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 .
The relationships between the basic trigonometric functions - sine, cosine, tangent and cotangent - are given trigonometric formulas. And since there are quite a lot of connections between trigonometric functions, this explains the abundance of trigonometric formulas. Some formulas connect trigonometric functions of the same angle, others - functions of a multiple angle, others - allow you to reduce the degree, fourth - express all functions through the tangent of a half angle, etc.
In this article we will list in order all the basic trigonometric formulas, which are sufficient to solve the vast majority of trigonometry problems. For ease of memorization and use, we will group them by purpose and enter them into tables.
Page navigation.
Basic trigonometric identities
Basic trigonometric identities define the relationship between 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 in terms of any other.
For a detailed description of these trigonometry formulas, their derivation and examples of application, see the article.
Reduction formulas
Reduction formulas follow from the properties of sine, cosine, tangent and cotangent, that is, they reflect the property of periodicity trigonometric functions, the property of symmetry, as well as the property of shift by 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 trigonometric functions of the sum or difference of two angles are expressed in terms of trigonometric functions of those angles. These formulas serve as the basis for deriving the following trigonometric formulas.
Formulas for double, triple, etc. angle
Formulas for double, triple, etc. angle (they are also called multiple angle formulas) show how 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 trigonometric functions of a half angle are expressed in terms of the cosine of a whole angle. These trigonometric formulas follow from the double angle formulas.
Their conclusion and examples of application can be found in the article.
Degree reduction formulas
Trigonometric formulas for reducing 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 you to reduce the powers of trigonometric functions to the first.
Formulas for the sum and difference of trigonometric functions
Main purpose formulas for the sum and difference of trigonometric functions is to go 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 you to factor 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 a sum or difference is carried out using the formulas for the product of sines, cosines and sine by cosine.
Universal trigonometric substitution
We complete our review of the basic formulas of trigonometry with formulas expressing trigonometric functions in terms of the tangent of a half angle. This replacement was called universal trigonometric substitution. Its convenience lies in the fact that all trigonometric functions are expressed rationally in terms of the tangent of a half angle without roots.
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.
Copyright by cleverstudents
All rights reserved.
Protected by copyright law. No part of the site, including internal materials and appearance, may be reproduced in any form or used without the prior written permission of the copyright holder.
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 converting 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 the 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.
Still have questions? Not sure how to simplify trigonometric expressions?
To get help from a tutor, register.
The first lesson is free!
website, when copying material in full or in part, a link to the source is required.
