Sine, cosine, tangent and cotangent: definitions in trigonometry, examples, formulas. Sine, cosine, tangent, cotangent of an acute angle

Instruction

A triangle is called a right triangle if one of its angles is 90 degrees. It consists of two legs and a hypotenuse. The hypotenuse is the longest side of this triangle. It lies against a right angle. The legs, respectively, are called its smaller sides. They can be either equal to each other or have different sizes. Equality of the legs that you are working with a right triangle. Its beauty is that it combines two figures: a right-angled and an isosceles triangle. If the legs are not equal, then the triangle is arbitrary and according to the basic law: the larger the angle, the more the one lying opposite it rolls.

There are several ways to find the hypotenuse by and angle. But before using one of them, you should determine which and the angle are known. Given an angle and the leg adjacent to it, it is easier to find the hypotenuse by the cosine of the angle. The cosine of an acute angle (cos a) in a right triangle is the ratio of the adjacent leg to the hypotenuse. This implies that the hypotenuse (c) will be equal to the ratio of the adjacent leg (b) to the cosine of the angle a (cos a). This can be written like this: cos a=b/c => c=b/cos a.

If an angle and an opposite leg are given, then work should be done. The sine of an acute angle (sin a) in a right triangle is the ratio of the opposite leg (a) to the hypotenuse (c). Here the principle is the same as in the previous example, only the sine is taken instead of the cosine function. sin a=a/c => c=a/sin a.

You can also use a trigonometric function such as . But finding the desired value is slightly more complicated. The tangent of an acute angle (tg a) in a right triangle is the ratio of the opposite leg (a) to the adjacent one (b). Having found both legs, apply the Pythagorean theorem (the square of the hypotenuse is equal to the sum of the squares of the legs) and the larger one will be found.

note

When working with the Pythagorean theorem, do not forget that you are dealing with a degree. Having found the sum of the squares of the legs, to get the final answer, you should take the square root.

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• how to find the leg and hypotenuse

The hypotenuse is the side in a right triangle that is opposite the 90 degree angle. In order to calculate its length, it is enough to know the length of one of the legs and the value of one of the acute angles of the triangle.

Instruction

With a known and acute right angle, then the size of the hypotenuse is the ratio of the leg to / of this angle, if the given angle is opposite / adjacent to it:

h = C1(or C2)/sinα;

h = С1(or С2)/cosα.

Example: Let ABC be given with hypotenuse AB and C. Let angle B be 60 degrees and angle A 30 degrees The length of the leg BC is 8 cm. You need the length of the hypotenuse AB. To do this, you can use any of the methods suggested above:

AB=BC/cos60=8 cm.

AB = BC/sin30 = 8 cm.

Word " leg” comes from the Greek words “perpendicular” or “vertical” - this explains why both sides of a right-angled triangle, which make up its ninety-degree angle, were named that way. Find the length of any of leg ov is not difficult if the value of the angle adjacent to it and any other of the parameters are known, since in this case the values ​​of all three angles will actually become known.

Instruction

If, in addition to the value of the adjacent angle (β), the length of the second leg a (b), then the length leg and (a) can be defined as the quotient of the length of the known leg and at a known angle: a=b/tg(β). This follows from the definition of this trigonometric. You can do without the tangent if you use the theorem. It follows from it that the length of the desired to the sine of the opposite angle to the ratio of the length of the known leg but to the sine of a known angle. Opposite to the desired leg y an acute angle can be expressed in terms of a known angle as 180°-90°-β = 90°-β, since the sum of all angles of any triangle must be 180°, and one of its angles is equal to 90°. So the desired length leg and can be calculated by the formula a=sin(90°-β)∗b/sin(β).

If the magnitude of the adjacent angle (β) and the length of the hypotenuse (c) are known, then the length leg and (a) can be calculated as the product of the length of the hypotenuse and the cosine of the known angle: a=c∗cos(β). This follows from the definition of cosine as a trigonometric function. But you can use, as in the previous step, the sine theorem and then the length of the desired leg a will be equal to the product of the sine between 90° and the known angle times the ratio of the length of the hypotenuse to the sine of the right angle. And since the sine of 90° is equal to one, it can be written as follows: a=sin(90°-β)∗c.

Practical calculations can be performed, for example, using the software calculator included in the Windows operating system. To run it, you can select the "Run" item in the main menu on the "Start" button, type the calc command and click the "OK" button. The simplest version of the interface of this program that opens by default does not provide trigonometric functions, therefore, after launching it, you need to click the “View” section in the menu and select the “Scientific” or “Engineering” line (depending on the version of the operating system you are using).

Related videos

The word "katet" came into Russian from Greek. In exact translation, it means a plumb line, that is, perpendicular to the surface of the earth. In mathematics, legs are called sides that form a right angle of a right triangle. The side opposite this angle is called the hypotenuse. The term "leg" is also used in architecture and welding technology.

Draw a right triangle ACB. Label its legs a and b, and label its hypotenuse c. All sides and angles of a right triangle are defined to each other. The ratio of the leg opposite one of the acute angles to the hypotenuse is called the sine of this angle. In this triangle sinCAB=a/c. Cosine is the ratio to the hypotenuse of the adjacent leg, i.e. cosCAB=b/c. The inverse relationships are called secant and cosecant.

The secant of this angle is obtained by dividing the hypotenuse by the adjacent leg, that is, secCAB=c/b. It turns out the reciprocal of the cosine, that is, it can be expressed by the formula secCAB=1/cosSAB.
The cosecant is equal to the quotient of dividing the hypotenuse by the opposite leg and is the reciprocal of the sine. It can be calculated using the formula cosecCAB=1/sinCAB

Both legs are interconnected and cotangent. In this case, the tangent will be the ratio of side a to side b, that is, the opposite leg to the adjacent one. This ratio can be expressed by the formula tgCAB=a/b. Accordingly, the inverse ratio will be the cotangent: ctgCAB=b/a.

The ratio between the sizes of the hypotenuse and both legs was determined by the ancient Greek Pythagoras. The theorem, his name, people still use. It says that the square of the hypotenuse is equal to the sum of the squares of the legs, that is, c2 \u003d a2 + b2. Accordingly, each leg will be equal to the square root of the difference between the squares of the hypotenuse and the other leg. This formula can be written as b=√(c2-a2).

The length of the leg can also be expressed through the relationships you know. According to the theorems of sines and cosines, the leg is equal to the product of the hypotenuse and one of these functions. You can express it and or cotangent. The leg a can be found, for example, by the formula a \u003d b * tan CAB. In exactly the same way, depending on the given tangent or , the second leg is determined.

In architecture, the term "leg" is also used. It is applied to an Ionic capital and plumb through the middle of its back. That is, in this case, by this term, the perpendicular to the given line.

In welding technology, there is a “leg of a fillet weld”. As in other cases, this is the shortest distance. Here we are talking about the gap between one of the parts to be welded to the border of the seam located on the surface of the other part.

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• what is the leg and hypotenuse in 2019

In life, we often have to face math problems: at school, at university, and then helping our child with homework. People of certain professions will encounter mathematics on a daily basis. Therefore, it is useful to memorize or recall mathematical rules. In this article, we will analyze one of them: finding the leg of a right triangle.

What is a right triangle

First, let's remember what a right triangle is. A right triangle is a geometric figure of three segments that connect points that do not lie on the same straight line, and one of the angles of this figure is 90 degrees. The sides that form a right angle are called the legs, and the side that lies opposite the right angle is called the hypotenuse.

Finding the leg of a right triangle

There are several ways to find out the length of the leg. I would like to consider them in more detail.

Pythagorean theorem to find the leg of a right triangle

If we know the hypotenuse and the leg, then we can find the length of the unknown leg using the Pythagorean theorem. It sounds like this: “The square of the hypotenuse is equal to the sum of the squares of the legs.” Formula: c²=a²+b², where c is the hypotenuse, a and b are the legs. We transform the formula and get: a²=c²-b².

Example. The hypotenuse is 5 cm, and the leg is 3 cm. We transform the formula: c²=a²+b² → a²=c²-b². Next, we decide: a²=5²-3²; a²=25-9; a²=16; a=√16; a=4 (cm).

Trigonometric relations to find the leg of a right triangle

It is also possible to find an unknown leg if any other side and any acute angle of a right triangle are known. There are four options for finding the leg using trigonometric functions: by sine, cosine, tangent, cotangent. To solve the problems, the table below will help us. Let's consider these options.

Find the leg of a right triangle using the sine

The sine of an angle (sin) is the ratio of the opposite leg to the hypotenuse. Formula: sin \u003d a / c, where a is the leg opposite the given angle, and c is the hypotenuse. Next, we transform the formula and get: a=sin*c.

Example. The hypotenuse is 10 cm and angle A is 30 degrees. According to the table, we calculate the sine of angle A, it is equal to 1/2. Then, using the transformed formula, we solve: a=sin∠A*c; a=1/2*10; a=5 (cm).

Find the leg of a right triangle using cosine

The cosine of an angle (cos) is the ratio of the adjacent leg to the hypotenuse. Formula: cos \u003d b / c, where b is the leg adjacent to the given angle, and c is the hypotenuse. Let's transform the formula and get: b=cos*c.

Example. Angle A is 60 degrees, the hypotenuse is 10 cm. According to the table, we calculate the cosine of angle A, it is equal to 1/2. Next, we solve: b=cos∠A*c; b=1/2*10, b=5 (cm).

Find the leg of a right triangle using the tangent

The tangent of an angle (tg) is the ratio of the opposite leg to the adjacent one. Formula: tg \u003d a / b, where a is the leg opposite to the corner, and b is adjacent. Let's transform the formula and get: a=tg*b.

Example. Angle A is 45 degrees, the hypotenuse is 10 cm. According to the table, we calculate the tangent of angle A, it is equal to Solve: a=tg∠A*b; a=1*10; a=10 (cm).

Find the leg of a right triangle using the cotangent

The cotangent of an angle (ctg) is the ratio of the adjacent leg to the opposite leg. Formula: ctg \u003d b / a, where b is the leg adjacent to the corner, and is opposite. In other words, the cotangent is the "inverted tangent". We get: b=ctg*a.

Example. Angle A is 30 degrees, the opposite leg is 5 cm. According to the table, the tangent of angle A is √3. Calculate: b=ctg∠A*a; b=√3*5; b=5√3 (cm).

So, now you know how to find the leg in a right triangle. As you can see, it is not so difficult, the main thing is to remember the formulas.

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

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

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

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

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

Rules:

Basic trigonometric identities in a right triangle:

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

As the acute angle increasessinα andtg α increase, andcos α decreases.

For any acute angle α:

sin (90° - α) = cos α

cos (90° - α) = sin α

Explanatory example:

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

Find 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 acute angles is 90º, then angle B \u003d 60º:

B \u003d 90º - 30º \u003d 60º.

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

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

3) Now we 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 into 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.

From this it follows that in a right triangle the sine of one acute angle is equal to the cosine of another 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 check it out 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 on trigonometry, see the Algebra section)

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

Sine of an angle- this is the ratio of the opposite (far) leg to the hypotenuse.

In our triangle:

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

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

In our triangle:

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

Angle tangent- this is the ratio of the opposite (far) leg to the adjacent (close).

In our triangle:

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

Cotangent of an 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 by 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, it is necessary to remember that the sine, cosine, tangent and cotangent as ratios of the sides of a triangle do not depend on the lengths of these sides (at one angle). Do not trust? 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 fix 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 degree and radian, we considered a circle with a radius equal to \ (1 \) . Such a circle is called single. It is very useful in the study of trigonometry. Therefore, we dwell on it in a little more detail.

As you can see, this circle is built in the Cartesian coordinate system. The radius of the circle is equal to one, while the center of the circle lies at the origin, 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 axis \(x \) and the coordinate along the axis \(y \) . What are these coordinate numbers? And in general, what do they have to do with the topic at hand? To do this, remember about the considered right-angled triangle. In the figure above, you can see two whole right triangles. Consider the triangle \(ACG \) . It's rectangular because \(CG \) is perpendicular to the \(x \) axis.

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

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

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

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

So, can you tell me what are the coordinates of the point \(C \) , which belongs to the circle? Well, no way? But 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, the \(y \) coordinate! So the point \(C(x;y)=C(\cos \alpha ;\sin \alpha) \).

What then are \(tg \alpha \) and \(ctg \alpha \) ? That's right, let's use the appropriate 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? Here, for example, as in this picture:

What has changed in this example? Let's figure it out. To do this, we again turn to a right-angled triangle. Consider a right triangle \(((A)_(1))((C)_(1))G \) : an angle (as adjacent to the 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 - the coordinate \ (x \) ; and the values ​​of tangent and cotangent to the corresponding ratios. Thus, these relations are applicable to any rotations 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, you will also get an 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, \(390()^\circ =360()^\circ +30()^\circ \), so the radius vector will make one full rotation and stop at \(30()^\circ \) or \(\dfrac(\pi )(6) \) .

In the second case, \(-1140()^\circ =-360()^\circ \cdot 3-60()^\circ \), that is, the radius vector will make three complete revolutions and stop at the 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 with 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 ​​\u200b\u200bare equal to:

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

Here's a unit circle to help you:

Any 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 measures of the angle. 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, 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(Need to remember or be able to output!! \) !}

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

No need to be scared, now we will show one of the examples of a fairly simple memorization of the corresponding values:

To use this method, it is vital to remember the sine values ​​\u200b\u200bfor all three angle measures ( \(30()^\circ =\dfrac(\pi )(6),\ 45()^\circ =\dfrac(\pi )(4),\ 60()^\circ =\dfrac(\pi )(3) \)), as well as the value of the tangent of the angle in \(30()^\circ \) . Knowing these \(4\) values, it is quite easy 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, it is possible to restore the values ​​for \(\text(tg)\ 45()^\circ , \text(tg)\ 60()^\circ \). The numerator “\(1 \) ” will match \(\text(tg)\ 45()^\circ \ \) , and the denominator “\(\sqrt(\text(3)) \) ” will match \(\text (tg)\ 60()^\circ \ \) . Cotangent values ​​are transferred in accordance with the arrows shown in the figure. If you understand this and remember the scheme with 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 derive a general formula for finding the coordinates of a point. Here, for example, we have such a circle:

We are given that point \(K(((x)_(0));((y)_(0)))=K(3;2) \) is the 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 \(x=((x)_(0))+r\cdot \cos \ \delta =3+1,5\cdot \cos \ \delta \).

By the same logic, we find the value of the y coordinate for the point \(P\) . In this way,

\(y=((y)_(0))+r\cdot \sin \ \delta =2+1,5\cdot \sin \delta \).

So, in general terms, the 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,

\(r\) - circle radius,

\(\delta \) - 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 zero, and the radius is equal to one:

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

We begin our study of trigonometry with a right triangle. Let's define what the sine and cosine are, as well as the tangent and cotangent of an acute angle. These are the basics of trigonometry.

Recall that right angle is an angle equal to 90 degrees. In other words, half of the unfolded corner.

Sharp corner- less than 90 degrees.

Obtuse angle- greater than 90 degrees. In relation to such an angle, "blunt" is not an insult, but a mathematical term :-)

Let's draw a right triangle. A right angle is usually denoted . Note that the side opposite the corner is denoted by the same letter, only small. So, the side lying opposite the angle A is denoted.

An angle is denoted by the corresponding Greek letter.

Hypotenuse A right triangle is the side opposite the right angle.

Legs- sides opposite sharp corners.

The leg opposite the corner is called opposite(relative to angle). The other leg, which lies on one side of the corner, is called adjacent.

Sinus acute angle in a right triangle is the ratio of the opposite leg to the hypotenuse:

Cosine acute angle in a right triangle - the ratio of the adjacent leg to the hypotenuse:

Tangent acute angle in a right triangle - the ratio of the opposite leg to the adjacent:

Another (equivalent) definition: the tangent of an acute angle is the ratio of the sine of an angle to its cosine:

Cotangent acute angle in a right triangle - the ratio of the adjacent leg to the opposite (or, equivalently, the ratio of cosine to sine):

Pay attention to the basic ratios for sine, cosine, tangent and cotangent, which are given below. They will be useful to us in solving problems.

Let's prove some of them.

Okay, we have given definitions and written formulas. But why do we need sine, cosine, tangent and cotangent?

We know that the sum of the angles of any triangle is.

We know the relationship between parties right triangle. This is the Pythagorean theorem: .

It turns out that knowing two angles in a triangle, you can find the third one. Knowing two sides in a right triangle, you can find the third. So, for angles - their ratio, for sides - their own. But what to do if in a right triangle one angle (except for a right one) and one side are known, but you need to find other sides?

This is what people faced in the past, making maps of the area and the starry sky. After all, it is not always possible to directly measure all the sides of a triangle.

Sine, cosine and tangent - they are also called trigonometric functions of the angle- give the ratio between parties and corners triangle. Knowing the angle, you can find all its trigonometric functions using special tables. And knowing the sines, cosines and tangents of the angles of a triangle and one of its sides, you can find the rest.

We will also draw a table of sine, cosine, tangent and cotangent values ​​for "good" angles from to.

Notice the two red dashes in the table. For the corresponding values ​​of the angles, the tangent and cotangent do not exist.

Let's analyze several problems in trigonometry from the Bank of FIPI tasks.

1. In a triangle, the angle is , . Find .

The problem is solved in four seconds.

Because the , .

2. In a triangle, the angle is , , . Find .

Let's find by the Pythagorean theorem.

Problem solved.

Often in problems there are triangles with angles and or with angles and . Memorize the basic ratios for them by heart!

For a triangle with angles and the leg opposite the angle at is equal to half of the hypotenuse.

A triangle with angles and is isosceles. In it, the hypotenuse is times larger than the leg.

We considered problems for solving right triangles - that is, for finding unknown sides or angles. But that's not all! In the variants of the exam in mathematics, there are many tasks where the sine, cosine, tangent or cotangent of the outer angle of the triangle appears. More on this in the next article.