2 60 °) = cos 90 ° = 0, we obtain that if one of the angles α, β or γ is equal to 60 °, then

cos (3α / 2) cos (3β / 2) cos (3γ / 2) = 0 and, therefore, sin 3α + sin 3β + sin 3γ = 0.

Let us now prove adequacy the specified condition.

If sin 3α + sin 3β + sin 3γ = 0, then cos (3α / 2) cos (3β / 2) cos (3γ / 2) = 0, and therefore

either cos (3α / 2) = 0, or cos (3β / 2) = 0, or cos (3γ / 2) = 0.

Hence,

or 3α / 2 = π / 2 + πk, i.e. α = π / 3 + 2πk / 3,

or 3β / 2 = π / 2 + πk, i.e. β = π / 3 + 2πk / 3,

or 3γ / 2 = π / 2 + πk,

those. γ = π / 3 + 2πk / 3, where k ϵ Z.

Since α, β, γ are the angles of the triangle, we have

0 < α < π, 0 < β < π, 0 < γ < π.

Therefore, for α = π / 3 + 2πk / 3 or β = π / 3 + 2πk / 3 or

γ = π / 3 + 2πk / 3 of all kϵZ only k = 0 fits.

Whence it follows that either α = π / 3 = 60 °, or β = π / 3 = 60 °, or γ = π / 3 = 60 °.

The statement is proven.

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