In this article we will show how to give definitions of sine, cosine, tangent and cotangent of an angle and number in trigonometry. Here we will talk about notations, give examples of entries, and give graphic illustrations. In conclusion, let us draw a parallel between the definitions of sine, cosine, tangent and cotangent in trigonometry and geometry.

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Definition of sine, cosine, tangent and cotangent

Let's see how the idea of ​​sine, cosine, tangent and cotangent is formed in school course mathematics. In geometry lessons, the definition of sine, cosine, tangent and cotangent of an acute angle in right triangle. And later trigonometry is studied, which talks about sine, cosine, tangent and cotangent of the angle of rotation and number. Let us present all these definitions, give examples and give the necessary comments.

Acute angle in a right triangle

From the geometry course we know the definitions of sine, cosine, tangent and cotangent of an acute angle in a right triangle. They are given as the ratio of the sides of a right triangle. Let us give their formulations.

Definition.

Sine of an acute angle in a right triangle is the ratio of the opposite side to the hypotenuse.

Definition.

Cosine of an acute angle in a right triangle is the ratio of the adjacent leg to the hypotenuse.

Definition.

Tangent of an acute angle in a right triangle– this is the ratio of the opposite side to the adjacent side.

Definition.

Cotangent of an acute angle in a right triangle- this is the ratio of the adjacent side to the opposite side.

The designations for sine, cosine, tangent and cotangent are also introduced there - sin, cos, tg and ctg, respectively.

For example, if ABC is a right triangle with right angle C, then the sine of the acute angle A is equal to the ratio of the opposite side BC to the hypotenuse AB, that is, sin∠A=BC/AB.

These definitions allow you to calculate the values ​​of sine, cosine, tangent and cotangent of an acute angle from known lengths sides of a right triangle, as well as along known values find the lengths of the other sides using sine, cosine, tangent, cotangent and the length of one of the sides. For example, if we knew that in a right triangle the leg AC is equal to 3 and the hypotenuse AB is equal to 7, then we could calculate the value of the cosine of the acute angle A by definition: cos∠A=AC/AB=3/7.

Rotation angle

In trigonometry, they begin to look at the angle more broadly - they introduce the concept of angle of rotation. The magnitude of the rotation angle, unlike an acute angle, is not limited to 0 to 90 degrees; the rotation angle in degrees (and in radians) can be expressed by any real number from −∞ to +∞.

In this light, the definitions of sine, cosine, tangent and cotangent are given not of an acute angle, but of an angle of arbitrary size - the angle of rotation. They are given through the x and y coordinates of the point A 1, to which the so-called starting point A(1, 0) goes after its rotation by an angle α around the point O - the beginning of the rectangular Cartesian coordinate system and the center of the unit circle.

Definition.

Sine of rotation angleα is the ordinate of point A 1, that is, sinα=y.

Definition.

Cosine of the rotation angleα is called the abscissa of point A 1, that is, cosα=x.

Definition.

Tangent of rotation angleα is the ratio of the ordinate of point A 1 to its abscissa, that is, tanα=y/x.

Definition.

Cotangent of the rotation angleα is the ratio of the abscissa of point A 1 to its ordinate, that is, ctgα=x/y.

Sine and cosine are defined for any angle α, since we can always determine the abscissa and ordinate of the point, which is obtained by rotating the starting point by angle α. But tangent and cotangent are not defined for any angle. The tangent is not defined for angles α at which the starting point goes to a point with zero abscissa (0, 1) or (0, −1), and this occurs at angles 90°+180° k, k∈Z (π /2+π·k rad). Indeed, at such angles of rotation, the expression tgα=y/x does not make sense, since it contains division by zero. As for the cotangent, it is not defined for angles α at which the starting point goes to the point with the zero ordinate (1, 0) or (−1, 0), and this occurs for angles 180° k, k ∈Z (π·k rad).

So, sine and cosine are defined for any rotation angles, tangent is defined for all angles except 90°+180°k, k∈Z (π/2+πk rad), and cotangent is defined for all angles except 180° ·k , k∈Z (π·k rad).

The definitions include the designations already known to us sin, cos, tg and ctg, they are also used to designate sine, cosine, tangent and cotangent of the angle of rotation (sometimes you can find the designations tan and cotcorresponding to tangent and cotangent). So the sine of a rotation angle of 30 degrees can be written as sin30°, the entries tg(−24°17′) and ctgα correspond to the tangent of the rotation angle −24 degrees 17 minutes and the cotangent of the rotation angle α. Recall that when writing the radian measure of an angle, the designation “rad” is often omitted. For example, the cosine of a rotation angle of three pi rad is usually denoted cos3·π.

In conclusion of this point, it is worth noting that when talking about sine, cosine, tangent and cotangent of the angle of rotation, the phrase “angle of rotation” or the word “rotation” is often omitted. That is, instead of the phrase “sine of the rotation angle alpha,” the phrase “sine of the alpha angle” or even shorter, “sine alpha,” is usually used. The same applies to cosine, tangent, and cotangent.

We will also say that the definitions of sine, cosine, tangent and cotangent of an acute angle in a right triangle are consistent with the definitions just given for sine, cosine, tangent and cotangent of an angle of rotation ranging from 0 to 90 degrees. We will justify this.

Numbers

Definition.

Sine, cosine, tangent and cotangent of a number t is a number equal to the sine, cosine, tangent and cotangent of the rotation angle in t radians, respectively.

For example, the cosine of the number 8·π by definition is a number equal to the cosine of the angle of 8·π rad. And the cosine of an angle of 8·π rad is equal to one, therefore, the cosine of the number 8·π is equal to 1.

There is another approach to determining the sine, cosine, tangent and cotangent of a number. It consists in the fact that everyone real number t is matched to the dot unit circle centered at the origin of the rectangular coordinate system, and sine, cosine, tangent and cotangent are determined through the coordinates of this point. Let's look at this in more detail.

Let us show how a correspondence is established between real numbers and points on a circle:

• the number 0 is assigned the starting point A(1, 0);
• positive number t is associated with the point of the unit circle, which we will get to if we move along the circle from the starting point in a counterclockwise direction and walk a path of length t;
• negative number t is associated with the point of the unit circle, which we will get to if we move along the circle from the starting point in a clockwise direction and walk a path of length |t| .

Now we move on to the definitions of sine, cosine, tangent and cotangent of the number t. Let us assume that the number t corresponds to a point on the circle A 1 (x, y) (for example, the number &pi/2; corresponds to the point A 1 (0, 1) ).

Definition.

Sine of the number t is the ordinate of the point on the unit circle corresponding to the number t, that is, sint=y.

Definition.

Cosine of the number t is called the abscissa of the point of the unit circle corresponding to the number t, that is, cost=x.

Definition.

Tangent of the number t is the ratio of the ordinate to the abscissa of a point on the unit circle corresponding to the number t, that is, tgt=y/x. In another equivalent formulation, the tangent of a number t is the ratio of the sine of this number to the cosine, that is, tgt=sint/cost.

Definition.

Cotangent of the number t is the ratio of the abscissa to the ordinate of a point on the unit circle corresponding to the number t, that is, ctgt=x/y. Another formulation is this: the tangent of the number t is the ratio of the cosine of the number t to the sine of the number t: ctgt=cost/sint.

Here we note that the definitions just given are consistent with the definition given at the beginning of this paragraph. Indeed, the point on the unit circle corresponding to the number t coincides with the point obtained by rotating the starting point by an angle of t radians.

It is still worth clarifying this point. Let's say we have the entry sin3. How can we understand whether we are talking about the sine of the number 3 or the sine of the rotation angle of 3 radians? This is usually clear from the context, otherwise it is likely not of fundamental importance.

Trigonometric functions of angular and numeric argument

According to the definitions given in the previous paragraph, each angle of rotation α corresponds to a very specific value sinα, as well as the value cosα. In addition, all rotation angles other than 90°+180°k, k∈Z (π/2+πk rad) correspond to tgα values, and values ​​other than 180°k, k∈Z (πk rad ) – values ​​of ctgα . Therefore sinα, cosα, tanα and ctgα are functions of the angle α. In other words, these are functions of the angular argument.

We can speak similarly about the functions sine, cosine, tangent and cotangent of a numerical argument. Indeed, each real number t corresponds to a very specific value sint, as well as cost. In addition, all numbers other than π/2+π·k, k∈Z correspond to values ​​tgt, and numbers π·k, k∈Z - values ​​ctgt.

The functions sine, cosine, tangent and cotangent are called basic trigonometric functions.

It is usually clear from the context whether we are dealing with trigonometric functions of an angular argument or a numerical argument. Otherwise, we can think of the independent variable as both a measure of the angle (angular argument) and a numeric argument.

However, at school we mainly study numerical functions, that is, functions whose arguments, as well as their corresponding function values, are numbers. Therefore, if we're talking about specifically about functions, it is advisable to consider trigonometric functions as functions of numerical arguments.

Relationship between definitions from geometry and trigonometry

If we consider the rotation angle α ranging from 0 to 90 degrees, then the definitions of sine, cosine, tangent and cotangent of the rotation angle in the context of trigonometry are fully consistent with the definitions of sine, cosine, tangent and cotangent of an acute angle in a right triangle, which are given in the geometry course. Let's justify this.

Let us depict the unit circle in the rectangular Cartesian coordinate system Oxy. Let's mark the starting point A(1, 0) . Let's rotate it by an angle α ranging from 0 to 90 degrees, we get point A 1 (x, y). Let us drop the perpendicular A 1 H from point A 1 to the Ox axis.

It is easy to see that in a right triangle angle A 1 OH equal to angle rotation α, the length of the leg OH adjacent to this angle is equal to the abscissa of point A 1, that is, |OH|=x, the length of the leg A 1 H opposite to the corner is equal to the ordinate of point A 1, that is, |A 1 H|=y, and the length of the hypotenuse OA 1 is equal to one, since it is the radius of the unit circle. Then, by definition from geometry, the sine of an acute angle α in a right triangle A 1 OH is equal to the ratio of the opposite leg to the hypotenuse, that is, sinα=|A 1 H|/|OA 1 |=y/1=y. And by definition from trigonometry, the sine of the rotation angle α is equal to the ordinate of point A 1, that is, sinα=y. This shows that determining the sine of an acute angle in a right triangle is equivalent to determining the sine of the rotation angle α when α is from 0 to 90 degrees.

Similarly, it can be shown that the definitions of cosine, tangent and cotangent of an acute angle α are consistent with the definitions of cosine, tangent and cotangent of the rotation angle α.

Bibliography.

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3. Algebra and elementary functions: Tutorial for 9th grade students high school/ E. S. Kochetkov, E. S. Kochetkova; Edited by Doctor of Physical and Mathematical Sciences O. N. Golovin. - 4th ed. M.: Education, 1969.
4. Algebra: Textbook for 9th grade. avg. school/Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova; Ed. S. A. Telyakovsky. - M.: Education, 1990. - 272 pp.: ill. - ISBN 5-09-002727-7
5. Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
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The ratio of the opposite side to the hypotenuse is called sinus of an acute angle right triangle.

\sin \alpha = \frac(a)(c)

Cosine of an acute angle of a right triangle

The ratio of the adjacent leg to the hypotenuse is called cosine of an acute angle right triangle.

\cos \alpha = \frac(b)(c)

Tangent of an acute angle of a right triangle

The ratio of the opposite side to the adjacent side is called tangent of an acute angle right triangle.

tg \alpha = \frac(a)(b)

Cotangent of an acute angle of a right triangle

The ratio of the adjacent side to the opposite side is called cotangent of an acute angle right triangle.

ctg \alpha = \frac(b)(a)

Sine of an arbitrary angle

The ordinate of a point on the unit circle to which the angle \alpha corresponds is called sine of an arbitrary angle rotation \alpha .

\sin \alpha=y

Cosine of an arbitrary angle

The abscissa of a point on the unit circle to which the angle \alpha corresponds is called cosine of an arbitrary angle rotation \alpha .

\cos \alpha=x

Tangent of an arbitrary angle

The ratio of the sine of an arbitrary rotation angle \alpha to its cosine is called tangent of an arbitrary angle rotation \alpha .

tan \alpha = y_(A)

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

Cotangent of an arbitrary angle

The ratio of the cosine of an arbitrary rotation angle \alpha to its sine is called cotangent of an arbitrary angle rotation \alpha .

ctg\alpha =x_(A)

ctg \alpha = \frac(\cos \alpha)(\sin \alpha)

An example of finding an arbitrary angle

If \alpha is some angle AOM, where M is a point of the unit circle, then

\sin \alpha=y_(M) , \cos \alpha=x_(M) , tg \alpha=\frac(y_(M))(x_(M)), ctg \alpha=\frac(x_(M))(y_(M)).

For example, if \angle AOM = -\frac(\pi)(4), then: the ordinate of point M is equal to -\frac(\sqrt(2))(2), abscissa is equal to \frac(\sqrt(2))(2) and that's why

\sin \left (-\frac(\pi)(4) \right)=-\frac(\sqrt(2))(2);

\cos \left (\frac(\pi)(4) \right)=\frac(\sqrt(2))(2);

tg;

ctg \left (-\frac(\pi)(4) \right)=-1.

Table of values ​​of sines of cosines of tangents of cotangents

The values ​​of the main frequently occurring angles are given in the table:

	0^(\circ) (0)	30^(\circ)\left(\frac(\pi)(6)\right)	45^(\circ)\left(\frac(\pi)(4)\right)	60^(\circ)\left(\frac(\pi)(3)\right)	90^(\circ)\left(\frac(\pi)(2)\right)	180^(\circ)\left(\pi\right)	270^(\circ)\left(\frac(3\pi)(2)\right)	360^(\circ)\left(2\pi\right)
\sin\alpha	0	\frac12	\frac(\sqrt 2)(2)	\frac(\sqrt 3)(2)	1	0	−1	0
\cos\alpha	1	\frac(\sqrt 3)(2)	\frac(\sqrt 2)(2)	\frac12	0	−1	0	1
tg\alpha	0	\frac(\sqrt 3)(3)	1	\sqrt3	—	0	—	0
ctg\alpha	—	\sqrt3	1	\frac(\sqrt 3)(3)	0	—	0	—

What is sine, cosine, tangent, cotangent of an angle will help you understand a right triangle.

What are the sides of a right triangle called? That's right, hypotenuse and legs: the hypotenuse is the side that lies opposite right angle(in our example this is the side \(AC\) ); legs are the two remaining sides \(AB\) and \(BC\) (those adjacent to the right angle), and if we consider the legs relative to the angle \(BC\), then leg \(AB\) is the adjacent leg, and leg \(BC\) is opposite. So, now let’s answer the question: what are sine, cosine, tangent and cotangent of an angle?

Sine of angle– this is the ratio of the opposite (distant) leg to the hypotenuse.

In our triangle:

\[ \sin \beta =\dfrac(BC)(AC) \]

Cosine of angle– this is the ratio of the adjacent (close) leg to the hypotenuse.

In our triangle:

\[ \cos \beta =\dfrac(AB)(AC) \]

Tangent of the angle– this is the ratio of the opposite (distant) side to the adjacent (close).

In our triangle:

\[ tg\beta =\dfrac(BC)(AB) \]

Cotangent of angle– this is the ratio of the adjacent (close) leg to the opposite (far).

In our triangle:

\[ ctg\beta =\dfrac(AB)(BC) \]

These definitions are necessary remember! To make it easier to remember which leg to divide into what, you need to clearly understand that in tangent And cotangent only the legs sit, and the hypotenuse appears only in sinus And cosine. And then you can come up with a chain of associations. For example, this one:

Cosine→touch→touch→adjacent;

Cotangent→touch→touch→adjacent.

First of all, you need to remember that sine, cosine, tangent and cotangent as the ratios of the sides of a triangle do not depend on the lengths of these sides (at the same angle). Do not believe? Then make sure by looking at the picture:

Consider, for example, the cosine of the angle \(\beta \) . By definition, from a triangle \(ABC\) : \(\cos \beta =\dfrac(AB)(AC)=\dfrac(4)(6)=\dfrac(2)(3) \), but we can calculate the cosine of the angle \(\beta \) from the triangle \(AHI \) : \(\cos \beta =\dfrac(AH)(AI)=\dfrac(6)(9)=\dfrac(2)(3) \). You see, the lengths of the sides are different, but the value of the cosine of one angle is the same. Thus, the values ​​of sine, cosine, tangent and cotangent depend solely on the magnitude of the angle.

If you understand the definitions, then go ahead and consolidate them!

For the triangle \(ABC \) shown in the figure below, we find \(\sin \ \alpha ,\ \cos \ \alpha ,\ tg\ \alpha ,\ ctg\ \alpha \).

\(\begin(array)(l)\sin \ \alpha =\dfrac(4)(5)=0.8\\\cos \ \alpha =\dfrac(3)(5)=0.6\\ tg\ \alpha =\dfrac(4)(3)\\ctg\ \alpha =\dfrac(3)(4)=0.75\end(array) \)

Well, did you get it? Then try it yourself: calculate the same for the angle \(\beta \) .

Answers: \(\sin \ \beta =0.6;\ \cos \ \beta =0.8;\ tg\ \beta =0.75;\ ctg\ \beta =\dfrac(4)(3) \).

Unit (trigonometric) circle

Understanding the concepts of degrees and radians, we considered a circle with a radius equal to \(1\) . Such a circle is called single. It will be very useful when studying trigonometry. Therefore, let's look at it in a little more detail.

As you can see, given circle constructed in a Cartesian coordinate system. The radius of the circle is equal to one, while the center of the circle lies at the origin of coordinates, the initial position of the radius vector is fixed along the positive direction of the \(x\) axis (in our example, this is the radius \(AB\)).

Each point on the circle corresponds to two numbers: the coordinate along the \(x\) axis and the coordinate along the \(y\) axis. What are these coordinate numbers? And in general, what do they have to do with the topic at hand? To do this, we need to remember about the considered right triangle. In the figure above, you can see two whole right triangles. Consider the triangle \(ACG\) . It is rectangular because \(CG\) is perpendicular to the \(x\) axis.

What is \(\cos \ \alpha \) from the triangle \(ACG \)? That's right \(\cos \ \alpha =\dfrac(AG)(AC) \). In addition, we know that \(AC\) is the radius of the unit circle, which means \(AC=1\) . Let's substitute this value into our formula for cosine. Here's what happens:

\(\cos \ \alpha =\dfrac(AG)(AC)=\dfrac(AG)(1)=AG \).

What is \(\sin \ \alpha \) from the triangle \(ACG \) equal to? Well, of course, \(\sin \alpha =\dfrac(CG)(AC)\)! Substitute the value of the radius \(AC\) into this formula and get:

\(\sin \alpha =\dfrac(CG)(AC)=\dfrac(CG)(1)=CG \)

So, can you tell what coordinates the point \(C\) belonging to the circle has? Well, no way? What if you realize that \(\cos \ \alpha \) and \(\sin \alpha \) are just numbers? What coordinate does \(\cos \alpha \) correspond to? Well, of course, the coordinate \(x\)! And what coordinate does \(\sin \alpha \) correspond to? That's right, coordinate \(y\)! So the point \(C(x;y)=C(\cos \alpha ;\sin \alpha) \).

What then are \(tg \alpha \) and \(ctg \alpha \) equal to? That’s right, let’s use the corresponding definitions of tangent and cotangent and get that \(tg \alpha =\dfrac(\sin \alpha )(\cos \alpha )=\dfrac(y)(x) \), A \(ctg \alpha =\dfrac(\cos \alpha )(\sin \alpha )=\dfrac(x)(y) \).

What if the angle is larger? For example, like in this picture:

What has changed in in this example? Let's figure it out. To do this, let's turn again to a right triangle. Consider a right triangle \(((A)_(1))((C)_(1))G \) : angle (as adjacent to angle \(\beta \) ). What is the value of sine, cosine, tangent and cotangent for an angle \(((C)_(1))((A)_(1))G=180()^\circ -\beta \ \)? That's right, we adhere to the corresponding definitions of trigonometric functions:

\(\begin(array)(l)\sin \angle ((C)_(1))((A)_(1))G=\dfrac(((C)_(1))G)(( (A)_(1))((C)_(1)))=\dfrac(((C)_(1))G)(1)=((C)_(1))G=y; \\\cos \angle ((C)_(1))((A)_(1))G=\dfrac(((A)_(1))G)(((A)_(1)) ((C)_(1)))=\dfrac(((A)_(1))G)(1)=((A)_(1))G=x;\\tg\angle ((C )_(1))((A)_(1))G=\dfrac(((C)_(1))G)(((A)_(1))G)=\dfrac(y)( x);\\ctg\angle ((C)_(1))((A)_(1))G=\dfrac(((A)_(1))G)(((C)_(1 ))G)=\dfrac(x)(y)\end(array) \)

Well, as you can see, the value of the sine of the angle still corresponds to the coordinate \(y\) ; the value of the cosine of the angle - coordinate \(x\) ; and the values ​​of tangent and cotangent to the corresponding ratios. Thus, these relations apply to any rotation of the radius vector.

It has already been mentioned that the initial position of the radius vector is along the positive direction of the \(x\) axis. So far we have rotated this vector counterclockwise, but what happens if we rotate it clockwise? Nothing extraordinary, it will turn out to be the same angle of a certain size, but only it will be negative. Thus, when rotating the radius vector counterclockwise, we get positive angles, and when rotating clockwise – negative.

So, we know that the whole revolution of the radius vector around the circle is \(360()^\circ \) or \(2\pi \) . Is it possible to rotate the radius vector by \(390()^\circ \) or by \(-1140()^\circ \)? Well, of course you can! In the first case, \(390()^\circ =360()^\circ +30()^\circ \), thus, the radius vector will make one full revolution and stop at the position \(30()^\circ \) or \(\dfrac(\pi )(6) \) .

In the second case, \(-1140()^\circ =-360()^\circ \cdot 3-60()^\circ \), that is, the radius vector will make three full revolutions and will stop at position \(-60()^\circ \) or \(-\dfrac(\pi )(3) \) .

Thus, from the above examples we can conclude that angles that differ by \(360()^\circ \cdot m \) or \(2\pi \cdot m \) (where \(m \) is any integer ), correspond to the same position of the radius vector.

The figure below shows the angle \(\beta =-60()^\circ \) . The same image corresponds to the corner \(-420()^\circ ,-780()^\circ ,\ 300()^\circ ,660()^\circ \) etc. This list can be continued indefinitely. All these angles can be written by the general formula \(\beta +360()^\circ \cdot m\) or \(\beta +2\pi \cdot m \) (where \(m \) is any integer)

\(\begin(array)(l)-420()^\circ =-60+360\cdot (-1);\\-780()^\circ =-60+360\cdot (-2); \\300()^\circ =-60+360\cdot 1;\\660()^\circ =-60+360\cdot 2.\end(array) \)

Now, knowing the definitions of the basic trigonometric functions and using the unit circle, try to answer what the values ​​are:

\(\begin(array)(l)\sin \ 90()^\circ =?\\\cos \ 90()^\circ =?\\\text(tg)\ 90()^\circ =? \\\text(ctg)\ 90()^\circ =?\\\sin \ 180()^\circ =\sin \ \pi =?\\\cos \ 180()^\circ =\cos \ \pi =?\\\text(tg)\ 180()^\circ =\text(tg)\ \pi =?\\\text(ctg)\ 180()^\circ =\text(ctg)\ \pi =?\\\sin \ 270()^\circ =?\\\cos \ 270()^\circ =?\\\text(tg)\ 270()^\circ =?\\\text (ctg)\ 270()^\circ =?\\\sin \ 360()^\circ =?\\\cos \ 360()^\circ =?\\\text(tg)\ 360()^ \circ =?\\\text(ctg)\ 360()^\circ =?\\\sin \ 450()^\circ =?\\\cos \ 450()^\circ =?\\\text (tg)\ 450()^\circ =?\\\text(ctg)\ 450()^\circ =?\end(array) \)

Here's a unit circle to help you:

Having difficulties? Then let's figure it out. So we know that:

\(\begin(array)(l)\sin \alpha =y;\\cos\alpha =x;\\tg\alpha =\dfrac(y)(x);\\ctg\alpha =\dfrac(x )(y).\end(array)\)

From here, we determine the coordinates of the points corresponding to certain angle measures. Well, let's start in order: the corner in \(90()^\circ =\dfrac(\pi )(2) \) corresponds to a point with coordinates \(\left(0;1 \right) \) , therefore:

\(\sin 90()^\circ =y=1 \) ;

\(\cos 90()^\circ =x=0 \) ;

\(\text(tg)\ 90()^\circ =\dfrac(y)(x)=\dfrac(1)(0)\Rightarrow \text(tg)\ 90()^\circ \)- does not exist;

\(\text(ctg)\ 90()^\circ =\dfrac(x)(y)=\dfrac(0)(1)=0 \).

Further, adhering to the same logic, we find out that the corners in \(180()^\circ ,\ 270()^\circ ,\ 360()^\circ ,\ 450()^\circ (=360()^\circ +90()^\circ)\ \ ) correspond to points with coordinates \(\left(-1;0 \right),\text( )\left(0;-1 \right),\text( )\left(1;0 \right),\text( )\left(0 ;1 \right) \), respectively. Knowing this, it is easy to determine the values ​​of trigonometric functions at the corresponding points. Try it yourself first, and then check the answers.

Answers:

\(\displaystyle \sin \180()^\circ =\sin \ \pi =0 \)

\(\displaystyle \cos \180()^\circ =\cos \ \pi =-1\)

\(\text(tg)\ 180()^\circ =\text(tg)\ \pi =\dfrac(0)(-1)=0 \)

\(\text(ctg)\ 180()^\circ =\text(ctg)\ \pi =\dfrac(-1)(0)\Rightarrow \text(ctg)\ \pi \)- does not exist

\(\sin \270()^\circ =-1\)

\(\cos \ 270()^\circ =0 \)

\(\text(tg)\ 270()^\circ =\dfrac(-1)(0)\Rightarrow \text(tg)\ 270()^\circ \)- does not exist

\(\text(ctg)\ 270()^\circ =\dfrac(0)(-1)=0 \)

\(\sin \360()^\circ =0\)

\(\cos \360()^\circ =1\)

\(\text(tg)\ 360()^\circ =\dfrac(0)(1)=0 \)

\(\text(ctg)\ 360()^\circ =\dfrac(1)(0)\Rightarrow \text(ctg)\ 2\pi \)- does not exist

\(\sin \ 450()^\circ =\sin \ \left(360()^\circ +90()^\circ \right)=\sin \ 90()^\circ =1 \)

\(\cos \ 450()^\circ =\cos \ \left(360()^\circ +90()^\circ \right)=\cos \ 90()^\circ =0 \)

\(\text(tg)\ 450()^\circ =\text(tg)\ \left(360()^\circ +90()^\circ \right)=\text(tg)\ 90() ^\circ =\dfrac(1)(0)\Rightarrow \text(tg)\ 450()^\circ \)- does not exist

\(\text(ctg)\ 450()^\circ =\text(ctg)\left(360()^\circ +90()^\circ \right)=\text(ctg)\ 90()^ \circ =\dfrac(0)(1)=0 \).

Thus, we can make the following table:

There is no need to remember all these values. It is enough to remember the correspondence between the coordinates of points on the unit circle and the values ​​of trigonometric functions:

\(\left. \begin(array)(l)\sin \alpha =y;\\cos \alpha =x;\\tg \alpha =\dfrac(y)(x);\\ctg \alpha =\ dfrac(x)(y).\end(array) \right\)\ \text(You must remember or be able to display it!! \) !}

But the values ​​of the trigonometric functions of angles in and \(30()^\circ =\dfrac(\pi )(6),\ 45()^\circ =\dfrac(\pi )(4)\) given in the table below, you must remember:

Don’t be scared, now we’ll show you one example of a fairly simple memorization of the corresponding values:

To use this method, it is vital to remember the sine values ​​for all three measures angle ( \(30()^\circ =\dfrac(\pi )(6),\ 45()^\circ =\dfrac(\pi )(4),\ 60()^\circ =\dfrac(\pi )(3)\)), as well as the value of the tangent of the angle in \(30()^\circ \) . Knowing these \(4\) values, it is quite simple to restore the entire table - the cosine values ​​are transferred in accordance with the arrows, that is:

\(\begin(array)(l)\sin 30()^\circ =\cos \ 60()^\circ =\dfrac(1)(2)\ \ \\\sin 45()^\circ = \cos \ 45()^\circ =\dfrac(\sqrt(2))(2)\\\sin 60()^\circ =\cos \ 30()^\circ =\dfrac(\sqrt(3 ))(2)\ \end(array) \)

\(\text(tg)\ 30()^\circ \ =\dfrac(1)(\sqrt(3)) \), knowing this, you can restore the values ​​for \(\text(tg)\ 45()^\circ , \text(tg)\ 60()^\circ \). The numerator "\(1 \)" will correspond to \(\text(tg)\ 45()^\circ \ \) and the denominator "\(\sqrt(\text(3)) \)" will correspond to \(\text (tg)\ 60()^\circ \ \) . Cotangent values ​​are transferred in accordance with the arrows indicated in the figure. If you understand this and remember the diagram with the arrows, then it will be enough to remember only \(4\) values ​​from the table.

Coordinates of a point on a circle

Is it possible to find a point (its coordinates) on a circle, knowing the coordinates of the center of the circle, its radius and angle of rotation? Well, of course you can! Let's get it out general formula to find the coordinates of a point. For example, here is a circle in front of us:

We are given that point \(K(((x)_(0));((y)_(0)))=K(3;2) \)- center of the circle. The radius of the circle is \(1.5\) . It is necessary to find the coordinates of the point \(P\) obtained by rotating the point \(O\) by \(\delta \) degrees.

As can be seen from the figure, the coordinate \(x\) of the point \(P\) corresponds to the length of the segment \(TP=UQ=UK+KQ\) . The length of the segment \(UK\) corresponds to the coordinate \(x\) of the center of the circle, that is, it is equal to \(3\) . The length of the segment \(KQ\) can be expressed using the definition of cosine:

\(\cos \ \delta =\dfrac(KQ)(KP)=\dfrac(KQ)(r)\Rightarrow KQ=r\cdot \cos \ \delta \).

Then we have that for the point \(P\) the coordinate \(x=((x)_(0))+r\cdot \cos \ \delta =3+1.5\cdot \cos \ \delta \).

Using the same logic, we find the value of the y coordinate for the point \(P\) . Thus,

\(y=((y)_(0))+r\cdot \sin \ \delta =2+1.5\cdot \sin \delta \).

So, in general view coordinates of points are determined by the formulas:

\(\begin(array)(l)x=((x)_(0))+r\cdot \cos \ \delta \\y=((y)_(0))+r\cdot \sin \ \delta \end(array) \), Where

\(((x)_(0)),((y)_(0)) \) - coordinates of the center of the circle,

\(r\) - radius of the circle,

\(\delta \) - rotation angle of the vector radius.

As you can see, for the unit circle we are considering, these formulas are significantly reduced, since the coordinates of the center are equal to zero and the radius is equal to one:

\(\begin(array)(l)x=((x)_(0))+r\cdot \cos \ \delta =0+1\cdot \cos \ \delta =\cos \ \delta \\y =((y)_(0))+r\cdot \sin \ \delta =0+1\cdot \sin \ \delta =\sin \ \delta \end(array) \)

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Average level

Right triangle. The Complete Illustrated Guide (2019)

RIGHT TRIANGLE. FIRST LEVEL.

In problems, the right angle is not at all necessary - the lower left, so you need to learn to recognize a right triangle in this form,

and in this

and in this

What's good about a right triangle? Well... first of all, there are special beautiful names for his sides.

Attention to the drawing!

Remember and don't confuse: there are two legs, and there is only one hypotenuse(one and only, unique and longest)!

Well, we’ve discussed the names, now the most important thing: the Pythagorean Theorem.

Pythagorean theorem.

This theorem is the key to solving many problems involving a right triangle. Pythagoras proved it completely time immemorial, and since then she has brought a lot of benefit to those who know her. And the best thing about it is that it is simple.

So, Pythagorean theorem:

Do you remember the joke: “Pythagorean pants are equal on all sides!”?

Let's draw these same Pythagorean pants and look at them.

Doesn't it look like some kind of shorts? Well, on which sides and where are they equal? Why and where did the joke come from? And this joke is connected precisely with the Pythagorean theorem, or more precisely with the way Pythagoras himself formulated his theorem. And he formulated it like this:

"Sum areas of squares, built on the legs, is equal to square area, built on the hypotenuse."

Does it really sound a little different? And so, when Pythagoras drew the statement of his theorem, this is exactly the picture that came out.

In this picture, the sum of the areas of the small squares is equal to the area of ​​the large square. And so that children can better remember that the sum of the squares of the legs is equal to the square of the hypotenuse, someone witty came up with this joke about Pythagorean pants.

Why are we now formulating the Pythagorean theorem?

Did Pythagoras suffer and talk about squares?

You see, in ancient times there was no... algebra! There were no signs and so on. There were no inscriptions. Can you imagine how terrible it was for the poor ancient students to remember everything in words??! And we can rejoice that we have a simple formulation of the Pythagorean theorem. Let's repeat it again to remember it better:

It should be easy now:

Square of the hypotenuse equal to the sum squares of legs.

Well, the most important theorem about right triangles has been discussed. If you are interested in how it is proven, read the following levels of theory, and now let's go further... into the dark forest... of trigonometry! To the terrible words sine, cosine, tangent and cotangent.

Sine, cosine, tangent, cotangent in a right triangle.

In fact, everything is not so scary at all. Of course, the “real” definition of sine, cosine, tangent and cotangent should be looked at in the article. But I really don’t want to, do I? We can rejoice: to solve problems about a right triangle, you can simply fill in the following simple things:

Why is everything just about the corner? Where is the corner? In order to understand this, you need to know how statements 1 - 4 are written in words. Look, understand and remember!

1.
Actually it sounds like this:

What about the angle? Is there a leg that is opposite the corner, that is, an opposite (for an angle) leg? Of course have! This is a leg!

What about the angle? Look carefully. Which leg is adjacent to the corner? Of course, the leg. This means that for the angle the leg is adjacent, and

Now, pay attention! Look what we got:

See how cool it is:

Now let's move on to tangent and cotangent.

How can I write this down in words now? What is the leg in relation to the angle? Opposite, of course - it “lies” opposite the corner. What about the leg? Adjacent to the corner. So what have we got?

See how the numerator and denominator have swapped places?

And now the corners again and made an exchange:

Summary

Let's briefly write down everything we've learned.

Pythagorean theorem:

The main theorem about right triangles is the Pythagorean theorem.

Pythagorean theorem

By the way, do you remember well what legs and hypotenuse are? If not very good, then look at the picture - refresh your knowledge

It is quite possible that you have already used the Pythagorean theorem many times, but have you ever wondered why such a theorem is true? How can I prove it? Let's do like the ancient Greeks. Let's draw a square with a side.

See how cleverly we divided its sides into lengths and!

Now let's connect the marked dots

Here we, however, noted something else, but you yourself look at the drawing and think why this is so.

What is the area of ​​the larger square? Right, . What about a smaller area? Certainly, . The total area of ​​the four corners remains. Imagine that we took them two at a time and leaned them against each other with their hypotenuses. What happened? Two rectangles. This means that the area of ​​the “cuts” is equal.

Let's put it all together now.

Let's transform:

So we visited Pythagoras - we proved his theorem in an ancient way.

Right triangle and trigonometry

For a right triangle, the following relations hold:

The sine of an acute angle is equal to the ratio of the opposite side to the hypotenuse

The cosine of an acute angle is equal to the ratio of the adjacent leg to the hypotenuse.

The tangent of an acute angle is equal to the ratio of the opposite side to the adjacent side.

The cotangent of an acute angle is equal to the ratio of the adjacent side to the opposite side.

And once again all this in the form of a tablet:

It is very comfortable!

Signs of equality of right triangles

I. On two sides

II. By leg and hypotenuse

III. By hypotenuse and acute angle

IV. Along the leg and acute angle

a)

b)

Attention! It is very important here that the legs are “appropriate”. For example, if it goes like this:

THEN TRIANGLES ARE NOT EQUAL, despite the fact that they have one identical acute angle.

Need to in both triangles the leg was adjacent, or in both it was opposite.

Have you noticed how the signs of equality of right triangles differ from the usual signs of equality of triangles? Take a look at the topic “and pay attention to the fact that for equality of “ordinary” triangles, three of their elements must be equal: two sides and the angle between them, two angles and the side between them, or three sides. But for the equality of right triangles, only two corresponding elements are enough. Great, right?

The situation is approximately the same with the signs of similarity of right triangles.

Signs of similarity of right triangles

I. Along an acute angle

II. On two sides

III. By leg and hypotenuse

Median in a right triangle

Why is this so?

Instead of a right triangle, consider a whole rectangle.

Let's draw a diagonal and consider a point - the point of intersection of the diagonals. What do you know about the diagonals of a rectangle?

And what follows from this?

So it turned out that

1. - median:

Remember this fact! Helps a lot!

What’s even more surprising is that the opposite is also true.

What good can be obtained from the fact that the median drawn to the hypotenuse is equal to half the hypotenuse? Let's look at the picture

Look carefully. We have: , that is, the distances from the point to all three vertices of the triangle turned out to be equal. But there is only one point in the triangle, the distances from which from all three vertices of the triangle are equal, and this is the CENTER OF THE CIRCLE. So what happened?

So let's start with this “besides...”.

Let's look at and.

But similar triangles have all equal angles!

The same can be said about and

Now let's draw it together:

What benefit can be derived from this “triple” similarity?

Well, for example - two formulas for the height of a right triangle.

Let us write down the relations of the corresponding parties:

To find the height, we solve the proportion and get the first formula "Height in a right triangle":

So, let's apply the similarity: .

What will happen now?

Again we solve the proportion and get the second formula:

You need to remember both of these formulas very well and use the one that is more convenient. Let's write them down again

Pythagorean theorem:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs: .

Signs of equality of right triangles:

• on two sides:
• by leg and hypotenuse: or
• along the leg and adjacent acute angle: or
• along the leg and the opposite acute angle: or
• by hypotenuse and acute angle: or.

Signs of similarity of right triangles:

• one acute corner: or
• from the proportionality of two legs:
• from the proportionality of the leg and hypotenuse: or.

Sine, cosine, tangent, cotangent in a right triangle

• The sine of an acute angle of a right triangle is the ratio of the opposite side to the hypotenuse:
• The cosine of an acute angle of a right triangle is the ratio of the adjacent leg to the hypotenuse:
• The tangent of an acute angle of a right triangle is the ratio of the opposite side to the adjacent side:
• The cotangent of an acute angle of a right triangle is the ratio of the adjacent side to the opposite side: .

Height of a right triangle: or.

In a right triangle, the median drawn from the vertex of the right angle is equal to half the hypotenuse: .

Area of ​​a right triangle:

• via legs:

Sinus acute angle α of a right triangle is the ratio opposite leg to hypotenuse.
It is denoted as follows: sin α.

Cosine The acute angle α of a right triangle is the ratio of the adjacent leg to the hypotenuse.
It is designated as follows: cos α.

Tangent acute angle α is the ratio of the opposite side to the adjacent side.
It is designated as follows: tg α.

Cotangent acute angle α is the ratio of the adjacent side to the opposite side.
It is designated as follows: ctg α.

The sine, cosine, tangent and cotangent of an angle depend only on the size of the angle.

Rules:

Basic trigonometric identities in a right triangle:

(α – acute angle opposite to the leg b and adjacent to the leg a . Side With – hypotenuse. β – second acute angle).

b sin α = - c	sin 2 α + cos 2 α = 1
a cos α = - c	1 1 + tan 2 α = -- cos 2 α
b tan α = - a	1 1 + ctg 2 α = -- sin 2 α
a ctg α = - b	1 1 1 + -- = -- tan 2 α sin 2 α
sin α tg α = -- cos α

As the acute angle increasessin α andtan α increase, andcos α decreases.

For any acute angle α:

sin (90° – α) = cos α

cos (90° – α) = sin α

Example-explanation:

Let in a right triangle ABC
AB = 6,
BC = 3,
angle A = 30º.

Let's find out the sine of angle A and the cosine of angle B.

Solution .

1) First, we find the value of angle B. Everything is simple here: since in a right triangle the sum of the acute angles is 90º, then angle B = 60º:

B = 90º – 30º = 60º.

2) Let's calculate sin A. We know that the sine is equal to the ratio of the opposite side to the hypotenuse. For angle A, the opposite side is side BC. So:

BC 3 1
sin A = -- = - = -
AB 6 2

3) Now let's calculate cos B. We know that the cosine is equal to the ratio of the adjacent leg to the hypotenuse. For angle B, the adjacent leg is the same side BC. This means that we again need to divide BC by AB - that is, perform the same actions as when calculating the sine of angle A:

BC 3 1
cos B = -- = - = -
AB 6 2

The result is:
sin A = cos B = 1/2.

sin 30º = cos 60º = 1/2.

It follows from this that in a right triangle, the sine of one acute angle is equal to the cosine of the other acute angle - and vice versa. This is exactly what our two formulas mean:
sin (90° – α) = cos α
cos (90° – α) = sin α

Let's make sure of this again:

1) Let α = 60º. Substituting the value of α into the sine formula, we get:
sin (90º – 60º) = cos 60º.
sin 30º = cos 60º.

2) Let α = 30º. Substituting the value of α into the cosine formula, we get:
cos (90° – 30º) = sin 30º.
cos 60° = sin 30º.

(For more information about trigonometry, see the Algebra section)