Inequalities containing trigonometric functions, when solved, are reduced to the simplest inequalities of the form cos(t)>a, sint(t)=a and similar ones. And already the simplest inequalities are solved. Let's look at various examples ways to solve simple trigonometric inequalities.
Example 1. Solve the inequality sin(t) > = -1/2.
Draw a unit circle. Since sin(t) by definition is the y coordinate, we mark the point y = -1/2 on the Oy axis. We draw a straight line through it parallel to the Ox axis. At the intersection of the line with the graph of the unit circle, mark points Pt1 and Pt2. We connect the origin of coordinates with points Pt1 and Pt2 by two segments.
The solution to this inequality will be all points of the unit circle located above these points. In other words, the solution will be the arc l. Now it is necessary to indicate the conditions under which an arbitrary point will belong to the arc l.
Pt1 lies in the right semicircle, its ordinate is -1/2, then t1=arcsin(-1/2) = - pi/6. To describe point Pt1, you can write the following formula:
t2 = pi - arcsin(-1/2) = 7*pi/6. As a result, we obtain the following inequality for t:
We preserve the inequalities. And since the sine function is periodic, it means that the solutions will be repeated every 2*pi. We add this condition to the resulting inequality for t and write down the answer.
Answer: -pi/6+2*pi*n< = t < = 7*pi/6 + 2*pi*n, при любом целом n.
Example 2. Solve cos(t) inequality<1/2.
Let's draw a unit circle. Since, according to the definition, cos(t) is the x coordinate, we mark the point x = 1/2 on the graph on the Ox axis.
We draw a straight line through this point parallel to the Oy axis. At the intersection of the line with the graph of the unit circle, mark points Pt1 and Pt2. We connect the origin of coordinates with points Pt1 and Pt2 by two segments.
The solutions will be all points of the unit circle that belong to the arc l. Let's find the points t1 and t2.
t1 = arccos(1/2) = pi/3.
t2 = 2*pi - arccos(1/2) = 2*pi-pi/3 = 5*pi/6.
We got the inequality for t: pi/3 Since cosine is a periodic function, the solutions will be repeated every 2*pi. We add this condition to the resulting inequality for t and write down the answer. Answer: pi/3+2*pi*n Example 3. Solve inequality tg(t)< = 1. The tangent period is equal to pi. Let's find solutions that belong to the interval (-pi/2;pi/2) right semicircle. Next, using the periodicity of the tangent, we write down all the solutions to this inequality. Let's draw a unit circle and mark a line of tangents on it. If t is a solution to the inequality, then the ordinate of the point T = tg(t) must be less than or equal to 1. The set of such points will make up the ray AT. The set of points Pt that will correspond to the points of this ray is the arc l. Moreover, the point P(-pi/2) does not belong to this arc. inequality solution in mode online solution almost any given inequality online. Mathematical inequalities online to solve mathematics. 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Trigonometric inequalities are expressions of the form: F(x) › a, F(x) ‹ a, F(x) ≤ a, F(x) ≥ a, in which F(x) is represented by one or more trigonometric functions. An example of the simplest trigonometric inequality is: sin x ‹ 1/2. It is customary to solve such problems graphically; two methods have been developed for this. To find an interval that satisfies the conditions inequality sin x ‹ 1/2, you must perform the following steps: When strict signs are present in an expression, the intersection points are not solutions. Since the smallest positive period of a sinusoid is 2π, we write the answer as follows: If the signs of the expression are not strict, then the solution interval must be enclosed in square brackets - . The answer to the problem can also be written as the following inequality: Similar problems can be easily solved using a trigonometric circle. The algorithm for finding answers is very simple: Let us analyze the stages of the solution using the example of the inequality sin x › 1/2. Points α and β are marked on the circle - values The points of the arc located above α and β are the interval for solving the given inequality. If you need to solve an example for cos, then the answer arc will be located symmetrically to the OX axis, not OY. You can consider the difference between the solution intervals for sin and cos in the diagrams below in the text. Graphical solutions for tangent and cotangent inequalities will differ from both sine and cosine. This is due to the properties of functions. Arctangent and arccotangent are tangents to a trigonometric circle, and the minimum positive period for both functions is π. To quickly and correctly use the second method, you need to remember on which axis the values of sin, cos, tg and ctg are plotted. The tangent tangent runs parallel to the OY axis. If we plot the value of arctan a on the unit circle, then the second required point will be located in the diagonal quarter. Angles They are break points for the function, since the graph tends to them, but never reaches them. In the case of cotangent, the tangent runs parallel to the OX axis, and the function is interrupted at points π and 2π. If the argument of the inequality function is represented not just by a variable, but by an entire expression containing an unknown, then we are talking about a complex inequality. The process and procedure for solving it are somewhat different from the methods described above. Suppose we need to find a solution to the following inequality: The graphical solution involves constructing an ordinary sinusoid y = sin x using arbitrarily selected values of x. Let's calculate a table with coordinates for the control points of the graph: The result should be a beautiful curve. To make finding a solution easier, let’s replace the complex function argument An algorithm for solving simple trigonometric inequalities and recognizing methods for solving trigonometric inequalities. Teachers of the highest qualification category: Shirko F.M. p. Progress, MOBU-SOSH No. 6 Sankina L.S. Armavir, private secondary school "New Way" There are no universal methods for teaching science and mathematics disciplines. Each teacher finds his own ways of teaching that are acceptable only to him. Our many years of teaching experience show that students more easily learn material that requires concentration and retention of a large amount of information in memory if they are taught to use algorithms in their activities at the initial stage of learning a complex topic. In our opinion, such a topic is the topic of solving trigonometric inequalities. So, before we begin with students to identify techniques and methods for solving trigonometric inequalities, we practice and consolidate an algorithm for solving the simplest trigonometric inequalities. Algorithm for solving simple trigonometric inequalities
Mark points on the corresponding axis ( For
sin
x– OA axis, forcos
x– OX axis) We restore a perpendicular to the axis that will intersect the circle at two points. The first point on the circle is a point that belongs to the interval of the arc function range by definition. Starting from the labeled point, shade the arc of the circle corresponding to the shaded part of the axis. We pay special attention to the direction of the detour. If the traversal is done clockwise (i.e. there is a transition through 0), then the second point on the circle will be negative, if counterclockwise it will be positive. We write the answer in the form of an interval, taking into account the periodicity of the function. Let's look at the operation of the algorithm using examples.
1)
sin
≥ 1/2;
Solution: We depict a unit circle.; We mark point ½ on the OU axis. We restore the perpendicular to the axis, which intersects the circle at two points. By definition of the arcsine, we first note point π/6. Shade the part of the axis that corresponds to given inequality, above the point ½. Shade the arc of the circle corresponding to the shaded part of the axis. The traversal is done counterclockwise, we get the point 5π/6. We write the answer in the form of an interval, taking into account the periodicity of the function; Answer:x;[π/6 + 2π n, 5π/6 + 2π n], n Z. The simplest inequality is solved using the same algorithm if the answer record does not contain a table value. Students, when solving inequalities at the board in their first lessons, recite each step of the algorithm out loud. 2) 5
cos
x
– 1 ≥ 0;
R 5 cos x – 1 ≥ 0; cos
x ≥ 1/5; Draw a unit circle. We mark a point with coordinate 1/5 on the OX axis. We restore the perpendicular to the axis, which intersects the circle at two points. The first point on the circle is a point that belongs to the interval of the arc cosine range by definition (0;π). We shade the part of the axis that corresponds to this inequality. Starting from the signed point
arccos 1/5, shade the arc of the circle corresponding to the shaded part of the axis. The traversal is done clockwise (i.e. there is a transition through 0), which means that the second point on the circle will be negative - arccos 1/5. We write the answer in the form of an interval, taking into account the periodicity of the function, from the smaller value to the larger one. Answer: x [-arccos 1/5 + 2π n, arccos 1/5 + 2π n], n Z. Improving the ability to solve trigonometric inequalities is facilitated by the following questions: “How will we solve a group of inequalities?”; “How does one inequality differ from another?”; “How is one inequality similar to another?”; How would the answer change if strict inequality were given?"; How would the answer change if instead of the sign "" there was a sign " The task of analyzing a list of inequalities from the standpoint of methods for solving them allows you to practice their recognition. Students are given inequalities that need to be solved in class. Question: Highlight the inequalities that require the use of equivalent transformations when reducing a trigonometric inequality to its simplest form? Answer 1, 3, 5. Question: What are the inequalities in which you need to consider a complex argument as a simple one? Answer: 1, 2, 3, 5, 6. Question: What are the inequalities where trigonometric formulas can be applied? Answer: 2, 3, 6. Question: Name the inequalities where the method of introducing a new variable can be applied? Answer: 6. The task of analyzing a list of inequalities from the standpoint of methods for solving them allows you to practice their recognition. When developing skills, it is important to identify the stages of its implementation and formulate them in a general form, which is presented in the algorithm for solving the simplest trigonometric inequalities.Method 1 - Solving inequalities by graphing a function
Method 2 - Solving trigonometric inequalities using the unit circle
Complex trigonometric inequalities
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