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 different trigonometric functions of one argument into a rational expression of one function tg (α /2):

Formulas for converting sums into products and products into 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 logarithmization, 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 = (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 lower powers. Reduction formulas for higher degrees are derived in the same way sin And cos.

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

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

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

The simplest trigonometric identities

The quotient of dividing the sine of an 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 an angle alpha by the sine of the same angle is equal to the cotangent of the same angle (Formula 2)
The secant of an angle 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 the squares of cosine and sine.
The sum of one and the tangent of an angle is equal to the ratio of one to the square of the cosine of this angle (Formula 5)
One plus the cotangent of an angle is equal to the quotient of one divided by the sine square of this angle (Formula 6)
The product of tangent and cotangent of the same angle is equal to one (Formula 7).

Converting negative angles of trigonometric functions (even and odd)

In order to get rid of the negative value of the degree measure of an angle when calculating the sine, cosine or tangent, you can use the following trigonometric transformations (identities) based on the principles of even or odd 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 value of the sine of the same positive angle (minus sine alpha).
The cosine minus alpha will give the same value as the cosine of the alpha angle.
Tangent minus alpha is equal to minus tangent alpha.

Formulas for reducing double angles (sine, cosine, tangent and cotangent of double angles)

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

Double Angle Conversion (sine of a double angle, cosine of a double angle and tangent of a double angle) in single occurs according to the following rules:

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

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

Cosine of double angle equal to one minus double sine squared single angle

Tangent of double angle is equal to a fraction whose numerator is twice the tangent of a single angle, and the denominator is equal to one minus the tangent squared of a single angle.

Cotangent of double angle is equal to a fraction whose numerator 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

Formulas for universal trigonometric substitution

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 the value of half an angle. From the value of α we obtain α/2.

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

Trigonometric identities for half-angle transformations

The following are the formulas for trigonometric conversion of half an angle to its whole 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 using the following rules for converting trigonometric functions:

Tangent of the sum of angles is equal to a fraction whose numerator is the sum of the tangent of the first and 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.

Tangent of angle difference is equal to a fraction whose numerator is equal to the difference between the tangent of the angle being reduced and the tangent of the angle being subtracted, and the denominator is one plus the product of the tangents of these angles.

Cotangent of the sum of angles is equal to a fraction whose numerator 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.

Cotangent of angle difference is equal to a fraction whose numerator is the product of the cotangents of these angles minus one, and the denominator is equal to the sum of 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 imagine it as tg (45 + 60), then you can use the given identical transformations of the tangent of the sum of angles, and then simply substitute the tabulated values ​​of tangent 45 and tangent 60 degrees.

Formulas for converting the sum or difference of trigonometric functions

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

Triple angle formulas - converting sin3α cos3α tan3α to sinα cosα tanα

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

Formulas for converting products of trigonometric functions

If there is a need to transform the product of sines of different angles, 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 the sine, cosine or tangent functions of different angles will be converted into a sum or difference.

Formulas for reducing trigonometric functions

You need to use the reduction table as follows. In the line we select the function that interests us. In the column there is an angle. 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.

IN identity transformations trigonometric expressions the following algebraic techniques can be used: adding and subtracting identical terms; putting the common factor out of brackets; multiplication and division by the same quantity; application of abbreviated multiplication formulas; selecting a complete square; factoring a quadratic trinomial; introduction of new variables to simplify transformations.

When converting trigonometric expressions that contain fractions, you can use the properties of proportion, reducing fractions, or reducing fractions to a common denominator. In addition, you can use the selection of the whole part of the fraction, multiplying the numerator and denominator of the fraction by the same amount, and also, 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 range of permissible values ​​of the expressions being converted.

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.

From the reduction formulas it follows:

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 adding arguments and the main trigonometric identity, we get

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 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 adding arguments and formulas for converting the sum of trigonometric functions into a product after appropriate grouping, we have

M = (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: M = 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 for all x from R and the same meaning. Find this value.

Solution.

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

A = (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

А´ = -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, due to 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 techniques for proving trigonometric identities are:

A) reducing the left side of the identity to the right through appropriate transformations;
b) reducing the right side of the identity to the left;
V) reducing the right and left sides of the identity to the same form;
G) reducing to zero 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 using 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.

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.

Considering that α, β, γ are the 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 – 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 has been proven.

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 involves proving both necessity and sufficiency.

First let's prove necessity.

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'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.

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.

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 of all kϵZ only k = 0 is suitable.

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

The statement has been proven.

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