Formulas for inverse trigonometric functions table. Let us express through all inverse trigonometric functions
Definitions of inverse trigonometric functions and their graphs are given. As well as formulas connecting inverses 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
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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.
Arccosine of some 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's write it down right side 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 given 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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Inverse trigonometric functions are mathematical functions that are the inverse of trigonometric functions.
Function y=arcsin(x)
The arcsine of a number α is a number α from the interval [-π/2;π/2] whose sine is equal to α.
Graph of a function
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 of the 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= cosx, 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= tanx, 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 cosine function
The range of values of the function y=cos x (see Fig. 2) is a segment. On the segment the function is continuous and monotonically decreasing.
Rice. 2
This means that the function inverse to the function y=cos x is defined on the segment. This inverse function is called arc cosine and is denoted y=arccos x.
Definition
The arccosine of a number a, if |a|1, is the angle whose cosine belongs to the segment; it is denoted by arccos a.
Thus, arccos a is an angle that satisfies the following two conditions: сos (arccos a)=a, |a|1; 0? arccos a ?р.
For example, arccos, since cos and; arccos, since cos and.
The function y = arccos x (Fig. 3) is defined on a segment; its range of values is the segment. On the segment, the function y=arccos x is continuous and monotonically decreases from p to 0 (since y=cos x is a continuous and monotonically decreasing function on the segment); at the ends of the segment it reaches its extreme values: arccos(-1)= p, arccos 1= 0. Note that arccos 0 = . The graph of the function y = arccos x (see Fig. 3) is symmetrical to the graph of the function y = cos x relative to the straight line y=x.
Rice. 3
Let us show that the equality arccos(-x) = p-arccos x holds.
In fact, by definition 0? arccos x? R. Multiplying by (-1) all parts of the last double inequality, we get - p? arccos x? 0. Adding p to all parts of the last inequality, we find that 0? p-arccos x? R.
Thus, the values of the angles arccos(-x) and p - arccos x belong to the same segment. Since the cosine decreases monotonically on a segment, there cannot be two different angles on it that have equal cosines. Let's find the cosines of the angles arccos(-x) and p-arccos x. By definition, cos (arccos x) = - x, according to the reduction formulas and by definition we have: cos (p - - arccos x) = - cos (arccos x) = - x. So, the cosines of the angles are equal, which means the angles themselves are equal.
Inverse sine function
Let's consider the function y=sin x (Fig. 6), which on the segment [-р/2;р/2] is increasing, continuous and takes values from the segment [-1; 1]. This means that on the segment [- p/2; p/2] the inverse function of the function y=sin x is defined.
Rice. 6
This inverse function is called the arcsine and is denoted y=arcsin x. Let us introduce the definition of the arcsine of a number.
The arcsine of a number is an angle (or arc) whose sine is equal to the number a and which belongs to the segment [-р/2; p/2]; it is denoted by arcsin a.
Thus, arcsin a is the angle satisfying following conditions: sin (arcsin a)=a, |a| ?1; -r/2 ? arcsin huh? r/2. For example, since sin and [- p/2; p/2]; arcsin, since sin = u [- p/2; p/2].
The function y=arcsin x (Fig. 7) is defined on the segment [- 1; 1], the range of its values is the segment [-р/2;р/2]. On the segment [- 1; 1] the function y=arcsin x is continuous and increases monotonically from -p/2 to p/2 (this follows from the fact that the function y=sin x on the segment [-p/2; p/2] is continuous and increases monotonically). Highest value it takes at x = 1: arcsin 1 = p/2, and the smallest at x = -1: arcsin (-1) = -p/2. At x = 0 the function is zero: arcsin 0 = 0.
Let us show that the function y = arcsin x is odd, i.e. arcsin(-x) = - arcsin x for any x [ - 1; 1].
Indeed, by definition, if |x| ?1, we have: - p/2 ? arcsin x ? ? r/2. Thus, the angles arcsin(-x) and - arcsin x belong to the same segment [ - p/2; p/2].
Let's find the sines of these angles: sin (arcsin(-x)) = - x (by definition); since the function y=sin x is odd, then sin (-arcsin x)= - sin (arcsin x)= - x. So, the sines of angles belonging to the same interval [-р/2; p/2], are equal, which means the angles themselves are equal, i.e. arcsin (-x)= - arcsin x. This means that the function y=arcsin x is odd. The graph of the function y=arcsin x is symmetrical about the origin.
Let us show that arcsin (sin x) = x for any x [-р/2; p/2].
Indeed, by definition -p/2? arcsin (sin x) ? p/2, and by condition -p/2? x? r/2. This means that the angles x and arcsin (sin x) belong to the same interval of monotonicity of the function y=sin x. If the sines of such angles are equal, then the angles themselves are equal. Let's find the sines of these angles: for angle x we have sin x, for angle arcsin (sin x) we have sin (arcsin(sin x)) = sin x. We found that the sines of the angles are equal, therefore, the angles are equal, i.e. arcsin(sin x) = x. .
Rice. 7
Rice. 8
The graph of the function arcsin (sin|x|) is obtained by the usual transformations associated with the modulus from the graph y=arcsin (sin x) (shown by the dashed line in Fig. 8). The desired graph y=arcsin (sin |x-/4|) is obtained from it by shifting by /4 to the right along the x-axis (shown as a solid line in Fig. 8)
Inverse function of tangent
The function y=tg x on the interval takes all numerical values: E (tg x)=. Over this interval it is continuous and increases monotonically. This means that a function inverse to the function y = tan x is defined on the interval. This inverse function is called the arctangent and is denoted y = arctan x.
The arctangent of a is an angle from an interval whose tangent is equal to a. Thus, arctg a is an angle that satisfies the following conditions: tg (arctg a) = a and 0? arctg a ? R.
So, any number x always corresponds to a single value of the function y = arctan x (Fig. 9).
It is obvious that D (arctg x) = , E (arctg x) = .
The function y = arctan x is increasing because the function y = tan x is increasing on the interval. It is not difficult to prove that arctg(-x) = - arctgx, i.e. that arctangent is an odd function.
Rice. 9
The graph of the function y = arctan x is symmetrical to the graph of the function y = tan x relative to the straight line y = x, the graph y = arctan x passes through the origin of coordinates (since arctan 0 = 0) and is symmetrical relative to the origin (like the graph of an odd function).
It can be proven that arctan (tan x) = x if x.
Cotangent inverse function
The function y = ctg x on an interval takes all numeric values from the interval. The range of its values coincides with the set of all real numbers. In the interval, the function y = cot x is continuous and increases monotonically. This means that on this interval a function is defined that is inverse to the function y = cot x. The inverse function of cotangent is called arccotangent and is denoted y = arcctg x.
The arc cotangent of a is an angle belonging to an interval whose cotangent is equal to a.
Thus, аrcctg a is an angle satisfying the following conditions: ctg (arcctg a)=a and 0? arcctg a ? R.
From the definition of the inverse function and the definition of arctangent it follows that D (arcctg x) = , E (arcctg x) = . The arc cotangent is a decreasing function because the function y = ctg x decreases in the interval.
The graph of the function y = arcctg x does not intersect the Ox axis, since y > 0 R. For x = 0 y = arcctg 0 =.
The graph of the function y = arcctg x is shown in Figure 11.
Rice. 11
Note that for all real values of x the identity is true: arcctg(-x) = p-arcctg x.
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:
- The function is defined on the interval [-1; 1].
- ODZ for arccos - .
- The graph is entirely located in the first and second quarters, and the function itself is neither even nor odd.
- Y = 0 at x = 1.
- 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 basic typical Unified State Exam assignments may lead students into confusion.
Exercise 1. Indicate the functions shown in the figure.
Answer: rice. 1 – 4, Fig. 2 – 1.
IN 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 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:
- The graph is infinite and defined on the interval (- ∞; + ∞).
- Arctangent is an odd function, therefore, arctan (- x) = - arctan x.
- Y = 0 at x = 0.
- The curve increases throughout the entire definition region.
Here's a short comparative analysis tg x and arctg x in table form.
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:
- The function definition interval is infinity.
- The range of acceptable values is the interval (0; π).
- F(x) is neither even nor odd.
- 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 appearance crooked. For solutions trigonometric equations and inequalities, identity tables are needed. 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 we correctly transform the expression and lead to the right type, 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.
