Methods for proving identities. Identity. Ways to prove identities Let's consider a few simple examples

During the learning process, students should develop the skills of proving identities in the following ways.

If you need to prove that A=B, then you can

1. prove that A - B = O,

2. prove that A/B = 1,

3. convert A to form B,

4. convert B to type A,

5. convert A and B to one form C.

The properties of arithmetic operations are used as the support on which proofs of identities are built. Sometimes geometric concepts and methods are used in the proof. Geometric proofs are not only instructive and visual, but also help strengthen interdisciplinary connections.

Proofs of identities can be divided into three types depending on the extent to which they satisfy the requirements of rigor:

a) Not completely rigorous reasoning, requiring the use of the method of mathematical induction to give it full rigor. These proofs are used to derive rules for operations with polynomials and properties of powers with natural exponents. For example,

a k a r = (a ·······a) (a ········a) = a ········a = a k+p

k times p times k+p times

b) Completely rigorous reasoning, based on the basic properties of arithmetic operations and not using other properties of the number system. The main area of application of such proofs is the identities of abbreviated multiplication. Many of the statements expressed by abbreviated multiplication formulas allow for visual geometric illustration.

Example For identity The teacher may suggest the following illustration:

c) Completely rigorous reasoning using conditions for the solvability of equations of the form Ψ(x) = a, where Ψ is the elementary function being studied. Such proofs are typical for deducing the properties of a power with a rational exponent and a logarithmic function. For example, when proving the property of the arithmetic root

(1)

we will rely on a reformulation of the definition of arithmetic square root: for non-negative numbers x and y equalities y =
And

y 2 = x are equivalent, therefore (1) is equivalent to (
) 2 = (
) 2 (2). From where it follows, and in = (
) 2 (
) 2 = a c.

The method of proof that was used here is used quite rarely, however, it must be emphasized that the main idea of the proof is to compare two operations (or functions) - direct and inverse to it, which will be used already in high school.

Technological chain of formation of algorithms and techniques

identity transformations of expressions in basic school

Algorithm and calculation methods

Whole expressions

Types of integer expressions (monomial, polynomial), their degree, standard form, special cases, abbreviated multiplication formulas. Actions with integer expressions: factoring a polynomial; identifying a perfect square in a trinomial.

1. Algorithms for performing basic actions with entire expressions.

2. Techniques for factoring a polynomial.

3. A special technique for isolating a complete square in a trinomial.

4. A generalized technique for simplifying an entire expression.

5. Techniques for proving identity.

Rational expressions

The main property of a fractional expression and its consequences. Reducing fractional expressions. Actions with rational

expressions.

6. Techniques for writing transformations of rational expressions.

7. Techniques for using analogies with actions on rational numbers in general and special cases.

8. Generalization of techniques 4 and 5.

Irrational

expressions

The main property of a root, the simplest transformations of roots. Actions with roots, raising an expression to a power with a fractional exponent.

9. Special techniques for basic transformations of arithmetic roots.

10.Techniques for converting expressions with powers with a rational exponent.

11. Receiving the proof of inequalities.

12. Generalization of techniques 2, 4, 5 and 11.

Assignment for the lecture

After analyzing school textbooks, create a table of identical equalities indicating the set on which it is true.

Example
, M 1 – those x for which f(x) makes sense.

LECTURE No. 3 Proof of identities

Purpose: 1. Repeat the definitions of identity and identically equal expressions.

2.Introduce the concept of identical transformation of expressions.

3. Multiplying a polynomial by a polynomial.

4. Factoring a polynomial using the grouping method.

Let every day and every hour

He'll get us something new,

Let our minds be good,

And the heart will be smart!

There are many concepts in mathematics. One of them is identity.

An identity is an equality that holds for all values of the variables included in it. We already know some identities.

For example, everyone abbreviated multiplication formulas are identities.

Abbreviated multiplication formulas

1. (a ± b)2 = a 2 ± 2 ab + b 2,

2. (a ± b)3 = a 3 ± 3 a 2b + 3ab 2 ± b 3,

3. a 2 - b 2 = (a - b)(a + b),

4. a 3 ± b 3 = (a ± b)(a 2 ab + b 2).

Prove identity- this means establishing that for any valid variable value, its left side is equal to the right side.

In algebra, there are several different ways to prove identities.

Methods for proving identities

left side of the identity.

the right side of the identity.

left and right sides of the identity.

From the right side of the identity we subtract the left side.

The right side is subtracted from the left side of the identity.

It should also be remembered that the identity is valid only for permissible values of the variables.

As you can see, there are quite a lot of ways. Which method to choose in a given case depends on the identity you need to prove. As you prove various identities, you will gain experience in choosing a method of proof.

An identity is an equation that is satisfied identically, that is, valid for any admissible values of the variables included in it. To prove an identity means to establish that for all admissible values of the variables, its left and right sides are equal.
Ways to prove identity:
1. Perform transformations on the left side and ultimately obtain the right side.
2. Perform transformations on the right side and ultimately obtain the left side.
3. Separately transform the right and left sides and obtain the same expression in both the first and second cases.
4. Compose the difference between the left and right sides and, as a result of its transformations, obtain zero.
Let's look at a few simple examples

Example 1. Prove the identity x·(a+b) + a·(b-x) = b·(a+x).

Solution.

Since the right side has a small expression, let's try to transform the left side of the equality.

x·(a+b) + a·(b-x) = x·a +x·b + a·b – a·x.

Let us present similar terms and take the common factor out of the bracket.

x a + x b + a b – a x = x b + a b = b (a + x).

We found that the left side after the transformations became the same as the right side. Therefore, this equality is an identity.

Example 2. Prove the identity: a² + 7·a + 10 = (a+5)·(a+2).

Solution:

In this example, you can proceed in the following way. Let's open the brackets on the right side of the equality.

(a+5)·(a+2) = (a²) + 5·a +2·a +10 = a²+7·a + 10.

We see that after the transformations, the right side of the equality became the same as the left side of the equality. Therefore, this equality is an identity.

“The replacement of one expression by another identically equal to it is called an identical transformation of the expression”

Find out which equality is an identity:

1. - (a – c) = - a – c;

2. 2 · (x + 4) = 2x – 4;

3. (x – 5) · (-3) = - 3x + 15.

4. рху (- р2 x2 y) = - р3 x3 y3.

“To prove that some equality is an identity, or, as they say differently, to prove an identity, identical transformations of expressions are used”

An equality that is true for any values of the variables is called identity. To prove that some equality is an identity, or, as they say differently, so that prove identity, use identical transformations of expressions.
Let's prove the identity:
xy - 3y - 5x + 16 = (x - 3)(y - 5) + 1 Transform the left side of this equality:
xy - 3y - 5x + 16 = (xy - 3y) + (- 5x + 15) +1 = y(x - 3) - 5(x -3) +1 = (y - 5)(x - 3) + 1 As a result identity transformation from the left side of the polynomial we obtained its right side and thereby proved that this equality is identity.
For proofs of identity transform its left side into the right or its right side into the left, or show that the left and right sides of the original equality are identically equal to the same expression.

Multiplying a polynomial by a polynomial

Let's multiply the polynomial a+b to a polynomial c + d. Let's compose the product of these polynomials:
(a+b)(c+d).
Let us denote the binomial a+b letter x and transform the resulting product according to the rule for multiplying a monomial by a polynomial:
(a+b)(c+d) = x(c+d) = xc + xd.
In expression xc + xd. let's substitute x polynomial a+b and again use the rule for multiplying a monomial by a polynomial:
xc + xd = (a+b)c + (a+b)d = ac + bc + ad + bd.
So: (a+b)(c+d) = ac + bc + ad + bd.
Product of polynomials a+b And c + d we represented it as a polynomial ac + bc + ad + bd. This polynomial is the sum of all monomials obtained by multiplying each term of the polynomial a+b for each term of the polynomial c + d.
Conclusion: the product of any two polynomials can be represented as a polynomial.
Rule: To multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of another polynomial and add the resulting products.
Note that when multiplying a polynomial containing m terms to a polynomial containing n terms in the product, before bringing similar terms, the result should be mn members. This can be used for control.

Factoring a polynomial using the grouping method:

Previously, we were introduced to factoring a polynomial by taking the common factor out of brackets. Sometimes it is possible to factor a polynomial using another method - grouping of its members.
Let's factor the polynomial
ab - 2b + 3a - 6 Let's group it so that the terms in each group have a common factor and take this factor out of brackets:
ab - 2b + 3a - 6 = (ab - 2b) + (3a - 6) = b(a - 2) + 3(a - 2) Each term of the resulting expression has a common factor (a - 2). Let's take this common factor out of brackets:
b(a - 2) + 3(a - 2) = (b +3)(a - 2) As a result, we factored the original polynomial:
ab - 2b + 3a - 6 = (b +3)(a - 2) The method we used to factor the polynomial is called grouping method.
Polynomial expansion ab - 2b + 3a - 6 factorization can be done by grouping its terms differently:
ab - 2b + 3a - 6 = (ab + 3a) + (- 2b - 6) = a(b + 3) -2(b + 3) = (a - 2)(b + 3)

Repeat:

1. Methods of proving identities.

2. What is called the identity transformation of an expression.

3. Multiplying a polynomial by a polynomial.

4. Factoring a polynomial using the grouping method

Back forward

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Goals:

Review the definitions of identity and identically equal expressions.
Introduce the concept of identity transformation of expressions.
To develop students' skills in proving identities using the method of identical transformation of expressions.
Bring up communicative culture students.

During the classes

Before the start of the lesson, students in the class are divided into six mixed study groups.

I

Teacher: Hello, guys, I propose to turn the classroom into a temporary research laboratory, and you and I in Masters of Science in Mathematical Sciences.

But every self-respecting scientist constantly decides some very important problem, so we, first of all, have to find out: what problem are we going to work on today?

To do this, we need to solve two problems: (Slide 1)

Factor the expression 4x – 8xy.(After completing the task, the word “Proof” appears on the slide)
Imagine the expression -5у(у – 2) in the form of a polynomial. (After completing the task, the word “Identities” appears on the slide)

Teacher: Today we will work on the “Proof of Identities”, and I propose to take these wonderful words as the motto of our work: (Slide 2)

Let every day and every hour
He'll get us something new,
Let our minds be good,
And the heart will be smart!

II

Teacher: Gentlemen, scientists, before solving the problem, we need to strengthen our theoretical base, because the concept of identity is already familiar to you. And therefore in the section (Slide 3) “Repetition is the mother of learning” I suggest you do the following:

In each scientific group there are formulations of three concepts on card 1, you must find two definitions among them: 1) Definition of identity, 2) Definition of identically equal expressions.

(Students study these definitions for 2-3 minutes, representatives of those groups that completed the task fastest are asked, the rest of the participants in other groups show agreement or disagreement using green and red signal cards)

Card 1

Once students give the correct definition, it is displayed on the screen.

Teacher: Okay, now let's test ourselves. Equalities will appear on the screen, if this equality is an identity, then I suggest you stand up, but if not, then you continue to sit: (Slide 4)

- (a – b) = - a + b
a (b + c) = ab - ac
a – (b + c) = a – b + c
(a + b) – c = a – c + c
- (a + b) = - b - a

III

Teacher: Okay, now it’s time for us to turn from theorists into practical scientists, but for this we need to find out what we need to use in order prove identity, and here we cannot do without scientific literature, we will find the answer to this question on the page ... of your textbook. Students find the answer in the textbook: “To prove that some equality is an identity, or, as they say differently, to prove an identity, they use identical transformations of expressions.” Participants in other groups indicate agreement or disagreement with the special signals discussed above. (Slide 5)

Teacher: Well done, but now it arises next question, and what is identity transformation of expressions? The answer can be found at card 1, this is the remaining third definition.

“The replacement of one expression with another, identically equal to it, is called an identical transformation of the expression” (the teacher invites one of the participants of any group to answer this question) (Slide 6)

Now we are already “ripe” for practical work, and I would ask you to turn your attention to card 2. Assignment: “Prove the identity,” each group of scientists received an example that they must solve independently; if difficulties arise, consultant cards will come to the rescue.

Card 2

Now we need to protect our work. (Presentation of completed work at the board, willing group members speak)

Teacher: Great, and now, dear colleagues, it’s time to sum up, what do we need to do to prove that equality is identity? Expected student answers: (Slide 7)

Write out the left side of the equality, transform it and make sure that it is equal to the right.
or
Write down the right side of the equality, transform it and make sure that it is equal to the left.
or
Transform both the left and right sides of the equality and make sure that they are equal to the same expression.

Teacher: What conclusion can be drawn in the case when everything we just said will not be fulfilled? Suggested student answer: Equality will not be identity.

IV

Teacher: To ensure that the knowledge gained is lasting, we will continue this work at home:

Homework: p. 30, 773, * Make up an equality that will be an identity.

V

Teacher: And now the hour has come for creativity: In the poem that you see, insert the missing words: (Slides 8-9)

There are all sorts of equalities, brothers,
And everyone, of course, knows about this.
There are – with variables, there are – (numeric),
Very, very complex (simple)
But among equalities there is a special class,
We will tell our story about him now.
This is called (identity) equality.
But we still have to prove this.
To do this we just need to take
And equality is (convert)
Of course, it won't be difficult for us to find out
What part will we have to change?
Or maybe we'll have to change both,
By equality of mind it is not difficult (to understand)
Hooray! We were able to apply our knowledge
Equality conversion completed.
And we boldly say the answer:
So (identity) is it, or not!

Proof of identities. There are many concepts in mathematics. One of them is identity.

An identity is an equality that holds for all values of the variables included in it.

We already know some identities. For example, all abbreviated multiplication formulas are identities.

Prove identity- this means establishing that for any valid variable value, its left side is equal to the right side.

In algebra, there are several different ways to prove identities.

Methods for proving identities

left side of the identity. If we end up with the right-hand side, then the identity is considered proven.
Perform equivalent conversions the right side of the identity. If we finally get the left side, then the identity is considered proven.
Perform equivalent conversions left and right sides of the identity. If we get the same result, then the identity is considered proven.
From the right side of the identity we subtract the left side.
The right side is subtracted from the left side of the identity. We perform equivalent transformations on the difference. And if in the end we get zero, then the identity is considered proven.

It should also be remembered that the identity is valid only for permissible values of the variables.

Let's look at some simple examples

Example 1.

Prove the identity x*(a+b) + a*(b-x) = b*(a+x).

Solution.

Since the right side has a small expression, let's try to transform the left side of the equality.

x*(a+b) + a*(b-x) = x*a+x*b+a*b – a*x.

Let us present similar terms and take the common factor out of the bracket.

x*a+x*b+a*b – a*x = x*b+a*b = b*(a+x).

We found that the left side after the transformations became the same as the right side. Therefore, this equality is an identity.

Example 2.

Prove the identity a^2 + 7*a + 10 = (a+5)*(a+2).

Solution.

In this example, you can proceed in the following way. Let's open the brackets on the right side of the equality.

(a+5)*(a+2) = (a^2) +5*a +2*a +10= a^2+7*a+10.

We see that after the transformations, the right side of the equality became the same as the left side of the equality. Therefore, this equality is an identity.

Example 2. Prove identity

We will prove this identity by transforming the expression on the right side.

Method 1.

That's why

Method 2.

First of all, note that ctg α =/= 0; otherwise the expression tg would not make sense α = 1/ctg α . But if ctg α =/= 0, then the numerator and denominator of the radical expression can be multiplied by ctg α , without changing the value of the fraction. Hence,

Using tg identities α ctg α = 1 and 1+ ctg 2 α = cosec 2 α , we get

That's why Q.E.D.

Comment. It should be noted that the left side of the proven identity (sin α ) is defined for all values α , and the right one - only when α =/= π / 2 n.

Therefore, only when all valid values α In general, these expressions are not equivalent to each other.

Example 3. Prove identity

sin (3 / 2 π + α ) + cos ( π - α ) = cos (2 π + α ) - 3sin ( π / 2 - α )

Let's transform the left and right sides of this identity using the reduction formulas:

sin (3 / 2 π + α ) + cos ( π - α ) = -cos α -cos α = - 2cos α ;

cos(2 π + α ) - 3sin ( π / 2 - α ) =cos α - 3cos α = - 2cos α .

So, the expressions appearing in both parts of this identity are reduced to the same form. This proves the identity.

Example 4. Prove identity

sin 4 α + cos 4 α - 1 = - 2 sin 2 α cos 2 α .

Let us show that the difference between the left and right sides. of this identity is equal to zero.

(sin 4 α + cos 4 α - 1) - (- 2 sin 2 α cos 2 α ) = (sin 4 α + 2sin 2 α cos 2 α + cos 4 α ) - 1 =

= (sin 2 α + cos 2 α ) 2 - 1 = 1 - 1 = 0.

This proves the identity.

Example 5. Prove identity

This identity can be considered as a proportion. But to prove the validity of the proportion a / b = c / d, it is enough to show that the product of its extreme terms ad equal to the product of its average terms bc. This is what we will do in in this case. Let us show that (1 - sin α ) (1+ sin α ) = cos α cos α .

Indeed, (1 - sin α ) (1 + sin α ) = 1 -sin 2 α = cos 2 α .