Properties and graphs of trigonometric functions. Trigonometric functions of a numerical argument. Graph and properties of the function y \u003d sin x Functions of a numeric argument

Whatever real number t is taken, it can be assigned a uniquely defined number sin t. True, the correspondence rule is rather complicated; as we saw above, it consists in the following.

To find the value of sin t by the number t, you need:

1) position the number circle in the coordinate plane so that the center of the circle coincides with the origin of coordinates, and the starting point A of the circle hits the point (1; 0);

2) find a point on the circle corresponding to the number t;

3) find the ordinate of this point.

This ordinate is sin t.

In fact, we are talking about the function u = sin t, where t is any real number.

All these functions are called trigonometric functions of the numerical argument t.

There are a number of relationships connecting the values ​​of various trigonometric functions, we have already received some of these relationships:

sin 2 t + cos 2 t = 1

From the last two formulas, it is easy to obtain a relation connecting tg t and ctg t:

All of these formulas are used in those cases when, knowing the value of any trigonometric function, it is required to calculate the values ​​of the remaining trigonometric functions.

The terms "sine", "cosine", "tangent" and "cotangent" were actually familiar, however, they were still used in a slightly different interpretation: in geometry and physics, they considered sine, cosine, tangent and cotangent g l a(but not

numbers, as it was in the previous paragraphs).

It is known from geometry that the sine (cosine) of an acute angle is the ratio of the leg of a right triangle to its hypotenuse, and the tangent (cotangent) of an angle is the ratio of the legs of a right triangle. A different approach to the concepts of sine, cosine, tangent and cotangent was developed in the previous paragraphs. In fact, these approaches are interrelated.

Let's take an angle with a degree measure b o and arrange it in the model "numerical circle in a rectangular coordinate system" as shown in Fig. 14

corner top compatible with center

circles (with the origin of the coordinate system),

and one side of the corner is compatible with

positive ray of the x-axis. Point

intersection of the other side of the angle with

the circle will be denoted by the letter M. Ordina-

Figure 14 b o , and the abscissa of this point is the cosine of the angle b o .

To find the sine or cosine of the angle b o it is not at all necessary to make these very complex constructions each time.

It suffices to note that the arc AM is the same part of the length of the numerical circle as the angle b o is from the angle of 360°. If the length of the arc AM is denoted by the letter t, then we get:

Thus,

For example,

It is believed that 30 ° is a degree measure of an angle, and is a radian measure of the same angle: 30 ° = rad. At all:

In particular, I'm glad from where, in turn, we get.

So what is 1 radian? There are various measures of segment lengths: centimeters, meters, yards, etc. There are also various measures to indicate the magnitude of the angles. We consider the central angles of the unit circle. An angle of 1° is a central angle based on an arc that is part of a circle. An angle of 1 radian is a central angle based on an arc of length 1, i.e. on an arc whose length is equal to the radius of the circle. From the formula, we get that 1 rad \u003d 57.3 °.

Considering the function u = sin t (or any other trigonometric function), we can consider the independent variable t as a numerical argument, as was the case in the previous paragraphs, but we can also consider this variable as a measure of the angle, i.e. angular argument. Therefore, speaking of a trigonometric function, in a certain sense it is indifferent to consider it a function of a numerical or angular argument.

Trigonometric functions of a numerical argument.

Trigonometric functions of a numeric argumentt are functions of the form y= cos t,
y= sint, y= tg t, y=ctgt.

Using these formulas, through the known value of one trigonometric function, you can find the unknown values ​​of other trigonometric functions.

Explanations.

1) Take the formula cos 2 t + sin 2 t = 1 and use it to derive a new formula.

To do this, we divide both parts of the formula by cos 2 t (for t ≠ 0, that is, t ≠ π/2 + π k). So:

cos 2 t sin 2 t 1
--- + --- = ---
cos 2 t cos 2 t cos 2 t

The first term is equal to 1. We know that the ratio of the sine to the conisus is the tangent, which means that the second term is equal to tg 2 t. As a result, we get a new (and already known to you) formula:

2) Now we divide cos 2 t + sin 2 t = 1 by sin 2 t (for t ≠ π k):

cos 2 t sin 2 t 1
--- + --- = ---, where t ≠ π k + π k, k- integer
sin 2 t sin 2 t sin 2 t

The ratio of cosine to sine is the cotangent. Means:

Knowing the elementary foundations of mathematics and having learned the basic formulas of trigonometry, you can easily derive most of the other trigonometric identities on your own. And this is even better than just memorizing them: what is learned by heart is quickly forgotten, and what is understood is remembered for a long time, if not forever. For example, it is not necessary to memorize what the sum of one and the square of the tangent is. Forgotten - you can easily remember if you know the simplest thing: tangent is the ratio of sine to cosine. In addition, apply a simple rule for adding fractions with different denominators - and get the result:

sin 2 t 1 sin 2 t cos 2 t + sin 2 t 1
1 + tg 2 t = 1 + --- = - + --- = ------ = ---
cos 2 t 1 cos 2 t cos 2 t cos 2 t

It is just as easy to find the sum of unity and the square of the cotangent, as well as many other identities.

Trigonometric functions of angular argument.

In functionsat = cost, at = sint, at = tgt, at = ctgt variablet can be more than just a numeric argument. It can also be considered a measure of an angle - that is, an angular argument.

With the help of a numerical circle and a coordinate system, you can easily find the sine, cosine, tangent, cotangent of any angle. For this, two important conditions must be met:
1) the vertex of the corner must be the center of the circle, which is also the center of the coordinate axis;

2) one of the sides of the angle must be a positive axis beam x.

In this case, the ordinate of the point at which the circle and the second side of the angle intersect is the sine of this angle, and the abscissa of this point is the cosine of the given angle.

Explanation. Let's draw an angle, one side of which is a positive ray of the axis x, and the second side comes out from the origin of the coordinate axis (and from the center of the circle) at an angle of 30º (see figure). Then the point of intersection of the second side with the circle corresponds to π/6. We know the ordinate and abscissa of this point. They are the cosine and sine of our angle:

√3 1
--; --
2 2

And knowing the sine and cosine of an angle, you can easily find its tangent and cotangent.

Thus, a number circle located in a coordinate system is a convenient way to find the sine, cosine, tangent, or cotangent of an angle.

But there is an easier way. It is possible not to draw a circle and a coordinate system. You can use simple and convenient formulas:

Example: find the sine and cosine of an angle equal to 60º.

Solution :

π 60 π √3
sin 60º = sin --- = sin -- = --
180 3 2

π 1
cos 60º = cos -- = -
3 2

Explanation: we found out that the sine and cosine of the angle 60º correspond to the values ​​​​of the circle point π / 3. Further, we simply find the values ​​of this point in the table - and thus solve our example. The table of sines and cosines of the main points of the numerical circle is in the previous section and on the "Tables" page.

Definition. Trigonometric functions of a numerical argument are the trigonometric functions of the same name of an angle equal to radians.

Let us clarify this definition with concrete examples.

Example 1. Calculate the value of . Here by we mean an abstract irrational number. By definition. So, .

Example 2. Calculate the value of . Here by 1.5 we mean an abstract number. As defined (see annex II).

Example 3. Calculate the value Similarly to the previous one, we obtain (see Appendix II).

So, in the future, under the argument of trigonometric functions, we will understand the angle (arc) or just a number, depending on the problem that we are solving. And in some cases, the argument can be a value that has another dimension, such as time, etc. Calling the argument an angle (arc), we can mean by it the number with which it is measured in radians.

Lesson and presentation on the topic: "Trigonometric function of a numerical argument, definition, identities"

Additional materials
Dear users, do not forget to leave your comments, feedback, suggestions. All materials are checked by an antivirus program.

Teaching aids and simulators in the online store "Integral" for grade 10
Algebraic problems with parameters, grades 9–11
Software environment "1C: Mathematical constructor 6.1"

What will we study:
1. Definition of a numeric argument.
2. Basic formulas.
3. Trigonometric identities.
4. Examples and tasks for independent solution.

Definition of the trigonometric function of a numeric argument

Let's remember the basic formulas:
$sin^2(t)+cos^2(t)=1$. By the way, what is the name of this formula?

$tg(t)=\frac(sin(t))(cos(t))$, for $t≠\frac(π)(2)+πk$.
$ctg(t)=\frac(cos(t))(sin(t))$, for $t≠πk$.

Let's derive new formulas.

Trigonometric identities

Now we divide both sides of the identity by $sin^2(t)$.
We get: $\frac(sin^2(t))(sin^2(t))+\frac(cos^2(t))(sin^2(t))=\frac(1)(sin^2 (t))$.
Let's transform: $1+(\frac(cos(t))(sin(t)))^2=\frac(1)(sin^2(t)).$
We get a new identity that is worth remembering:
$ctg^2(t)+1=\frac(1)(sin^2(t))$, for $t≠πk$.

We managed to get two new formulas. Remember them.
These formulas are used if, by some known value of a trigonometric function, it is required to calculate the value of another function.

Solving examples for trigonometric functions of a numerical argument

$cos(t) =\frac(5)(7)$, find $sin(t)$; $tg(t)$; $ctg(t)$ for all t.

Solution:

$sin^2(t)+cos^2(t)=1$.
Then $sin^2(t)=1-cos^2(t)$.
$sin^2(t)=1-(\frac(5)(7))^2=1-\frac(25)(49)=\frac(49-25)(49)=\frac(24) (49)$.
$sin(t)=±\frac(\sqrt(24))(7)=±\frac(2\sqrt(6))(7)$.
$tg(t)=±\sqrt(\frac(1)(cos^2(t))-1)=±\sqrt(\frac(1)(\frac(25)(49))-1)= ±\sqrt(\frac(49)(25)-1)=±\sqrt(\frac(24)(25))=±\frac(\sqrt(24))(5)$.
$ctg(t)=±\sqrt(\frac(1)(sin^2(t))-1)=±\sqrt(\frac(1)(\frac(24)(49))-1)= ±\sqrt(\frac(49)(24)-1)=±\sqrt(\frac(25)(24))=±\frac(5)(\sqrt(24))$.

Example 2

$tg(t) = \frac(5)(12)$, find $sin(t)$; $cos(t)$; $ctg(t)$, for all $0

Solution:
$tg^2(t)+1=\frac(1)(cos^2(t))$.
Then $\frac(1)(cos^2(t))=1+\frac(25)(144)=\frac(169)(144)$.
We get that $cos^2(t)=\frac(144)(169)$.
Then $cos^2(t)=±\frac(12)(13)$, but $0 The cosine in the first quadrant is positive. Then $cos(t)=\frac(12)(13)$.
We get: $sin(t)=tg(t)*cos(t)=\frac(5)(12)*\frac(12)(13)=\frac(5)(13)$.
$ctg(t)=\frac(1)(tg(t))=\frac(12)(5)$.

Tasks for independent solution