Executed for all values ​​of the argument (from the general scope).

Universal substitution formulas.

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

Formulas for converting sums to products and products to sums.

Previously, the above formulas were used to simplify calculations. They calculated using logarithmic tables, and later - a slide rule, since logarithms are best suited for multiplying numbers. That is why each original expression was reduced to a form that would be convenient for logarithms, 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 obtained from the above.

Degree reduction formulas.

sin 2 α \u003d (1 - cos 2α) / 2;	cos2α = (1 + cos2α)/2;
sin 3α = (3 sinα -sin 3α )/4;	cos 3 a = (3 cosα + cos 3α )/4.

With the help of these formulas, trigonometric equations are easily reduced to equations with lower degrees. In the same way, lowering formulas are derived for higher degrees sin and cos.

Expression of trigonometric functions through one of them of the same argument.

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

The ratios between the main trigonometric functions - sine, cosine, tangent and cotangent - are given trigonometric formulas. And since there are quite a lot of connections between trigonometric functions, this also explains the abundance of trigonometric formulas. Some formulas connect the trigonometric functions of the same angle, others - the functions of a multiple angle, others - allow you to lower the degree, the fourth - to express all functions through the tangent of a half angle, etc.

In this article, we list in order all the basic trigonometric formulas, which are enough to solve the vast majority of trigonometry problems. For ease of memorization and use, we will group them according to their purpose, and enter them into tables.

Page navigation.

Basic trigonometric identities

Basic trigonometric identities set the relationship between the sine, cosine, tangent and cotangent of one angle. They follow from the definition of sine, cosine, tangent and cotangent, as well as the concept of the unit circle. They allow you to express one trigonometric function through any other.

For a detailed description of these trigonometry formulas, their derivation and application examples, see the article.

Cast formulas

Cast formulas follow from the properties of sine, cosine, tangent and cotangent, that is, they reflect the property of periodicity of trigonometric functions, the property of symmetry, and also the property of shift by a given angle. These trigonometric formulas allow you to move from working with arbitrary angles to working with angles ranging from zero to 90 degrees.

The rationale for these formulas, a mnemonic rule for memorizing them, and examples of their application can be studied in the article.

Addition Formulas

Trigonometric addition formulas show how the trigonometric functions of the sum or difference of two angles are expressed in terms of the trigonometric functions of these angles. These formulas serve as the basis for the derivation of the following trigonometric formulas.

Formulas for double, triple, etc. corner

Formulas for double, triple, etc. angle (they are also called multiple angle formulas) show how the trigonometric functions of double, triple, etc. angles () are expressed in terms of trigonometric functions of a single angle. Their derivation is based on addition formulas.

More detailed information is collected in the article formulas for double, triple, etc. angle .

Half Angle Formulas

Half Angle Formulas show how the trigonometric functions of a half angle are expressed in terms of the cosine of an integer angle. These trigonometric formulas follow from the double angle formulas.

Their conclusion and examples of application can be found in the article.

Reduction Formulas

Trigonometric formulas for decreasing degrees are designed to facilitate the transition from natural powers of trigonometric functions to sines and cosines in the first degree, but multiple angles. In other words, they allow one to reduce the powers of trigonometric functions to the first.

Formulas for the sum and difference of trigonometric functions

main destination sum and difference formulas for trigonometric functions consists in the transition to the product of functions, which is very useful when simplifying trigonometric expressions. These formulas are also widely used in solving trigonometric equations, as they allow factoring the sum and difference of sines and cosines.

Formulas for the product of sines, cosines and sine by cosine

The transition from the product of trigonometric functions to the sum or difference is carried out through the formulas for the product of sines, cosines and sine by cosine.

Universal trigonometric substitution

We complete the review of the basic formulas of trigonometry with formulas expressing trigonometric functions in terms of the tangent of a half angle. This replacement is called universal trigonometric substitution. Its convenience lies in the fact that all trigonometric functions are expressed in terms of the tangent of a half angle rationally without roots.

Bibliography.

• Algebra: Proc. for 9 cells. avg. school / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova; Ed. S. A. Telyakovsky.- M.: Enlightenment, 1990.- 272 p.: Ill.- ISBN 5-09-002727-7
• Bashmakov M.I. Algebra and the beginning of analysis: Proc. for 10-11 cells. avg. school - 3rd ed. - M.: Enlightenment, 1993. - 351 p.: ill. - ISBN 5-09-004617-4.
• Algebra and the beginning of the analysis: Proc. for 10-11 cells. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorova.- 14th ed.- M.: Enlightenment, 2004.- 384 p.: ill.- ISBN 5-09-013651-3.
• Gusev V. A., Mordkovich A. G. Mathematics (a manual for applicants to technical schools): Proc. allowance.- M.; Higher school, 1984.-351 p., ill.

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AT identical transformations trigonometric expressions the following algebraic tricks can be used: adding and subtracting identical terms; taking the common factor out of brackets; multiplication and division by the same value; application of abbreviated multiplication formulas; selection of a full square; factorization of a square trinomial; introduction of new variables to simplify transformations.

When converting trigonometric expressions containing fractions, you can use the properties of proportion, reduction of fractions, or reduction of 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, and also, if possible, take into account the uniformity of the numerator or denominator. If necessary, you can represent a fraction as a 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 range of permissible values ​​of the converted expressions.

Let's look at a few examples.

Example 1

Calculate A = (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) \u003d -sin 2x; cos (3π - x) \u003d -cos x;

sin (2x - 9π / 2) \u003d -cos 2x; cos (x + π/2) = -sin x;

cos (x - π / 2) \u003d sin x; cos (2x - 7π/2) = -sin 2x;

sin (3π / 2 - x) \u003d -cos x; sin (2x - 5π / 2) \u003d -cos 2x.

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

A \u003d (sin 2x cos x + cos 2x sin x) 2 + (-sin x sin 2x + cos x cos 2x) 2 \u003d sin 2 (2x + x) + cos 2 (x + 2x) \u003d
= sin 2 3x + cos 2 3x = 1

Answer: 1.

Example 2

Convert the expression M = 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 converting the sum of trigonometric functions into a product, after the appropriate 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 \u003d cos 2 (x + π / 6) - cos (x + π / 6) cos (x - π / 6) + cos 2 (x - π / 6) takes for all x from R one and the same value. Find this value.

Solution.

We present two methods for solving this problem. Applying the first method, by isolating the full square and using the corresponding basic trigonometric formulas, we obtain

A \u003d (cos (x + π / 6) - cos (x - π / 6)) 2 + cos (x - π / 6) cos (x - π / 6) \u003d

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

А´ \u003d -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 of 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) reduction of the left side of the identity to the right side by appropriate transformations;
b) reduction of the right side of the identity to the left;
in) reduction of the right and left parts of the identity to the same form;
G) reduction to zero of the difference between the left and right parts 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 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 is reduced to the left side.

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 hence γ = π – α – β.

sin 2 α + sin 2 β + sin 2 γ – 2cos 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

Prove that in order 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 the proof of both necessity and sufficiency.

First we 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 hence sin 3α + sin 3β + sin 3γ = 0.

Let's prove now 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.

Consequently,

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.

From the fact that α, β, γ are the angles of a 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 out 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 assertion has been proven.

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