Inverse trigonometric functions. Let us express through all inverse trigonometric functions

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. sin x If we limit the argument x to the interval , then on it the function y = 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 Arcsine ( Arc cosine () arccos x is the inverse function of sine ( is the inverse function of cosine ( cos y

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

tg y Arcsine ( arccotangent () arcctg x is the inverse function of sine ( is the inverse function of cotangent ( cos y

ctg y

Graphs of inverse trigonometric functions

Arcsine ( y =

Arcsine ( Arc cosine (

Arcsine ( Arctangent (

Arcsine ( arccotangent (

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.

Basic formulas Here you should pay special attention to the intervals for which the formulas are valid.
arcsin(sin x) = x
at Here you should pay special attention to the intervals for which the formulas are valid.
sin(arcsin x) = x

arccos(cos x) = x Here you should pay special attention to the intervals for which the formulas are valid.
cos(arccos x) = x
arctan(tg x) = x Here you should pay special attention to the intervals for which the formulas are valid.
tg(arctg x) = x

arcctg(ctg x) = x

ctg(arcctg x) = x

Formulas relating inverse trigonometric functions

Sum and difference formulas

at or

at and

at and

Inverse trigonometric functions are mathematical functions that are the inverse of trigonometric functions.
Function y=arcsin(x)
The function у= sin⁡(x) on the interval [-π/2;π/2], is strictly increasing and continuous; therefore, it has an inverse function, strictly increasing and continuous.
The inverse function for the function y= sin⁡(x), where x ∈[-π/2;π/2], is called the arcsine and is denoted y=arcsin(x), where x∈[-1;1].
So, according to the definition inverse function, the domain of definition of the arcsine is the segment [-1;1], and the set of values is the segment [-π/2;π/2].
Note that the graph of the function y=arcsin(x), where x ∈[-1;1], is symmetrical to the graph of the function y= sin(⁡x), where x∈[-π/2;π/2], with respect to the bisector of the coordinate angles first and third quarters.

Function range y=arcsin(x).

Example No. 1.

Find arcsin(1/2)?

Since the range of values of the function arcsin(x) belongs to the interval [-π/2;π/2], then only the value π/6 is suitable. Therefore, arcsin(1/2) =π/6.
Answer:π/6

Example No. 2.
Find arcsin(-(√3)/2)?

Since the range of values arcsin(x) x ∈[-π/2;π/2], then only the value -π/3 is suitable. Therefore, arcsin(-(√3)/2) =- π/3.

Function y=arccos(x)

The arc cosine of a number α is a number α from the interval whose cosine is equal to α.

Graph of a function

The function y= cos(⁡x) on the segment is strictly decreasing and continuous; therefore, it has an inverse function, strictly decreasing and continuous.
The inverse function for the function y= cos⁡x, where x ∈, is called arc cosine and is denoted by y=arccos(x),where x ∈[-1;1].
So, according to the definition of the inverse function, the domain of definition of the arc cosine is the segment [-1;1], and the set of values is the segment.
Note that the graph of the function y=arccos(x), where x ∈[-1;1] is symmetrical to the graph of the function y= cos(⁡x), where x ∈, with respect to the bisector of the coordinate angles of the first and third quarters.

Function range y=arccos(x).

Example No. 3.

Find arccos(1/2)?

Since the range of values is arccos(x) x∈, then only the value π/3 is suitable. Therefore, arccos(1/2) =π/3.
Example No. 4.
Find arccos(-(√2)/2)?

Since the range of values of the function arccos(x) belongs to the interval, then only the value 3π/4 is suitable. Therefore, arccos(-(√2)/2) = 3π/4.

Answer: 3π/4

Function y=arctg(x)

The arctangent of a number α is a number α from the interval [-π/2;π/2] whose tangent is equal to α.

Graph of a function

The tangent function is continuous and strictly increasing on the interval (-π/2;π/2); therefore, it has an inverse function that is continuous and strictly increasing.
The inverse function for the function y= tan⁡(x), where x∈(-π/2;π/2); is called the arctangent and is denoted by y=arctg(x), where x∈R.
So, according to the definition of the inverse function, the domain of definition of the arctangent is the interval (-∞;+∞), and the set of values is the interval
(-π/2;π/2).
Note that the graph of the function y=arctg(x), where x∈R, is symmetrical to the graph of the function y= tan⁡x, where x ∈ (-π/2;π/2), relative to the bisector of the coordinate angles of the first and third quarters.

The range of the function y=arctg(x).

Example No. 5?

Find arctan((√3)/3).

Since the range of values arctg(x) x ∈(-π/2;π/2), then only the value π/6 is suitable. Therefore, arctg((√3)/3) =π/6.
Example No. 6.
Find arctg(-1)?

Since the range of values arctg(x) x ∈(-π/2;π/2), then only the value -π/4 is suitable. Therefore, arctg(-1) = - π/4.

Function y=arcctg(x)

The arc cotangent of a number α is a number α from the interval (0;π) whose cotangent is equal to α.

Graph of a function

On the interval (0;π), the cotangent function strictly decreases; in addition, it is continuous at every point of this interval; therefore, on the interval (0;π), this function has an inverse function, which is strictly decreasing and continuous.
The inverse function for the function y=ctg(x), where x ∈(0;π), is called arccotangent and is denoted y=arcctg(x), where x∈R.
So, according to the definition of the inverse function, the domain of definition of the arc cotangent will be R, and by a set values – interval (0;π).The graph of the function y=arcctg(x), where x∈R is symmetrical to the graph of the function y=ctg(x) x∈(0;π),relative to the bisector of the coordinate angles of the first and third quarters.

Function range y=arcctg(x).

Example No. 7.
Find arcctg((√3)/3)?

Since the range of values arcctg(x) x ∈(0;π), then only the value π/3 is suitable. Therefore arccos((√3)/3) =π/3.

Example No. 8.
Find arcctg(-(√3)/3)?

Since the range of values is arcctg(x) x∈(0;π), then only the value 2π/3 is suitable. Therefore, arccos(-(√3)/3) = 2π/3.

Editors: Ageeva Lyubov Aleksandrovna, Gavrilina Anna Viktorovna

Inverse trigonometric functions have wide application 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 in many textbooks and textbooks Problems of this type are given too little attention. 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.

Let's express 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 and 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's solve the result quadratic equation using a discriminant, having previously brought it into a 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 successful completion exams. That is why training in solving such problems is simply necessary and mandatory when preparing for the Unified State Exam.

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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 the angle by known value 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
Function Properties	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's give a few 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 gap 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 changes within the specified limits, from the table data we obtain thatThus, 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 plot 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 plot 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 highest value functions 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