Transformations of trigonometric expressions. Lesson "Simplifying Trigonometric Expressions"

Lesson 1

Subject: 11th grade (preparation for the Unified State Exam)

Simplification trigonometric expressions.

Solving simple trigonometric equations. (2 hours)

Goals:

• Systematize, generalize, expand students’ knowledge and skills related to the use of trigonometry formulas and solving simple trigonometric equations.

Equipment for the lesson:

Lesson structure:

1. Organizational moment
2. Testing on laptops. The discussion of the results.
3. Simplifying trigonometric expressions
4. Solving simple trigonometric equations
5. Independent work.
6. Lesson summary. Explanation of homework assignment.

1. Organizational moment. (2 minutes.)

The teacher greets the audience, announces the topic of the lesson, reminds them that they were previously given the task of repeating trigonometry formulas, and prepares students for testing.

2. Testing. (15 min + 3 min discussion)

The goal is to test knowledge of trigonometric formulas and the ability to apply them. Each student has a laptop on their desk with a version of the test.

There can be any number of options, I will give an example of one of them:

I option.

Simplify expressions:

a) basic trigonometric identities

1. sin 2 3y + cos 2 3y + 1;

b) addition formulas

3. sin5x - sin3x;

c) converting a product into a sum

6. 2sin8y cos3y;

d) double angle formulas

7. 2sin5x cos5x;

e) formulas for half angles

f) triple angle formulas

and) universal substitution

h) reduction in degree

16. cos 2 (3x/7);

Students see their answers on the laptop next to each formula.

The work is instantly checked by the computer. The results are displayed on a large screen for everyone to see.

Also, after finishing the work, the correct answers are shown on the students’ laptops. Each student sees where the mistake was made and what formulas he needs to repeat.

3. Simplification of trigonometric expressions. (25 min.)

The goal is to repeat, practice and consolidate the use of basic trigonometry formulas. Solving problems B7 from the Unified State Exam.

On at this stage It is advisable to divide the class into groups of strong people (they work independently with subsequent testing) and weak students who work with the teacher.

Assignment for strong students (prepared in advance for printed basis). The main emphasis is on reduction formulas and double angle, according to the Unified State Examination 2011.

Simplify expressions (for strong students):

At the same time, the teacher works with weak students, discussing and solving tasks on the screen under the students’ dictation.

Calculate:

5) sin(270º - α) + cos (270º + α)

6)

Simplify:

It was time to discuss the results of the work of the strong group.

The answers appear on the screen, and also, using a video camera, the work of 5 different students is displayed (one task for each).

The weak group sees the condition and method of solution. Discussion in progress and analysis. Using technical means it happens quickly.

4. Solving simple trigonometric equations. (30 min.)

The goal is to repeat, systematize and generalize the solution of the simplest trigonometric equations and write down their roots. Solution of problem B3.

Any trigonometric equation, no matter how we solve it, leads to the simplest.

When completing the task, students should pay attention to writing down the roots of equations of special cases and general view and on the selection of roots in the last equation.

Solve equations:

Write down the smallest positive root as your answer.

5. Independent work (10 min.)

The goal is to test the acquired skills, identify problems, errors and ways to eliminate them.

Multi-level work is offered to the student's choice.

Option "3"

1) Find the value of the expression

2) Simplify the expression 1 - sin 2 3α - cos 2 3α

3) Solve the equation

Option for "4"

1) Find the value of the expression

2) Solve the equation Write down the smallest positive root in your answer.

Option "5"

1) Find tanα if

2) Find the root of the equation Write down the smallest positive root as your answer.

6. Lesson summary (5 min.)

The teacher summarizes what was repeated and reinforced in the lesson trigonometric formulas, solving simple trigonometric equations.

Homework is assigned (prepared on a printed basis in advance) with spot check in the next lesson.

Solve equations:

9)

10) In your answer, indicate the smallest positive root.

Lesson 2

Subject: 11th grade (preparation for the Unified State Exam)

Methods for solving trigonometric equations. Root selection. (2 hours)

Goals:

• Generalize and systematize knowledge on solving trigonometric equations of various types.
• Promote development mathematical thinking students, the ability to observe, compare, generalize, classify.
• Encourage students to overcome difficulties in the process mental activity, to self-control, introspection of one’s activities.

Equipment for the lesson: KRMu, laptops for each student.

Lesson structure:

1. Organizational moment
2. Discussion of d/z and self. work from last lesson
3. Review of methods for solving trigonometric equations.
4. Solving trigonometric equations
5. Selection of roots in trigonometric equations.
6. Independent work.
7. Lesson summary. Homework.

1. Organizational moment (2 min.)

The teacher greets the audience, announces the topic of the lesson and the work plan.

2. a) Analysis homework(5 minutes.)

The goal is to check execution. One work is displayed on the screen using a video camera, the rest are selectively collected for teacher checking.

b) Analysis independent work(3 min.)

The goal is to analyze mistakes and indicate ways to overcome them.

Answers and solutions are on the screen; students have their work given out in advance. Analysis proceeds quickly.

3. Review of methods for solving trigonometric equations (5 min.)

The goal is to recall methods for solving trigonometric equations.

Ask students what methods of solving trigonometric equations they know. Emphasize that there are so-called basic (frequently used) methods:

• variable replacement,
• factorization,
• homogeneous equations,

and there is applied methods:

• using the formulas for converting a sum into a product and a product into a sum,
• according to formulas reduction in degree,
• universal trigonometric substitution
• introduction auxiliary angle,
• multiplication by some trigonometric function.

It should also be recalled that one equation can be solved in different ways.

4. Solving trigonometric equations (30 min.)

The goal is to generalize and consolidate knowledge and skills on this topic, to prepare for the C1 solution from the Unified State Exam.

I consider it advisable to solve equations for each method together with students.

The student dictates the solution, the teacher writes it down on the tablet, and the whole process is displayed on the screen. This will allow you to quickly and effectively recall previously covered material in your memory.

Solve equations:

1) replacing the variable 6cos 2 x + 5sinx - 7 = 0

2) factorization 3cos(x/3) + 4cos 2 (x/3) = 0

3) homogeneous equations sin 2 x + 3cos 2 x - 2sin2x = 0

4) converting the sum into a product cos5x + cos7x = cos(π + 6x)

5) converting the product into the sum 2sinx sin2x + cos3x = 0

6) reduction of the degree sin2x - sin 2 2x + sin 2 3x = 0.5

7) universal trigonometric substitution sinx + 5cosx + 5 = 0.

When solving this equation, it should be noted that using this method leads to a narrowing of the definition range, since sine and cosine are replaced by tg(x/2). Therefore, before writing out the answer, you need to check whether the numbers from the set π + 2πn, n Z are horses of this equation.

8) introduction of an auxiliary angle √3sinx + cosx - √2 = 0

9) multiplication by some trigonometric cosx function cos2x cos4x = 1/8.

5. Selection of roots of trigonometric equations (20 min.)

Since in conditions of fierce competition when entering universities, solving the first part of the exam alone is not enough, most students should pay attention to the tasks of the second part (C1, C2, C3).

Therefore, the goal of this stage of the lesson is to remember previously studied material and prepare to solve problem C1 from the Unified State Exam 2011.

Exist trigonometric equations, in which it is necessary to select roots when writing out the answer. This is due to some restrictions, for example: the denominator of the fraction is not equal to zero, the expression under the even root is non-negative, the expression under the logarithm sign is positive, etc.

Such equations are considered equations increased complexity and in version of the Unified State Exam are in the second part, namely C1.

Solve the equation:

A fraction is equal to zero if then by using unit circle let's select the roots (see Figure 1)

Picture 1.

we get x = π + 2πn, n Z

Answer: π + 2πn, n Z

On the screen, the selection of roots is shown on a circle in a color image.

The product is equal to zero when at least one of the factors is equal to zero, and the arc does not lose its meaning. Then

Using the unit circle, we select the roots (see Figure 2)

The video lesson “Simplification of trigonometric expressions” is intended to develop students’ skills in solving trigonometric problems using basic trigonometric identities. During the video lesson, types of trigonometric identities and examples of solving problems using them are discussed. Applying visual material, it is easier for the teacher to achieve the lesson objectives. Vivid presentation of material promotes memorization important points. The use of animation effects and voice-over allows you to completely replace the teacher at the stage of explaining the material. Thus, by using this visual aid in mathematics lessons, the teacher can increase the effectiveness of teaching.

At the beginning of the video lesson, its topic is announced. Then we recall the trigonometric identities studied earlier. The screen displays the equalities sin 2 t+cos 2 t=1, tg t=sin t/cos t, where t≠π/2+πk for kϵZ, ctg t=cos t/sin t, correct for t≠πk, where kϵZ, tg t· ctg t=1, for t≠πk/2, where kϵZ, called the basic trigonometric identities. It is noted that these identities are often used in solving problems where it is necessary to prove equality or simplify an expression.

Below we consider examples of the application of these identities in solving problems. First, it is proposed to consider solving problems of simplifying expressions. In example 1, it is necessary to simplify the expression cos 2 t- cos 4 t+ sin 4 t. To solve the example, first put it out of brackets common multiplier cos 2 t. As a result of this transformation in parentheses, the expression 1- cos 2 t is obtained, the value of which from the main identity of trigonometry is equal to sin 2 t. After transforming the expression, it is obvious that one more common factor sin 2 t can be taken out of brackets, after which the expression takes the form sin 2 t(sin 2 t+cos 2 t). From the same basic identity we derive the value of the expression in brackets equal to 1. As a result of simplification, we obtain cos 2 t- cos 4 t+ sin 4 t= sin 2 t.

In example 2, the expression cost/(1- sint)+ cost/(1+ sint) needs to be simplified. Since the numerators of both fractions contain the expression cost, it can be taken out of brackets as a common factor. The fractions in parentheses are then reduced to common denominator multiplying (1- sint)(1+ sint). After bringing similar terms the numerator remains 2, and the denominator 1 - sin 2 t. On the right side of the screen is a reminder of basic trigonometry identity sin 2 t+cos 2 t=1. Using it, we find the denominator of the fraction cos 2 t. After reducing the fraction, we obtain a simplified form of the expression cost/(1- sint)+ cost/(1+ sint)=2/cost.

Next, we consider examples of proofs of identities that use the acquired knowledge about the basic identities of trigonometry. In example 3, it is necessary to prove the identity (tg 2 t-sin 2 t)·ctg 2 t=sin 2 t. The right side of the screen displays three identities that will be needed for the proof - tg t·ctg t=1, ctg t=cos t/sin t and tg t=sin t/cos t with restrictions. To prove the identity, the brackets are first opened, after which a product is formed that reflects the expression of the main trigonometric identity tg t·ctg t=1. Then, according to the identity from the definition of cotangent, ctg 2 t is transformed. As a result of the transformations, the expression 1-cos 2 t is obtained. Using the main identity, we find the meaning of the expression. Thus, it has been proven that (tg 2 t-sin 2 t)·ctg 2 t=sin 2 t.

In example 4, you need to find the value of the expression tg 2 t+ctg 2 t if tg t+ctg t=6. To calculate the expression, first square the right and left sides of the equality (tg t+ctg t) 2 =6 2. The abbreviated multiplication formula is recalled on the right side of the screen. After opening the brackets on the left side of the expression, the sum tg 2 t+2· tg t·ctg t+ctg 2 t is formed, to transform which you can apply one of the trigonometric identities tg t·ctg t=1, the form of which is recalled on the right side of the screen. After the transformation, the equality tg 2 t+ctg 2 t=34 is obtained. The left side of the equality coincides with the condition of the problem, so the answer is 34. The problem is solved.

The video lesson “Simplification of trigonometric expressions” is recommended for use in traditional school lesson mathematics. The material will also be useful to the teacher implementing distance learning. In order to develop skills in solving trigonometric problems.

TEXT DECODING:

"Simplification of trigonometric expressions."

Equalities

1) sin 2 t + cos 2 t = 1 (sine square te plus cosine square te equals one)

2)tgt =, for t ≠ + πk, kϵZ (tangent te is equal to the ratio of sine te to cosine te with te not equal to pi by two plus pi ka, ka belongs to zet)

3)ctgt = , for t ≠ πk, kϵZ (cotangent te is equal to the ratio of cosine te to sine te with te not equal to pi ka, ka belongs to zet).

4) tgt ∙ ctgt = 1 for t ≠ , kϵZ (the product of tangent te by cotangent te is equal to one when te is not equal to peak ka, divided by two, ka belongs to zet)

are called basic trigonometric identities.

They are often used in simplifying and proving trigonometric expressions.

Let's look at examples of using these formulas to simplify trigonometric expressions.

EXAMPLE 1. Simplify the expression: cos 2 t - cos 4 t + sin 4 t. (expression a cosine squared te minus cosine of the fourth degree te plus sine of the fourth degree te).

Solution. cos 2 t - cos 4 t + sin 4 t = cos 2 t∙ (1 - cos 2 t) + sin 4 t =cos 2 t ∙ sin 2 t + sin 4 t = sin 2 t (cos 2 t + sin 2 t) = sin 2 t 1= sin 2 t

(we take out the common factor cosine square te, in brackets we get the difference between unity and the squared cosine te, which is equal to the squared sine te by the first identity. We get the sum of the fourth power sine te of the product cosine square te and sine square te. We take out the common factor sine square te outside the brackets, in brackets we get the sum of the squares of the cosine and sine, which is basically trigonometric identity equals one. As a result, we obtain the square of the sine te).

EXAMPLE 2. Simplify the expression: + .

(expression be the sum of two fractions in the numerator of the first cosine te in the denominator one minus sine te, in the numerator of the second cosine te in the denominator of the second one plus sine te).

(Let’s take the common factor cosine te out of brackets, and in brackets we bring it to a common denominator, which is the product of one minus sine te by one plus sine te.

In the numerator we get: one plus sine te plus one minus sine te, we give similar ones, the numerator is equal to two after bringing similar ones.

In the denominator, you can apply the abbreviated multiplication formula (difference of squares) and obtain the difference between unity and the square of the sine te, which, according to the basic trigonometric identity

equal to the square of the cosine te. After reducing by cosine te we get the final answer: two divided by cosine te).

Let's look at examples of using these formulas when proving trigonometric expressions.

EXAMPLE 3. Prove the identity (tg 2 t - sin 2 t) ∙ ctg 2 t = sin 2 t (the product of the difference between the squares of tangent te and sine te by the square of cotangent te is equal to the square of sine te).

Proof.

Let's transform left side equality:

(tg 2 t - sin 2 t) ∙ ctg 2 t = tg 2 t ∙ ctg 2 t - sin 2 t ∙ ctg 2 t = 1 - sin 2 t ∙ ctg 2 t =1 - sin 2 t ∙ = 1 - cos 2 t = sin 2 t

(Let's open the brackets; from the previously obtained relation it is known that the product of the squares of tangent te by cotangent te is equal to one. Recall that cotangent te equal to the ratio cosine te by sine te, which means the square of the cotangent is the ratio of the square of the cosine te by the square of the sine te.

After reduction by sine square te we obtain the difference between unity and cosine square te, which is equal to sine square te). Q.E.D.

EXAMPLE 4. Find the value of the expression tg 2 t + ctg 2 t if tgt + ctgt = 6.

(the sum of the squares of tangent te and cotangent te, if the sum of tangent and cotangent is six).

Solution. (tgt + ctgt) 2 = 6 2

tg 2 t + 2 ∙ tgt ∙ctgt + ctg 2 t = 36

tg 2 t + 2 + ctg 2 t = 36

tg 2 t + ctg 2 t = 36-2

tg 2 t + ctg 2 t = 34

Let's square both sides of the original equality:

(tgt + ctgt) 2 = 6 2 (the square of the sum of tangent te and cotangent te is equal to six squared). Let us recall the formula for abbreviated multiplication: Square of the sum of two quantities equal to square the first plus twice the product of the first and the second plus the square of the second. (a+b) 2 =a 2 +2ab+b 2 We get tg 2 t + 2 ∙ tgt ∙ctgt + ctg 2 t = 36 (tangent squared te plus double the product of tangent te by cotangent te plus cotangent squared te equals thirty-six) .

Since the product of tangent te and cotangent te is equal to one, then tg 2 t + 2 + ctg 2 t = 36 (the sum of the squares of tangent te and cotangent te and two is equal to thirty-six),