Trigonometric identities of addition. Basic trigonometric formulas

This is the last and most main lesson, necessary to solve problems B11. We already know how to convert angles from a radian measure to a degree measure (see the lesson “Radian and degree measure of an angle”), and we also know how to determine the sign of a trigonometric function, focusing on the coordinate quarters (see the lesson “Signs of trigonometric functions”).

The only thing left to do is calculate the value of the function itself - the very number that is written in the answer. This is where the basic trigonometric identity comes to the rescue.

Basic trigonometric identity. For any angle α the following statement is true:

sin 2 α + cos 2 α = 1.

This formula relates the sine and cosine of one angle. Now, knowing the sine, we can easily find the cosine - and vice versa. It is enough to take the square root:

Note the "±" sign in front of the roots. The fact is that from the basic trigonometric identity it is not clear what the original sine and cosine were: positive or negative. After all, squaring is an even function that “burns” all the minuses (if there were any).

That is why in all problems B11, which appear in the Unified State Exam in mathematics, there must be additional conditions, which help get rid of uncertainty with signs. Usually this is an indication of the coordinate quarter, by which the sign can be determined.

An attentive reader will probably ask: “What about tangent and cotangent?” It is impossible to directly calculate these functions from the above formulas. However, there are important consequences from the basic trigonometric identity, which already contain tangents and cotangents. Namely:

An important corollary: for any angle α, the basic trigonometric identity can be rewritten as follows:

These equations are easily derived from the main identity - it is enough to divide both sides by cos 2 α (to obtain the tangent) or by sin 2 α (to obtain the cotangent).

Let's look at all this at specific examples. Below are the real B11 problems which are taken from the mock ones Unified State Exam options in mathematics 2012.

We know the cosine, but we don't know the sine. The main trigonometric identity (in its “pure” form) connects just these functions, so we will work with it. We have:

sin 2 α + cos 2 α = 1 ⇒ sin 2 α + 99/100 = 1 ⇒ sin 2 α = 1/100 ⇒ sin α = ±1/10 = ±0.1.

To solve the problem, it remains to find the sign of the sine. Since the angle α ∈ (π /2; π ), then in degree measure it is written as follows: α ∈ (90°; 180°).

Consequently, angle α lies in the second coordinate quarter - all sines there are positive. Therefore sin α = 0.1.

So, we know the sine, but we need to find the cosine. Both of these functions are in the basic trigonometric identity. Let's substitute:

sin 2 α + cos 2 α = 1 ⇒ 3/4 + cos 2 α = 1 ⇒ cos 2 α = 1/4 ⇒ cos α = ±1/2 = ±0.5.

All that remains is to figure out the sign in front of the fraction. What to choose: plus or minus? By condition, angle α belongs to the interval (π 3π /2). Let's convert the angles from radian measures to degrees - we get: α ∈ (180°; 270°).

Obviously, this is the III coordinate quarter, where all cosines are negative. Therefore cos α = −0.5.

Task. Find tan α if the following is known:

Tangent and cosine are related by the equation following from the basic trigonometric identity:

We get: tan α = ±3. The sign of the tangent is determined by the angle α. It is known that α ∈ (3π /2; 2π ). Let's convert the angles from radian measures to degrees - we get α ∈ (270°; 360°).

Obviously, this is the IV coordinate quarter, where all tangents are negative. Therefore tan α = −3.

Task. Find cos α if the following is known:

Again the sine is known and the cosine is unknown. Let us write down the main trigonometric identity:

sin 2 α + cos 2 α = 1 ⇒ 0.64 + cos 2 α = 1 ⇒ cos 2 α = 0.36 ⇒ cos α = ±0.6.

The sign is determined by the angle. We have: α ∈ (3π /2; 2π ). Let's convert the angles from degrees to radians: α ∈ (270°; 360°) is the IV coordinate quarter, the cosines there are positive. Therefore, cos α = 0.6.

Task. Find sin α if the following is known:

Let us write down a formula that follows from the basic trigonometric identity and directly connects sine and cotangent:

From here we get that sin 2 α = 1/25, i.e. sin α = ±1/5 = ±0.2. It is known that angle α ∈ (0; π /2). In degree measure, this is written as follows: α ∈ (0°; 90°) - I coordinate quarter.

So, the angle is in the I coordinate quadrant - all trigonometric functions there are positive, so sin α = 0.2.

Reference data for tangent (tg x) and cotangent (ctg x). Geometric definition, properties, graphs, formulas. Table of tangents and cotangents, derivatives, integrals, series expansions. Expressions through complex variables. Connection with hyperbolic functions.

Geometric definition

|BD| - length of the arc of a circle with center at point A.
α is the angle expressed in radians.

Tangent ( tan α) is a trigonometric function depending on the angle α between the hypotenuse and the leg of a right triangle, equal to the ratio of the length of the opposite leg |BC| to the length of the adjacent leg |AB| .

Cotangent ( ctg α) is a trigonometric function depending on the angle α between the hypotenuse and the leg of a right triangle, equal to the ratio of the length of the adjacent leg |AB| to the length of the opposite leg |BC| .

Tangent

Where n- whole.

In Western literature, tangent is denoted as follows:
.
;
;
.

Graph of the tangent function, y = tan x

Cotangent

Where n- whole.

In Western literature, cotangent is denoted as follows:
.
The following notations are also accepted:
;
;
.

Graph of the cotangent function, y = ctg x

Properties of tangent and cotangent

Periodicity

Functions y = tg x and y = ctg x are periodic with period π.

Parity

The tangent and cotangent functions are odd.

Areas of definition and values, increasing, decreasing

The tangent and cotangent functions are continuous in their domain of definition (see proof of continuity). The main properties of tangent and cotangent are presented in the table ( n- whole).

	y= tg x	y= ctg x
Scope and continuity
Range of values	-∞ < y < +∞	-∞ < y < +∞
Increasing		-
Descending	-
Extremes	-	-
Zeros, y = 0
Intercept points with the ordinate axis, x = 0	y= 0	-

Formulas

Expressions using sine and cosine

; ;
; ;
;

Formulas for tangent and cotangent from sum and difference

The remaining formulas are easy to obtain, for example

Product of tangents

Formula for the sum and difference of tangents

This table presents the values of tangents and cotangents for certain values of the argument.

Expressions using complex numbers

Expressions through hyperbolic functions

;
;

Derivatives

; .

.
Derivative of the nth order with respect to the variable x of the function:
.
Deriving formulas for tangent > > > ; for cotangent > > >

Integrals

Series expansions

To obtain the expansion of the tangent in powers of x, you need to take several terms of the expansion in a power series for the functions sin x And cos x and divide these polynomials by each other, . This produces the following formulas.

At .

at .
Where Bn- Bernoulli numbers. They are determined either from the recurrence relation:
;
;
Where .
Or according to Laplace's formula:

Inverse functions

Inverse functions to tangent and cotangent are arctangent and arccotangent, respectively.

Arctangent, arctg

, Where n- whole.

Arccotangent, arcctg

, Where n- whole.

Used literature:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
G. Korn, Handbook of Mathematics for Scientists and Engineers, 2012.

Trigonometric identities- these are equalities that establish a connection between sine, cosine, tangent and cotangent of one angle, which allows you to find any of these functions, provided that any other is known.

tg \alpha = \frac(\sin \alpha)(\cos \alpha), \enspace ctg \alpha = \frac(\cos \alpha)(\sin \alpha)

tg \alpha \cdot ctg \alpha = 1

This identity says that the sum of the square of the sine of one angle and the square of the cosine of one angle is equal to one, which in practice makes it possible to calculate the sine of one angle when its cosine is known and vice versa.

When converting trigonometric expressions, this identity is very often used, which allows you to replace the sum of the squares of the cosine and sine of one angle with one and also perform the replacement operation in the reverse order.

Finding tangent and cotangent using sine and cosine

tg \alpha = \frac(\sin \alpha)(\cos \alpha),\enspace

These identities are formed from the definitions of sine, cosine, tangent and cotangent. After all, if you look at it, then by definition the ordinate y is a sine, and the abscissa x is a cosine. Then the tangent will be equal to the ratio \frac(y)(x)=\frac(\sin \alpha)(\cos \alpha), and the ratio \frac(x)(y)=\frac(\cos \alpha)(\sin \alpha)- will be a cotangent.

Let us add that only for such angles \alpha at which the trigonometric functions included in them make sense, the identities will hold, ctg \alpha=\frac(\cos \alpha)(\sin \alpha).

For example: tg \alpha = \frac(\sin \alpha)(\cos \alpha) is valid for angles \alpha that are different from \frac(\pi)(2)+\pi z, A ctg \alpha=\frac(\cos \alpha)(\sin \alpha)- for an angle \alpha other than \pi z, z is an integer.

Relationship between tangent and cotangent

tg \alpha \cdot ctg \alpha=1

This identity is valid only for angles \alpha that are different from \frac(\pi)(2) z. Otherwise, either cotangent or tangent will not be determined.

Based on the above points, we obtain that tg \alpha = \frac(y)(x), A ctg \alpha=\frac(x)(y). It follows that tg \alpha \cdot ctg \alpha = \frac(y)(x) \cdot \frac(x)(y)=1. Thus, the tangent and cotangent of the same angle at which they make sense are mutually inverse numbers.

Relationships between tangent and cosine, cotangent and sine

tg^(2) \alpha + 1=\frac(1)(\cos^(2) \alpha)- the sum of the square of the tangent of the angle \alpha and 1 is equal to the inverse square of the cosine of this angle. This identity is valid for all \alpha other than \frac(\pi)(2)+ \pi z.

1+ctg^(2) \alpha=\frac(1)(\sin^(2)\alpha)- the sum of 1 and the square of the cotangent of the angle \alpha is equal to the inverse square of the sine of the given angle. This identity is valid for any \alpha different from \pi z.

Examples with solutions to problems using trigonometric identities

Example 1

Find \sin \alpha and tg \alpha if \cos \alpha=-\frac12 And \frac(\pi)(2)< \alpha < \pi ;

Show solution

Solution

The functions \sin \alpha and \cos \alpha are related by the formula \sin^(2)\alpha + \cos^(2) \alpha = 1. Substituting into this formula \cos \alpha = -\frac12, we get:

\sin^(2)\alpha + \left (-\frac12 \right)^2 = 1

This equation has 2 solutions:

\sin \alpha = \pm \sqrt(1-\frac14) = \pm \frac(\sqrt 3)(2)

By condition \frac(\pi)(2)< \alpha < \pi . In the second quarter the sine is positive, so \sin \alpha = \frac(\sqrt 3)(2).

In order to find tan \alpha, we use the formula tg \alpha = \frac(\sin \alpha)(\cos \alpha)

tg \alpha = \frac(\sqrt 3)(2) : \frac12 = \sqrt 3

Example 2

Find \cos \alpha and ctg \alpha if and \frac(\pi)(2)< \alpha < \pi .

Show solution

Solution

Substituting into the formula \sin^(2)\alpha + \cos^(2) \alpha = 1 given number \sin \alpha=\frac(\sqrt3)(2), we get \left (\frac(\sqrt3)(2)\right)^(2) + \cos^(2) \alpha = 1. This equation has two solutions \cos \alpha = \pm \sqrt(1-\frac34)=\pm\sqrt\frac14.

By condition \frac(\pi)(2)< \alpha < \pi . In the second quarter the cosine is negative, so \cos \alpha = -\sqrt\frac14=-\frac12.

In order to find ctg \alpha , we use the formula ctg \alpha = \frac(\cos \alpha)(\sin \alpha). We know the corresponding values.

ctg \alpha = -\frac12: \frac(\sqrt3)(2) = -\frac(1)(\sqrt 3).

At the very beginning of this article, we examined the concept of trigonometric functions. Their main purpose is to study the basics of trigonometry and study periodic processes. And it was not in vain that we drew the trigonometric circle, because in most cases trigonometric functions are defined as the ratio of the sides of a triangle or its certain segments in a unit circle. I also mentioned the undeniably enormous importance of trigonometry in modern life. But science does not stand still, as a result we can significantly expand the scope of trigonometry and transfer its provisions to real, and sometimes to complex numbers.

Trigonometry formulas There are several types. Let's look at them in order.

Ratios of trigonometric functions of the same angle

Here we come to consider such a concept as basic trigonometric identities.

A trigonometric identity is an equality that consists of trigonometric relations and which holds for all values of the angles that are included in it.

Let's look at the most important trigonometric identities and their proofs:

The first identity follows from the very definition of tangent.

Let's take right triangle, in which there is an acute angle x at vertex A.

To prove the identities, you need to use the Pythagorean theorem:

(BC) 2 + (AC) 2 = (AB) 2

Now we divide both sides of the equality by (AB) 2 and recalling the definitions of sin and cos angle, we obtain the second identity:

(BC) 2 /(AB) 2 + (AC) 2 /(AB) 2 = 1

sin x = (BC)/(AB)

cos x = (AC)/(AB)

sin 2 x + cos 2 x = 1

To prove the third and fourth identities, we will use the previous proof.

To do this, divide both sides of the second identity by cos 2 x:

sin 2 x/ cos 2 x + cos 2 x/ cos 2 x = 1/ cos 2 x

sin 2 x/ cos 2 x + 1 = 1/ cos 2 x

Based on the first identity tg x = sin x /cos x we obtain the third:

1 + tan 2 x = 1/cos 2 x

Now let's divide the second identity by sin 2 x:

sin 2 x/ sin 2 x + cos 2 x/ sin 2 x = 1/ sin 2 x

1+ cos 2 x/ sin 2 x = 1/ sin 2 x

cos 2 x/ sin 2 x is nothing more than 1/tg 2 x, so we get the fourth identity:

1 + 1/tg 2 x = 1/sin 2 x

It's time to remember the sum theorem internal corners triangle, which states that the sum of the angles of a triangle = 180 0. It turns out that at vertex B of the triangle there is an angle whose value is 180 0 – 90 0 – x = 90 0 – x.

Let us again recall the definitions for sin and cos and obtain the fifth and sixth identities:

sin x = (BC)/(AB)

cos(90 0 – x) = (BC)/(AB)

cos(90 0 – x) = sin x

Now let's do the following:

cos x = (AC)/(AB)

sin(90 0 – x) = (AC)/(AB)

sin(90 0 – x) = cos x

As you can see, everything is elementary here.

There are other identities that are used in solving mathematical identities, I will give them simply in the form reference information, because they all stem from the above.

Expressing trigonometric functions through each other

(the choice of sign in front of the root is determined by which of the quarters of the circle the corner is located in?)