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6. Simplify the expression:
Because cofunctions of angles complementary to each other up to 90° are equal, then we replace sin50° in the numerator of the fraction with cos40° and apply the formula for the sine of a double argument to the numerator. We get 5sin80° in the numerator. Let's replace sin80° with cos10°, which will allow us to reduce the fraction.
Formulas applied: 1) sinα=cos(90°-α); 2) sin2α=2sinαcosα.
7. In an arithmetic progression whose difference is 12 and whose eighth term is 54, find the number of negative terms.
Solution plan. Let's create a formula for the general term of this progression and find out at what values of n negative terms will be obtained. To do this, we will need to find the first term of the progression.
We have d=12, a 8 =54. Using the formula a n =a 1 +(n-1)∙d we write:
a 8 =a 1 +7d. Let's substitute the available data. 54=a 1 +7∙12;
a 1 =-30. Substitute this value into the formula a n =a 1 +(n-1)∙d
a n =-30+(n-1)∙12 or a n =-30+12n-12. Let's simplify: a n =12n-42.
We are looking for the number of negative terms, so we need to solve the inequality:
a n<0, т.е. неравенство: 12n-42<0;
12n<42 ⇒ n<3,5. Из чего заключаем, что в данной прогрессии всего три отрицательных члена, т.е. n=3.
8. Find the range of values of the following function: y=x-|x|.
Let's open the modular brackets. If x≥0, then y=x-x ⇒ y=0. The graph will be the Ox axis to the right of the origin. If x<0, то у=х+х ⇒ у=2х. Графиком будет та часть прямой у=2х, которая лежит ниже оси Ох. Таким образом, график данной функции y=x-|x| есть объединение полупрямых. Областью значений служат все неположительные числа, т.е. E(y)=(-∞; 0].
9. Find the lateral surface area of a right circular cone if its generatrix is 18 cm and the area of its base is 36 cm 2 .
Given is a cone with an axial section MAV. Generator VM=18, S main. =36π. We calculate the area of the lateral surface of the cone using the formula: S side. =πRl, where l is the generator and according to the condition is equal to 18 cm, R is the radius of the base, we will find it using the formula: S cr. = πR 2 . We have S cr. = S basic = 36π. Hence πR 2 =36π ⇒ R=6.
Then S side. =π∙6∙18 ⇒ S side. =108π cm 2.
12. Solving a logarithmic equation. A fraction is equal to 1 if its numerator is equal to its denominator, i.e.
log(x 2 +5x+4)=2logx for logx≠0. We apply to the right side of the equality the property of the power of a number under the logarithm sign: lg(x 2 +5x+4)=lgx 2. These decimal logarithms are equal, therefore the numbers under the logarithm signs are equal, therefore:
x 2 +5x+4=x 2, hence 5x=-4; we get x=-0.8. However, this value cannot be taken, since only positive numbers can be under the sign of the logarithm, therefore this equation has no solutions. Note. You shouldn’t find ODZ at the beginning of the decision (waste your time!), it’s better to check (as we are doing now) at the end.
13. Find the value of the expression (x o – y o), where (x o; y o) is the solution to the system of equations:
14. Solve the equation:
If you divide by 2 and the numerator and denominator of the fraction, you will learn the formula for the tangent of a double angle. The result is a simple equation: tg4x=1.
15. Find the derivative of the function: f(x)=(6x 2 -4x) 5.
We are given a complex function. We define it in one word - this is degree. Therefore, according to the rule of differentiation of a complex function, we find the derivative of the degree and multiply it by the derivative of the base of this degree according to the formula:
(u n)’ = n ∙ u n -1 ∙ u'.
f ‘(x)= 5(6x 2 -4x) 4 ∙ (6x 2 -4x)’ = 5(6x 2 -4x) 4 ∙ (12x-4)= 5(6x 2 -4x) 4 ∙ 4(3x-1)=20(3x-1)(6x 2 -4x) 4 .
16. It is required to find f ‘(1) if the function
17. In an equilateral triangle, the sum of all bisectors is 33√3 cm. Find the area of the triangle.
The bisector of an equilateral triangle is both the median and the altitude. Thus, the length of the altitude BD of this triangle is equal to
Let's find the side AB from the rectangular Δ ABD. Since sin60° = BD : AB, then AB = BD : sin60°.
18. A circle is inscribed in an equilateral triangle whose height is 12 cm. Find the area of the circle.
The circle (O; OD) is inscribed in the equilateral Δ ABC. The altitude BD is also a bisector and a median, and the center of the circle, point O, lies on BD.
O – the point of intersection of heights, bisectors and medians divides the median BD in a ratio of 2:1, counting from the vertex. Therefore, OD=(1/3)BD=12:3=4. Radius of the circle R=OD=4 cm. Area of the circle S=πR 2 =π∙4 2 ⇒ S=16π cm 2.
19. The lateral edges of a regular quadrangular pyramid are 9 cm, and the side of the base is 8 cm. Find the height of the pyramid.
The base of a regular quadrangular pyramid is the square ABCD, the base of the height MO is the center of the square.
20. Simplify:
In the numerator, the square of the difference is folded.
We factorize the denominator using the method of grouping terms.
21. Calculate:
In order to be able to extract an arithmetic square root, the radical expression must be a perfect square. Let us represent the expression under the root sign in the form of the squared difference of two expressions using the formula:
a 2 -2ab+b 2 =(a-b) 2, assuming that a 2 +b 2 =10.
22. Solve the inequality:
Let us represent the left side of the inequality as a product. The sum of the sines of two angles is equal to twice the product of the sine of the half-sum of these angles and the cosine of the half-difference of these angles:
We get:
Let's solve this inequality graphically. We select those points of the y=cost graph that lie above the straight line and determine the abscissas of these points (shown by shading).
23. Find all antiderivatives for the function: h(x)=cos 2 x.
Let's transform this function by lowering its degree using the formula:
1+cos2α=2cos 2 α. We get the function:
24. Find the coordinates of the vector
25. Insert arithmetic signs instead of asterisks so that you get the correct equality: (3*3)*(4*4) = 31 – 6.
We reason: the number should be 25 (31 – 6 = 25). How to get this number from two “threes” and two “fours” using action signs?
Of course it is: 3 ∙ 3 + 4 ∙ 4 = 9 + 16 = 25. Answer E).
Lesson 1
Subject: 11th grade (preparation for the Unified State Exam)
Simplifying 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:
- Organizational moment
- Testing on laptops. The discussion of the results.
- Simplifying trigonometric expressions
- Solving simple trigonometric equations
- Independent work.
- 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
g) 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.
At this stage, it is advisable to divide the class into groups of strong students (work independently with subsequent testing) and weak students who work with the teacher.
Assignment for strong students (prepared in advance on a printed basis). The main emphasis is on the formulas of reduction and double angle, according to the Unified State Exam 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 and analysis are underway. With the use of technical means this 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 the roots of equations of special cases and general form and to selecting the 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 sums up the fact that during the lesson they repeated and reinforced trigonometric formulas and solving the simplest trigonometric equations.
Homework is assigned (prepared on a printed basis in advance) with a random check at 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.
- To promote the development of students’ mathematical thinking, the ability to observe, compare, generalize, and classify.
- Encourage students to overcome difficulties in the process of mental activity, to self-control, and introspection of their activities.
Equipment for the lesson: KRMu, laptops for each student.
Lesson structure:
- Organizational moment
- Discussion of d/z and self. work from last lesson
- Review of methods for solving trigonometric equations.
- Solving trigonometric equations
- Selection of roots in trigonometric equations.
- Independent work.
- 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 of homework (5 min.)
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 of 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 are applied methods:
- using the formulas for converting a sum into a product and a product into a sum,
- according to the formulas for reducing the degree,
- universal trigonometric substitution
- introduction of an 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 the use of 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 function cosx 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.
There are trigonometric equations in which you need 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 of increased complexity and in the Unified State Exam version they are found in the second part, namely C1.
Solve the equation:
A fraction is equal to zero if then 
using the unit circle we will 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 “Simplifying Trigonometric Expressions” is designed 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. By using visual aids, it is easier for the teacher to achieve the lesson objectives. Vivid presentation of the material helps to remember 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 take the common factor cos 2 t out of brackets. 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. Then the fractions in brackets are reduced to a common denominator by 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, the basic trigonometric identity sin 2 t+cos 2 t=1 is recalled. 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 a traditional school mathematics lesson. The material will also be useful to teachers providing 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, according to the basic trigonometric identity, is equal to 1. As a result, we get 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 the left side of the 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 parentheses; from the previously obtained relationship it is known that the product of the squares of tangent te by cotangent te is equal to one. Let us recall that cotangent te is equal to the ratio of cosine te by sine te, which means that the square of cotangent is the ratio of the square of cosine te by the square of 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: The square of the sum of two quantities is equal to the square of the first plus twice the product of the first by 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),
