Graph of tangent and arctangent. Inverse trigonometric functions

Inverse trigonometric functions are widely used in mathematical analysis. However, for most high school students, tasks associated with this type of function cause significant difficulties. This is mainly due to the fact that many textbooks and teaching aids pay too little attention to tasks of this type. And if students at least somehow cope with problems of calculating the values ​​of inverse trigonometric functions, then equations and inequalities containing such functions, for the most part, baffle the children. In fact, this is not surprising, because practically no textbook explains how to solve even the simplest equations and inequalities containing inverse trigonometric functions.

Let's look at several equations and inequalities involving inverse trigonometric functions and solve them with detailed explanations.

Example 1.

Solve the equation: 3arccos (2x + 3) = 5π/2.

Solution.

Expressing the inverse trigonometric function from the equation, we get:

arccos (2x + 3) = 5π/6. Now let's use the definition of arc cosine.

The arc cosine of a certain number a, belonging to the segment from -1 to 1, is an angle y from the segment from 0 to π such that its cosine is equal to the number x. Therefore we can write it like this:

2x + 3 = cos 5π/6.

Let us write the right side of the resulting equation using the reduction formula:

2x + 3 = cos (π – π/6).

2x + 3 = -cos π/6;

2x + 3 = -√3/2;

2x = -3 – √3/2.

Let's reduce the right side to a common denominator.

2x = -(6 + √3) / 2;

x = -(6 + √3) / 4.

Answer: -(6 + √3) / 4 .

Example 2.

Solve the equation: cos (arccos (4x – 9)) = x 2 – 5x + 5.

Solution.

Since cos (arcсos x) = x with x belonging to [-1; 1], then this equation is equivalent to the system:

(4x – 9 = x 2 – 5x + 5,
(-1 ≤ 4x – 9 ≤ 1.

Let's solve the equation included in the system.

4x – 9 = x 2 – 5x + 5.

It is square, so we get that

x 2 – 9x + 14 = 0;

D = 81 – 4 14 = 25;

x 1 = (9 + 5) / 2 = 7;

x 2 = (9 – 5) / 2 = 2.

Let us solve the double inequality included in the system.

1 ≤ 4x – 9 ≤ 1. Add 9 to all parts, we have:

8 ≤ 4x ≤ 10. Divide each number by 4, we get:

2 ≤ x ≤ 2.5.

Now let's combine the answers we received. It is easy to see that the root x = 7 does not satisfy the answer to the inequality. Therefore, the only solution to the equation is x = 2.

Answer: 2.

Example 3.

Solve the equation: tg (arctg (0.5 – x)) = x 2 – 4x + 2.5.

Solution.

Since tg (arctg x) = x for all real numbers, this equation is equivalent to the equation:

0.5 – x = x 2 – 4x + 2.5.

Let us solve the resulting quadratic equation using the discriminant, having first brought it into standard form.

x 2 – 3x + 2 = 0;

D = 9 – 4 2 = 1;

x 1 = (3 + 1) / 2 = 2;

x 2 = (3 – 1) / 2 = 1.

Answer: 1; 2.

Example 4.

Solve the equation: arcctg (2x – 1) = arcctg (x 2 /2 + x/2).

Solution.

Since arcctg f(x) = arcctg g(x) if and only if f(x) = g(x), then

2x – 1 = x 2 /2 + x/2. Let's solve the resulting quadratic equation:

4x – 2 = x 2 + x;

x 2 – 3x + 2 = 0.

By Vieta's theorem we obtain that

x = 1 or x = 2.

Answer: 1; 2.

Example 5.

Solve the equation: arcsin (2x – 15) = arcsin (x 2 – 6x – 8).

Solution.

Since an equation of the form arcsin f(x) = arcsin g(x) is equivalent to the system

(f(x) = g(x),
(f(x) € [-1; 1],

then the original equation is equivalent to the system:

(2x – 15 = x 2 – 6x + 8,
(-1 ≤ 2x – 15 ≤ 1.

Let's solve the resulting system:

(x 2 – 8x + 7 = 0,
(14 ≤ 2x ≤ 16.

From the first equation, using Vieta’s theorem, we have that x = 1 or x = 7. Solving the second inequality of the system, we find that 7 ≤ x ≤ 8. Therefore, only the root x = 7 is suitable for the final answer.

Answer: 7.

Example 6.

Solve the equation: (arccos x) 2 – 6 arccos x + 8 = 0.

Solution.

Let arccos x = t, then t belongs to the segment and the equation takes the form:

t 2 – 6t + 8 = 0. Solve the resulting quadratic equation using Vieta’s theorem, we find that t = 2 or t = 4.

Since t = 4 does not belong to the segment, we obtain that t = 2, i.e. arccos x = 2, which means x = cos 2.

Answer: cos 2.

Example 7.

Solve the equation: (arcsin x) 2 + (arccos x) 2 = 5π 2 /36.

Solution.

Let's use the equality arcsin x + arccos x = π/2 and write the equation in the form

(arcsin x) 2 + (π/2 – arcsin x) 2 = 5π 2 /36.

Let arcsin x = t, then t belongs to the segment [-π/2; π/2] and the equation takes the form:

t 2 + (π/2 – t) 2 = 5π 2 /36.

Let's solve the resulting equation:

t 2 + π 2 /4 – πt + t 2 = 5π 2 /36;

2t 2 – πt + 9π 2 /36 – 5π 2 /36 = 0;

2t 2 – πt + 4π 2 /36 = 0;

2t 2 – πt + π 2 /9 = 0. Multiplying each term by 9 to get rid of fractions in the equation, we get:

18t 2 – 9πt + π 2 = 0.

Let's find the discriminant and solve the resulting equation:

D = (-9π) 2 – 4 · 18 · π 2 = 9π 2 .

t = (9π – 3π) / 2 18 or t = (9π + 3π) / 2 18;

t = 6π/36 or t = 12π/36.

After reduction we have:

t = π/6 or t = π/3. Then

arcsin x = π/6 or arcsin x = π/3.

Thus, x = sin π/6 or x = sin π/3. That is, x = 1/2 or x =√3/2.

Answer: 1/2; √3/2.

Example 8.

Find the value of the expression 5nx 0, where n is the number of roots, and x 0 is the negative root of the equation 2 arcsin x = - π – (x + 1) 2.

Solution.

Since -π/2 ≤ arcsin x ≤ π/2, then -π ≤ 2 arcsin x ≤ π. Moreover, (x + 1) 2 ≥ 0 for all real x,
then -(x + 1) 2 ≤ 0 and -π – (x + 1) 2 ≤ -π.

Thus, the equation can have a solution if both its sides are simultaneously equal to –π, i.e. the equation is equivalent to the system:

(2 arcsin x = -π,
(-π – (x + 1) 2 = -π.

Let us solve the resulting system of equations:

(arcsin x = -π/2,
((x + 1) 2 = 0.

From the second equation we have that x = -1, respectively n = 1, then 5nx 0 = 5 · 1 · (-1) = -5.

Answer: -5.

As practice shows, the ability to solve equations with inverse trigonometric functions is a necessary condition for successfully passing exams. That is why training in solving such problems is simply necessary and mandatory when preparing for the Unified State Exam.

Still have questions? Don't know how to solve equations?
To get help from a tutor -.
The first lesson is free!

blog.site, when copying material in full or in part, a link to the original source is required.

Lessons 32-33. Inverse trigonometric functions

09.07.2015 5917 0

Target: consider inverse trigonometric functions and their use for writing solutions to trigonometric equations.

I. Communicating the topic and purpose of the lessons

II. Learning new material

1. Inverse trigonometric functions

Let's begin our discussion of this topic with the following example.

Example 1

Let's solve the equation: a) sin x = 1/2; b) sin x = a.

a) On the ordinate axis we plot the value 1/2 and construct the angles x 1 and x2, for which sin x = 1/2. In this case x1 + x2 = π, whence x2 = π – x 1 . Using the table of values ​​of trigonometric functions, we find the value x1 = π/6, thenLet's take into account the periodicity of the sine function and write down the solutions to this equation:where k ∈ Z.

b) Obviously, the algorithm for solving the equation sin x = a is the same as in the previous paragraph. Of course, now the value a is plotted along the ordinate axis. There is a need to somehow designate the angle x1. We agreed to denote this angle with the symbol arcsin A. Then the solutions to this equation can be written in the formThese two formulas can be combined into one: wherein

The remaining inverse trigonometric functions are introduced in a similar way.

Very often it is necessary to determine the magnitude of an angle from the known value of its trigonometric function. Such a problem is multivalued - there are countless angles whose trigonometric functions are equal to the same value. Therefore, based on the monotonicity of trigonometric functions, the following inverse trigonometric functions are introduced to uniquely determine angles.

Arcsine of the number a (arcsin , whose sine is equal to a, i.e.

Arc cosine of a number a(arccos a) is an angle a from the interval whose cosine is equal to a, i.e.

Arctangent of a number a(arctg a) - such an angle a from the intervalwhose tangent is equal to a, i.e.tg a = a.

Arccotangent of a number a(arcctg a) is an angle a from the interval (0; π), the cotangent of which is equal to a, i.e. ctg a = a.

Example 2

Let's find:

Taking into account the definitions of inverse trigonometric functions, we obtain:


Example 3

Let's calculate

Let angle a = arcsin 3/5, then by definition sin a = 3/5 and . Therefore, we need to find cos A. Using the basic trigonometric identity, we get:It is taken into account that cos a ≥ 0. So,

Function properties

Function

y = arcsin x

y = arccos x

y = arctan x

y = arcctg x

Domain

x ∈ [-1; 1]

x ∈ [-1; 1]

x ∈ (-∞; +∞)

x ∈ (-∞ +∞)

Range of values

y ∈ [ -π/2 ; π /2 ]

y ∈

y ∈ (-π/2 ; π /2 )

y ∈ (0;π)

Parity

Odd

Neither even nor odd

Odd

Neither even nor odd

Function zeros (y = 0)

At x = 0

At x = 1

At x = 0

y ≠ 0

Intervals of sign constancy

y > 0 for x ∈ (0; 1],

at< 0 при х ∈ [-1; 0)

y > 0 for x ∈ [-1; 1)

y > 0 for x ∈ (0; +∞),

at< 0 при х ∈ (-∞; 0)

y > 0 for x ∈ (-∞; +∞)

Monotone

Increasing

Descending

Increasing

Descending

Relation to the trigonometric function

sin y = x

cos y = x

tg y = x

ctg y = x

Schedule



Let us give a number of more typical examples related to the definitions and basic properties of inverse trigonometric functions.

Example 4

Let's find the domain of definition of the function

In order for the function y to be defined, it is necessary to satisfy the inequalitywhich is equivalent to the system of inequalitiesThe solution to the first inequality is the interval x(-∞; +∞), second - This interval and is a solution to the system of inequalities, and therefore the domain of definition of the function

Example 5

Let's find the area of ​​change of the function

Let's consider the behavior of the function z = 2x - x2 (see picture).

It is clear that z ∈ (-∞; 1]. Considering that the argument z the arc cotangent function varies within the specified limits, from the table data we obtain thatSo the area of ​​change

Example 6

Let us prove that the function y = arctg x odd. LetThen tg a = -x or x = - tg a = tg (- a), and Therefore, - a = arctg x or a = - arctg X. Thus, we see thati.e. y(x) is an odd function.

Example 7

Let us express through all inverse trigonometric functions

Let It's obvious that Then since

Let's introduce the angle Because That

Likewise therefore And

So,

Example 8

Let's build a graph of the function y = cos(arcsin x).

Let us denote a = arcsin x, then Let's take into account that x = sin a and y = cos a, i.e. x 2 + y2 = 1, and restrictions on x (x[-1; 1]) and y (y ≥ 0). Then the graph of the function y = cos(arcsin x) is a semicircle.

Example 9

Let's build a graph of the function y = arccos (cos x ).

Since the cos function x changes on the interval [-1; 1], then the function y is defined on the entire numerical axis and varies on the segment . Let's keep in mind that y = arccos(cosx) = x on the segment; the function y is even and periodic with period 2π. Considering that the function has these properties cos x Now it's easy to create a graph.


Let us note some useful equalities:

Example 10

Let's find the smallest and largest values ​​of the function Let's denote Then Let's get the function This function has a minimum at the point z = π/4, and it is equal to The greatest value of the function is achieved at the point z = -π/2, and it is equal Thus, and

Example 11

Let's solve the equation

Let's take into account that Then the equation looks like:or where By definition of arctangent we get:

2. Solving simple trigonometric equations

Similar to example 1, you can obtain solutions to the simplest trigonometric equations.

The equation

Solution

tgx = a

ctg x = a

Example 12

Let's solve the equation

Since the sine function is odd, we write the equation in the formSolutions to this equation:where do we find it from?

Example 13

Let's solve the equation

Using the given formula, we write down the solutions to the equation:and we'll find

Note that in special cases (a = 0; ±1) when solving the equations sin x = a and cos x = and it’s easier and more convenient to use not general formulas, but to write down solutions based on the unit circle:

for the equation sin x = 1 solution

for the equation sin x = 0 solutions x = π k;

for the equation sin x = -1 solution

for the cos equation x = 1 solution x = 2π k ;

for the equation cos x = 0 solutions

for the equation cos x = -1 solution

Example 14

Let's solve the equation

Since in this example there is a special case of the equation, we will write the solution using the appropriate formula:where can we find it from?

III. Control questions (frontal survey)

1. Define and list the main properties of inverse trigonometric functions.

2. Give graphs of inverse trigonometric functions.

3. Solving simple trigonometric equations.

IV. Lesson assignment

§ 15, No. 3 (a, b); 4 (c, d); 7(a); 8(a); 12 (b); 13(a); 15 (c); 16(a); 18 (a, b); 19 (c); 21;

§ 16, No. 4 (a, b); 7(a); 8 (b); 16 (a, b); 18(a); 19 (c, d);

§ 17, No. 3 (a, b); 4 (c, d); 5 (a, b); 7 (c, d); 9 (b); 10 (a, c).

V. Homework

§ 15, No. 3 (c, d); 4 (a, b); 7 (c); 8 (b); 12(a); 13(b); 15 (g); 16 (b); 18 (c, d); 19 (g); 22;

§ 16, No. 4 (c, d); 7 (b); 8(a); 16 (c, d); 18 (b); 19 (a, b);

§ 17, No. 3 (c, d); 4 (a, b); 5 (c, d); 7 (a, b); 9 (g); 10 (b, d).

VI. Creative tasks

1. Find the domain of the function:


Answers:

2. Find the range of the function:

Answers:

3. Plot a graph of the function:


VII. Summing up the lessons

Definitions of inverse trigonometric functions and their graphs are given. As well as formulas connecting inverse trigonometric functions, formulas for sums and differences.

Definition of inverse trigonometric functions

Since trigonometric functions are periodic, their inverse functions are not unique. So, the equation y = sin x, for a given , has infinitely many roots. Indeed, due to the periodicity of the sine, if x is such a root, then so is x + 2πn(where n is an integer) will also be the root of the equation. Thus, inverse trigonometric functions are multivalued. To make it easier to work with them, the concept of their main meanings is introduced. Consider, for example, sine: y = sin x. If we limit the argument x to the interval , then on it the function y = sin x increases monotonically. Therefore, it has a unique inverse function, which is called the arcsine: x = arcsin y.

Unless otherwise stated, by inverse trigonometric functions we mean their main values, which are determined by the following definitions.

Arcsine ( y = arcsin x) is the inverse function of sine ( x = siny

Arc cosine ( y = arccos x) is the inverse function of cosine ( x = cos y), having a domain of definition and a set of values.

Arctangent ( y = arctan x) is the inverse function of tangent ( x = tg y), having a domain of definition and a set of values.

arccotangent ( y = arcctg x) is the inverse function of cotangent ( x = ctg y), having a domain of definition and a set of values.

Graphs of inverse trigonometric functions

Graphs of inverse trigonometric functions are obtained from graphs of trigonometric functions by mirror reflection with respect to the straight line y = x. See sections Sine, cosine, Tangent, cotangent.

y = arcsin x


y = arccos x


y = arctan x


y = arcctg x

Basic formulas

Here you should pay special attention to the intervals for which the formulas are valid.

arcsin(sin x) = x at
sin(arcsin x) = x
arccos(cos x) = x at
cos(arccos x) = x

arctan(tg x) = x at
tg(arctg x) = x
arcctg(ctg x) = x at
ctg(arcctg x) = x

Formulas relating inverse trigonometric functions

Sum and difference formulas


at or

at and

at and


at or

at and

at and


at

at


at

at

The functions sin, cos, tg and ctg are always accompanied by arcsine, arccosine, arctangent and arccotangent. One is a consequence of the other, and pairs of functions are equally important for working with trigonometric expressions.

Consider a drawing of a unit circle, which graphically displays the values ​​of trigonometric functions.

If we calculate arcs OA, arcos OC, arctg DE and arcctg MK, then they will all be equal to the value of angle α. The formulas below reflect the relationship between the basic trigonometric functions and their corresponding arcs.

To understand more about the properties of the arcsine, it is necessary to consider its function. Schedule has the form of an asymmetric curve passing through the coordinate center.

Properties of arcsine:

If we compare the graphs sin And arcsin, two trigonometric functions can have common principles.

arc cosine

Arccos of a number is the value of the angle α, the cosine of which is equal to a.

Curve y = arcos x mirrors the arcsin x graph, with the only difference being that it passes through the point π/2 on the OY axis.

Let's look at the arc cosine function in more detail:

  1. The function is defined on the interval [-1; 1].
  2. ODZ for arccos - .
  3. The graph is entirely located in the first and second quarters, and the function itself is neither even nor odd.
  4. Y = 0 at x = 1.
  5. The curve decreases along its entire length. Some properties of the arc cosine coincide with the cosine function.

Some properties of the arc cosine coincide with the cosine function.

Perhaps schoolchildren will find such a “detailed” study of “arches” unnecessary. However, otherwise, some elementary standard exam tasks can lead students into a dead end.

Exercise 1. Indicate the functions shown in the figure.

Answer: rice. 1 – 4, Fig. 2 – 1.

In this example, the emphasis is on the little things. Typically, students are very inattentive to the construction of graphs and the appearance of functions. Indeed, why remember the type of curve if it can always be plotted using calculated points. Do not forget that under test conditions, the time spent on drawing for a simple task will be required to solve more complex tasks.

Arctangent

Arctg the numbers a are the value of the angle α such that its tangent is equal to a.

If we consider the arctangent graph, we can highlight the following properties:

  1. The graph is infinite and defined on the interval (- ∞; + ∞).
  2. Arctangent is an odd function, therefore, arctan (- x) = - arctan x.
  3. Y = 0 at x = 0.
  4. The curve increases throughout the entire definition range.

Let us present a brief comparative analysis of tg x and arctg x in the form of a table.

Arccotangent

Arcctg of a number - takes a value α from the interval (0; π) such that its cotangent is equal to a.

Properties of the arc cotangent function:

  1. The function definition interval is infinity.
  2. The range of acceptable values ​​is the interval (0; π).
  3. F(x) is neither even nor odd.
  4. Throughout its entire length, the graph of the function decreases.

It is very simple to compare ctg x and arctg x; you just need to make two drawings and describe the behavior of the curves.

Task 2. Match the graph and the notation form of the function.

If we think logically, it is clear from the graphs that both functions are increasing. Therefore, both figures display a certain arctan function. From the properties of the arctangent it is known that y=0 at x = 0,

Answer: rice. 1 – 1, fig. 2 – 4.

Trigonometric identities arcsin, arcos, arctg and arcctg

Previously, we have already identified the relationship between arches and the basic functions of trigonometry. This dependence can be expressed by a number of formulas that allow one to express, for example, the sine of an argument through its arcsine, arccosine, or vice versa. Knowledge of such identities can be useful when solving specific examples.

There are also relationships for arctg and arcctg:

Another useful pair of formulas sets the value for the sum of arcsin and arcos, as well as arcctg and arcctg of the same angle.

Examples of problem solving

Trigonometry tasks can be divided into four groups: calculate the numerical value of a specific expression, construct a graph of a given function, find its domain of definition or ODZ and perform analytical transformations to solve the example.

When solving the first type of problem, you must adhere to the following action plan:

When working with function graphs, the main thing is knowledge of their properties and the appearance of the curve. Solving trigonometric equations and inequalities requires identity tables. The more formulas a student remembers, the easier it is to find the answer to the task.

Let’s say in the Unified State Examination you need to find the answer for an equation like:

If you correctly transform the expression and bring it to the desired form, then solving it is very simple and quick. First, let's move arcsin x to the right side of the equality.

If you remember the formula arcsin (sin α) = α, then we can reduce the search for answers to solving a system of two equations:

The restriction on the model x arose, again from the properties of arcsin: ODZ for x [-1; 1]. When a ≠0, part of the system is a quadratic equation with roots x1 = 1 and x2 = - 1/a. When a = 0, x will be equal to 1.