Trigonometric equations of degree 2. Lesson "homogeneous trigonometric equations"

If you follow the definition, then the derivative of a function at a point is the limit of the ratio of the increment of the function Δ y to the argument increment Δ x:

Everything seems to be clear. But try using this formula to calculate, say, the derivative of the function f(x) = x 2 + (2x+ 3) · e x sin x. If you do everything by definition, then after a couple of pages of calculations you will simply fall asleep. Therefore, there are simpler and more effective ways.

To begin with, we note that from the entire variety of functions we can distinguish the so-called elementary functions. These are relatively simple expressions, the derivatives of which have long been calculated and tabulated. Such functions are quite easy to remember - along with their derivatives.

Derivatives of elementary functions

Elementary functions are all those listed below. The derivatives of these functions must be known by heart. Moreover, it is not at all difficult to memorize them - that’s why they are elementary.

So, derivatives of elementary functions:

If an elementary function is multiplied by an arbitrary constant, then the derivative of the new function is also easily calculated:

(C · f)’ = C · f ’.

In general, constants can be taken out of the sign of the derivative. For example:

(2x 3)’ = 2 · ( x 3)’ = 2 3 x 2 = 6x 2 .

Obviously, elementary functions can be added to each other, multiplied, divided - and much more. This is how new functions will appear, no longer particularly elementary, but also differentiated according to certain rules. These rules are discussed below.

Derivative of sum and difference

Let the functions be given f(x) And g(x), the derivatives of which are known to us. For example, you can take the elementary functions discussed above. Then you can find the derivative of the sum and difference of these functions:

1. (f + g)’ = f ’ + g ’
2. (f − g)’ = f ’ − g ’

So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, ( f + g + h)’ = f ’ + g ’ + h ’.

Strictly speaking, there is no concept of “subtraction” in algebra. There is a concept of “negative element”. Therefore the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.

f(x) = x 2 + sin x; g(x) = x 4 + 2x 2 − 3.

Function f(x) is the sum of two elementary functions, therefore:

f ’(x) = (x 2 + sin x)’ = (x 2)’ + (sin x)’ = 2x+ cos x;

We reason similarly for the function g(x). Only there are already three terms (from the point of view of algebra):

g ’(x) = (x 4 + 2x 2 − 3)’ = (x 4 + 2x 2 + (−3))’ = (x 4)’ + (2x 2)’ + (−3)’ = 4x 3 + 4x + 0 = 4x · ( x 2 + 1).

Answer:
f ’(x) = 2x+ cos x;
g ’(x) = 4x · ( x 2 + 1).

Derivative of the product

Mathematics is a logical science, so many people believe that if the derivative of a sum is equal to the sum of derivatives, then the derivative of the product strike">equal to the product of derivatives. But screw you! The derivative of a product is calculated using a completely different formula. Namely:

(f · g) ’ = f ’ · g + f · g ’

The formula is simple, but it is often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.

Task. Find derivatives of functions: f(x) = x 3 cos x; g(x) = (x 2 + 7x− 7) · e x .

Function f(x) is the product of two elementary functions, so everything is simple:

f ’(x) = (x 3 cos x)’ = (x 3)’ cos x + x 3 (cos x)’ = 3x 2 cos x + x 3 (− sin x) = x 2 (3cos x − x sin x)

Function g(x) the first multiplier is a little more complicated, but the general scheme does not change. Obviously, the first factor of the function g(x) is a polynomial and its derivative is the derivative of the sum. We have:

g ’(x) = ((x 2 + 7x− 7) · e x)’ = (x 2 + 7x− 7)’ · e x + (x 2 + 7x− 7) ( e x)’ = (2x+ 7) · e x + (x 2 + 7x− 7) · e x = e x· (2 x + 7 + x 2 + 7x −7) = (x 2 + 9x) · e x = x(x+ 9) · e x .

Answer:
f ’(x) = x 2 (3cos x − x sin x);
g ’(x) = x(x+ 9) · e x .

Please note that in the last step the derivative is factorized. Formally, this does not need to be done, but most derivatives are not calculated on their own, but to examine the function. This means that further the derivative will be equated to zero, its signs will be determined, and so on. For such a case, it is better to have an expression factorized.

If there are two functions f(x) And g(x), and g(x) ≠ 0 on the set we are interested in, we can define a new function h(x) = f(x)/g(x). For such a function you can also find the derivative:

Not weak, huh? Where did the minus come from? Why g 2? And so! This is one of the most complex formulas - you can’t figure it out without a bottle. Therefore, it is better to study it with specific examples.

Task. Find derivatives of functions:

The numerator and denominator of each fraction contain elementary functions, so all we need is the formula for the derivative of the quotient:

According to tradition, let's factorize the numerator - this will greatly simplify the answer:

A complex function is not necessarily a half-kilometer-long formula. For example, it is enough to take the function f(x) = sin x and replace the variable x, say, on x 2 + ln x. It will work out f(x) = sin ( x 2 + ln x) - this is a complex function. It also has a derivative, but it will not be possible to find it using the rules discussed above.

What should I do? In such cases, replacing a variable and formula for the derivative of a complex function helps:

f ’(x) = f ’(t) · t', If x is replaced by t(x).

As a rule, the situation with understanding this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it using specific examples, with a detailed description of each step.

Task. Find derivatives of functions: f(x) = e 2x + 3 ; g(x) = sin ( x 2 + ln x)

Note that if in the function f(x) instead of expression 2 x+ 3 will be easy x, then we get an elementary function f(x) = e x. Therefore, we make a replacement: let 2 x + 3 = t, f(x) = f(t) = e t. We look for the derivative of a complex function using the formula:

f ’(x) = f ’(t) · t ’ = (e t)’ · t ’ = e t · t ’

And now - attention! We perform the reverse replacement: t = 2x+ 3. We get:

f ’(x) = e t · t ’ = e 2x+ 3 (2 x + 3)’ = e 2x+ 3 2 = 2 e 2x + 3

Now let's look at the function g(x). Obviously it needs to be replaced x 2 + ln x = t. We have:

g ’(x) = g ’(t) · t’ = (sin t)’ · t’ = cos t · t ’

Reverse replacement: t = x 2 + ln x. Then:

g ’(x) = cos ( x 2 + ln x) · ( x 2 + ln x)’ = cos ( x 2 + ln x) · (2 x + 1/x).

That's it! As can be seen from the last expression, the whole problem has been reduced to calculating the derivative sum.

Answer:
f ’(x) = 2 · e 2x + 3 ;
g ’(x) = (2x + 1/x) cos ( x 2 + ln x).

Very often in my lessons, instead of the term “derivative,” I use the word “prime.” For example, the stroke of the sum is equal to the sum of the strokes. Is that clearer? Well, that's good.

Thus, calculating the derivative comes down to getting rid of these same strokes according to the rules discussed above. As a final example, let's return to the derivative power with a rational exponent:

(x n)’ = n · x n − 1

Few people know that in the role n may well be a fractional number. For example, the root is x 0.5. What if there is something fancy under the root? Again, the result will be a complex function - they like to give such constructions in tests and exams.

Task. Find the derivative of the function:

First, let's rewrite the root as a power with a rational exponent:

f(x) = (x 2 + 8x − 7) 0,5 .

Now we make a replacement: let x 2 + 8x − 7 = t. We find the derivative using the formula:

f ’(x) = f ’(t) · t ’ = (t 0.5)’ · t’ = 0.5 · t−0.5 · t ’.

Let's do the reverse replacement: t = x 2 + 8x− 7. We have:

f ’(x) = 0.5 · ( x 2 + 8x− 7) −0.5 · ( x 2 + 8x− 7)’ = 0.5 · (2 x+ 8) ( x 2 + 8x − 7) −0,5 .

Finally, back to the roots:

“The greatness of a man lies in his ability to think.”
Blaise Pascal.

Lesson objectives:

1) Educational– introduce students to homogeneous equations, consider methods for solving them, and promote the development of skills in solving previously studied types of trigonometric equations.

2) Developmental– develop students’ creative activity, their cognitive activity, logical thinking, memory, the ability to work in a problem situation, achieve the ability to correctly, consistently, rationally express their thoughts, broaden students’ horizons, and increase the level of their mathematical culture.

3) Educational– to cultivate a desire for self-improvement, hard work, to develop the ability to competently and accurately perform mathematical notes, to cultivate activity, to help stimulate interest in mathematics.

Lesson type: combined.

Equipment:

1. Punch cards for six students.
2. Cards for independent and individual work of students.
3. Stands “Solving trigonometric equations”, “Numerical unit circle”.
4. Electrified trigonometry tables.
5. Presentation for the lesson (Appendix 1).

Lesson progress

1. Organizational stage (2 minutes)

Mutual greeting; checking students' preparedness for the lesson (workplace, appearance); organization of attention.

The teacher tells the students the topic of the lesson, goals (slide 2) and explains that during the lesson the handouts that are on the desks will be used.

2. Repetition of theoretical material (15 minutes)

Punch card tasks(6 people) . Working time using punched cards – 10 minutes (Appendix 2)

By solving problems, students will learn where trigonometric calculations are used. The following answers are obtained: triangulation (a technique that allows one to measure distances to nearby stars in astronomy), acoustics, ultrasound, tomography, geodesy, cryptography.

(slide 5)

Frontal survey.

1. What equations are called trigonometric?
2. What types of trigonometric equations do you know?
3. What equations are called the simplest trigonometric equations?
4. What equations are called quadratic trigonometric?
5. Formulate the definition of the arcsine of a.
6. Formulate the definition of the arc cosine of a.
7. Formulate the definition of the arctangent of a.
8. Formulate the definition of the arc cotangent of the number a.

Game "Guess the encrypted word"

Blaise Pascal once said that mathematics is such a serious science that one should not miss an opportunity to make it a little more entertaining. That's why I suggest playing. After solving the examples, determine the sequence of numbers used to compose the encrypted word. In Latin this word means "sine". (slide 3)

2) arc tg (-√3)

4) tg (arc cos (1/2))

5) tg (arc ctg √3)

Answer: "Bend"

Game "Abstract Mathematician"»

Tasks for oral work are projected on the screen:

Check that the equations are solved correctly.(the correct answer appears on the slide after the student's answer). (slide 4)

Checking homework.

The teacher establishes the correctness and awareness of homework completion by all students; identifies gaps in knowledge; improves students' knowledge, skills and abilities in the field of solving simple trigonometric equations.

1 equation. The student comments on the solution to the equation, the lines of which appear on the slide in the order of the comment). (slide 6)

√3tg2x = 1;

tg2x =1/√3;

2х= arctan 1/√3 +πn, n ∈Z.

2х= π/6 +πn, n ∈Z.

x= π/12 + π/2 n, n ∈Z.

2 equation. Solution h written to students on the board.

2 sin 2 x + 3 cosx = 0.

3. Updating new knowledge (3 minutes)

Students, at the request of the teacher, recall ways to solve trigonometric equations. They choose those equations that they already know how to solve, name the method for solving the equation and the resulting result. . The answers appear on the slide. (slide 7) .

Introducing a new variable:

No. 1. 2sin 2 x – 7sinx + 3 = 0.

Let sinx = t, then:

2t 2 – 7t + 3 = 0.

Factorization:

№2. 3sinx cos4x – cos4x = 0;

сos4x(3sinx – 1) = 0;

cos4x = 0 or 3 sinx – 1 = 0; ...

No. 3. 2 sinx – 3 cosx = 0,

No. 4. 3 sin 2 x – 4 sinx cosx + cos 2 x = 0.

Teacher: You still don’t know how to solve the last two types of equations. They are both the same species. They cannot be reduced to an equation for the functions sinx or cosx. Are called homogeneous trigonometric equations. But only the first is a homogeneous equation of the first degree, and the second is a homogeneous equation of the second degree. Today in the lesson we will get acquainted with such equations and learn how to solve them.

4. Explanation of new material (25 minutes)

The teacher gives students definitions of homogeneous trigonometric equations and introduces methods for solving them.

Definition. An equation of the form a sinx + b cosx =0, where a ≠ 0, b ≠ 0 is called homogeneous trigonometric equation of the first degree.(slide 8)

An example of such an equation is equation No. 3. Let us write down the general form of the equation and analyze it.

a sinx + b cosx = 0.

If cosx = 0, then sinx = 0.

– Could such a situation happen?

- No. We have obtained a contradiction to the basic trigonometric identity.

This means cosx ≠ 0. Let’s perform term-by-term division by cosx:

a tgx + b = 0

tgx = –b / a– the simplest trigonometric equation.

Conclusion: Homogeneous trigonometric equations of the first degree are solved by dividing both sides of the equation by cosx (sinx).

For example: 2 sinx – 3 cosx = 0,

Because cosx ≠ 0, then

tgx = 3/2 ;

x = arctan (3/2) +πn, n ∈Z.

Definition. An equation of the form a sin 2 x + b sinx cosx + c cos 2 x = 0, where a ≠ 0, b ≠ 0, c ≠ 0 is called trigonometric equation of the second degree. (slide 8)

An example of such an equation is equation No. 4. Let us write down the general form of the equation and analyze it.

a sin 2 x + b sinx cosx + c cos 2 x = 0.

If cosx = 0, then sinx = 0.

Again we got a contradiction to the basic trigonometric identity.

This means cosx ≠ 0. Let us perform term-by-term division by cos 2 x:

and tg 2 x + b tgx + c = 0 is an equation that reduces to a quadratic.

Conclusion: Oh homogeneous trigonometric equations of the second degree are solved by dividing both sides of the equation by cos 2 x (sin 2 x).

For example: 3 sin 2 x – 4 sinx cosx + cos 2 x = 0.

Because cos 2 x ≠ 0, then

3tg 2 x – 4 tgx + 1 = 0 (Invite the student to go to the board and complete the equation independently).

Replacement: tgx = y. 3y 2 – 4 y + 1 = 0

D = 16 – 12 = 4

y 1 = 1 or y 2 = 1/3

tgx = 1 or tgx = 1/3

x = arctan (1/3) + πn, n ∈Z.

x = arctan + πn, n ∈Z.

x = π/4 + πn, n ∈Z.

5. Stage of checking students' understanding of new material (1 min.)

Select the odd one out:

sinx = 2cosx; 2sinx + cosx = 2;

√3sinx + cosx = 0; sin 2 x – 2 sinx cosx + 4cos 2 x = 0;

4cosx + 5sinx = 0; √3sinx – cosx = 0.

(slide 9)

6. Consolidation of new material (24 min).

Students, together with their respondents, solve equations for new material at the board. The tasks are written on a slide in the form of a table. When solving an equation, the corresponding part of the picture on the slide opens. As a result of completing 4 equations, students are presented with a portrait of a mathematician who had a significant influence on the development of trigonometry. (students will recognize the portrait of François Vieta, a great mathematician who made a great contribution to trigonometry, discovered the property of the roots of the reduced quadratic equation and was involved in cryptography) . (slide 10)

1) √3sinx + cosx = 0,

Because cosx ≠ 0, then

√3tgx + 1 = 0;

tgx = –1/√3;

x = arctan (–1/√3) + πn, n ∈Z.

x = –π/6 + πn, n ∈Z.

2) sin 2 x – 10 sinx cosx + 21cos 2 x = 0.

Because cos 2 x ≠ 0, then tg 2 x – 10 tgx + 21 = 0

Replacement: tgx = y.

y 2 – 10 y + 21 = 0

y 1 = 7 or y 2 = 3

tgx = 7 or tgx = 3

x = arctan7 + πn, n ∈Z

x = arctan3 + πn, n ∈Z

3) sin 2 2x – 6 sin2x cos2x + 5cos 2 2x = 0.

Because cos 2 2x ≠ 0, then 3tg 2 2x – 6tg2x +5 = 0

Replacement: tg2x = y.

3y 2 – 6y + 5 = 0

D = 36 – 20 = 16

y 1 = 5 or y 2 = 1

tg2x = 5 or tg2x = 1

2х = arctan5 + πn, n ∈Z

x = 1/2 arctan5 + π/2 n, n ∈Z

2х = arctan1 + πn, n ∈Z

x = π/8 + π/2 n, n ∈Z

4) 6sin 2 x + 4 sin(π-x) cos(2π-x) = 1.

6sin 2 x + 4 sinx cosx = 1.

6sin 2 x + 4 sinx cosx – sin 2 x – cos 2 x = 0.

5sin 2 x + 4 sinx cosx – cos 2 x = 0.

Because cos 2 x ≠0, then 5tg 2 x + 4 tgx –1 = 0

Replacement: tg x = y.

5у 2 + 4у – 1 = 0

D = 16 + 20 = 36

y 1 = 1/5 or y 2 = –1

tg x = 1/5 or tg x = –1

x = arctan1/5 + πn, n ∈Z

x = arctan(–1) + πn, n ∈Z

x = –π/4 + πn, n ∈Z

Additionally (on the card):

Solve the equation and, choosing one option from the four proposed, guess the name of the mathematician who derived the reduction formulas:

2sin 2 x – 3 sinx cosx – 5cos 2 x = 0.

Possible answers:

x = arctan2 + 2πn, n ∈Z x = –π/2 + πn, n ∈Z – P. Chebyshev

x = arctan 12.5 + 2πn, n ∈Z x = –3π/4 + πn, n ∈Z – Euclid

x = arctan 5 + πn, n ∈Z x = –π/3 + πn, n ∈Z – Sofya Kovalevskaya

x = arctan2.5 + πn, n ∈Z x = –π/4 + πn, n ∈Z – Leonhard Euler

Correct answer: Leonhard Euler.

7. Differentiated independent work (8 min.)

The great mathematician and philosopher more than 2500 years ago suggested a way to develop thinking abilities. “Thinking begins with wonder,” he said. We have been convinced of the correctness of these words many times today. Having completed independent work on 2 options, you will be able to show how you have mastered the material and find out the name of this mathematician. For independent work, use the handouts that are on your tables. You can choose one of the three proposed equations yourself. But remember that by solving the equation corresponding to the color yellow, you can only get “3”, solving the equation corresponding to the color green - “4”, and the color red - “5”. (Appendix 3)

Whatever level of difficulty the students choose, after correctly solving the equation, the first option gets the word “ARIST”, the second - “HOTEL”. The word on the slide is: “ARIST-HOTEL.” (slide 11)

Worksheets with independent work are submitted for verification. (Appendix 4)

8. Recording homework (1 min)

D/z: §7.17. Compose and solve 2 homogeneous equations of the first degree and 1 homogeneous equation of the second degree (using Vieta’s theorem to compose). (slide 12)

9. Summing up the lesson, grading (2 minutes)

The teacher once again draws attention to those types of equations and those theoretical facts that were recalled in the lesson, talking about the need to learn them.

Students answer the questions:

1. What type of trigonometric equations are we familiar with?
2. How are these equations solved?

The teacher notes the most successful work of individual students in the lesson and gives grades.