Right triangle: sine, cosine, tangent, cotangent of an angle. Right triangle. Complete illustrated guide (2019)

The ratio of the opposite leg to the hypotenuse is called sinus acute angle right triangle.

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

Cosine of an acute angle of a right triangle

The ratio of the nearest 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 leg to the adjacent leg is called acute angle tangent right triangle.

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

Cotangent of an acute angle of a right triangle

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

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

Sine of an arbitrary angle

The ordinate of the 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 .

tg \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 on 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 the point M is -\frac(\sqrt(2))(2), the abscissa is \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 encountered angles are given in the table:

Trigonometric identities are equalities that establish a relationship between the 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 reverse order.

Finding tangent and cotangent through 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, then by definition, the ordinate of y is the sine, and the abscissa of x is the 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.

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

For example: tg \alpha = \frac(\sin \alpha)(\cos \alpha) is valid for \alpha angles 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 points above, we get that tg \alpha = \frac(y)(x), a ctg\alpha=\frac(x)(y). Hence it follows that tg \alpha \cdot ctg \alpha = \frac(y)(x) \cdot \frac(x)(y)=1. Thus, the tangent and cotangent of one angle at which they make sense are mutually reciprocal 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 , equals the inverse square of the sine of the given angle. This identity is valid for any \alpha other than \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

Decision

The functions \sin \alpha and \cos \alpha are linked 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).

To find tg \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

Decision

Substituting into the formula \sin^(2)\alpha + \cos^(2) \alpha = 1 conditional 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).

In this article, we will take a comprehensive look at . Main trigonometric identities are equalities that establish a relationship between the sine, cosine, tangent and cotangent of one angle, and allow you to find any of these trigonometric functions through a known other.

We immediately list the main trigonometric identities, which we will analyze in this article. We write them down in a table, and below we give the derivation of these formulas and give the necessary explanations.

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Relationship between sine and cosine of one angle

Sometimes they talk not about the main trigonometric identities listed in the table above, but about one single basic trigonometric identity kind . The explanation for this fact is quite simple: the equalities are obtained from the basic trigonometric identity after dividing both of its parts by and respectively, and the equalities and follow from the definitions of sine, cosine, tangent, and cotangent. We will discuss this in more detail in the following paragraphs.

That is, it is the equality that is of particular interest, which was given the name of the main trigonometric identity.

Before proving the basic trigonometric identity, we give its formulation: the sum of the squares of the sine and cosine of one angle is identically equal to one. Now let's prove it.

The basic trigonometric identity is very often used in transformation of trigonometric expressions. It allows the sum of the squares of the sine and cosine of one angle to be replaced by one. No less often, the basic trigonometric identity is used in reverse order: the unit is replaced by the sum of the squares of the sine and cosine of any angle.

Tangent and cotangent through sine and cosine

Identities connecting the tangent and cotangent with the sine and cosine of one angle of the form and immediately follow from the definitions of sine, cosine, tangent and cotangent. Indeed, by definition, the sine is the ordinate of y, the cosine is the abscissa of x, the tangent is the ratio of the ordinate to the abscissa, that is, , and the cotangent is the ratio of the abscissa to the ordinate, that is, .

Due to this obviousness of the identities and often the definitions of tangent and cotangent are given not through the ratio of the abscissa and the ordinate, but through the ratio of the sine and cosine. So the tangent of an angle is the ratio of the sine to the cosine of this angle, and the cotangent is the ratio of the cosine to the sine.

To conclude this section, it should be noted that the identities and hold for all such angles for which the trigonometric functions in them make sense. So the formula is valid for any other than (otherwise the denominator will be zero, and we did not define division by zero), and the formula - for all , different from , where z is any .

Relationship between tangent and cotangent

An even more obvious trigonometric identity than the two previous ones is the identity connecting the tangent and cotangent of one angle of the form . It is clear that it takes place for any angles other than , otherwise either the tangent or the cotangent is not defined.

Proof of the formula very simple. By definition and from where . The proof could have been carried out in a slightly different way. Since and , then .

So, the tangent and cotangent of one angle, at which they make sense, is.

I think you deserve more than that. Here is my key to trigonometry:

• Draw the dome, wall and ceiling
• Trigonometric functions are nothing but percentages of these three forms.

Metaphor for sine and cosine: dome

Instead of just looking at the triangles themselves, imagine them in action by finding some particular example from life.

Imagine that you are in the middle of a dome and want to hang up a movie projector screen. You point your finger at the dome at some "x" angle, and a screen should be hung from that point.

The angle you point to determines:

• sine(x) = sin(x) = screen height (floor to dome mounting point)
• cosine(x) = cos(x) = distance from you to the screen (by floor)
• hypotenuse, the distance from you to the top of the screen, always the same, equal to the radius of the dome

Do you want the screen to be as big as possible? Hang it right above you.

Do you want the screen to hang as far away from you as possible? Hang it straight perpendicular. The screen will have zero height at this position and will hang as far back as you requested.

The height and distance from the screen are inversely proportional: the closer the screen hangs, the higher its height will be.

Sine and cosine are percentages

No one in my years of study, alas, explained to me that the trigonometric functions sine and cosine are nothing but percentages. Their values ​​range from +100% to 0 to -100%, or from a positive maximum to zero to a negative maximum.

Let's say I paid a tax of 14 rubles. You don't know how much it is. But if you say that I paid 95% in tax, you will understand that I was simply skinned like a sticky.

Absolute height means nothing. But if the sine value is 0.95, then I understand that the TV is hanging almost on top of your dome. Very soon it will reach its maximum height in the center of the dome, and then begin to decline again.

How can we calculate this percentage? Very simple: divide the current screen height by the maximum possible (the radius of the dome, also called the hypotenuse).

That's why we are told that “cosine = opposite leg / hypotenuse”. This is all in order to get a percentage! The best way to define the sine is “the percentage of the current height from the maximum possible”. (The sine becomes negative if your angle points "underground". The cosine becomes negative if the angle points to a dome point behind you.)

Let's simplify the calculations by assuming we are at the center of the unit circle (radius = 1). We can skip the division and just take the sine equal to the height.

Each circle is essentially a single circle, scaled up or down to right size. So determine the relationships on the unit circle and apply the results to your particular circle size.

Experiment: take any corner and see what percentage of height to width it displays:

The graph of the growth of the value of the sine is not just a straight line. The first 45 degrees cover 70% of the height, and the last 10 degrees (from 80° to 90°) cover only 2%.

This will make it clear to you: if you go in a circle, at 0 ° you rise almost vertically, but as you approach the top of the dome, the height changes less and less.

Tangent and secant. Wall

One day a neighbor built a wall right back to back to your dome. Wept your view from the window and good price for resale!

But is it possible to somehow win in this situation?

Of course yes. What if we hang a movie screen right on the neighbor's wall? You aim at the corner (x) and get:

• tan(x) = tan(x) = screen height on the wall
• distance from you to the wall: 1 (this is the radius of your dome, the wall doesn't move anywhere from you, right?)
• secant(x) = sec(x) = “length of ladder” from you standing in the center of the dome to the top of the suspended screen

Let's clarify a couple of things about the tangent, or screen height.

• it starts at 0, and can go infinitely high. You can stretch the screen higher and higher on the wall to get just an endless canvas for watching your favorite movie! (For such a huge one, of course, you will have to spend a lot of money).
• tangent is just an enlarged version of sine! And while the growth of the sine slows down as you move towards the top of the dome, the tangent continues to grow!

Sekansu also has something to brag about:

• the secant starts at 1 (the ladder is on the floor, away from you towards the wall) and starts going up from there
• The secant is always longer than the tangent. The sloped ladder you hang your screen with needs to be longer than the screen itself, right? (For unrealistic sizes, when the screen is sooooo long and the ladder needs to be placed almost vertically, their sizes are almost the same. But even then the secant will be a little longer).

Remember the values ​​are percent. If you decide to hang the screen at a 50 degree angle, tan(50)=1.19. Your screen is 19% larger than the distance to the wall (dome radius).

(Enter x=0 and test your intuition - tan(0) = 0 and sec(0) = 1.)

Cotangent and cosecant. Ceiling

Incredibly, your neighbor has now decided to build a ceiling over your dome. (What's the matter with him? He apparently doesn't want you to peep on him while he walks around the yard naked...)

Well, it's time to build an exit to the roof and talk to the neighbor. You choose the angle of inclination, and start building:

• the vertical distance between the roof outlet and the floor is always 1 (radius of the dome)
• cotangent(x) = cot(x) = distance between dome top and exit point
• cosecant(x) = csc(x) = length of your path to the roof

The tangent and secant describe the wall, while the cotangent and cosecant describe the floor.

Our intuitive conclusions this time are similar to the previous ones:

• If you take an angle of 0°, your exit to the roof will take forever as it will never reach the ceiling. Problem.
• The shortest “stairway” to the roof will be obtained if you build it at an angle of 90 degrees to the floor. The cotangent will be equal to 0 (we do not move along the roof at all, we exit strictly perpendicularly), and the cosecant will be equal to 1 (“the length of the ladder” will be minimal).

Visualize Connections

If all three cases are drawn in a dome-wall-floor combination, the following will be obtained:

Well, wow, it's all the same triangle, enlarged in size to reach the wall and the ceiling. We have vertical sides (sine, tangent), horizontal sides (cosine, cotangent), and “hypotenuses” (secant, cosecant). (You can see from the arrows how far each element reaches. The cosecant is the total distance from you to the roof).

A little magic. All triangles share the same equalities:

From the Pythagorean theorem (a 2 + b 2 = c 2) we see how the sides of each triangle are connected. In addition, height-to-width ratios must also be the same for all triangles. (Just step back from the largest triangle to the smaller one. Yes, the size has changed, but the proportions of the sides will remain the same).

Knowing which side in each triangle is 1 (the radius of the dome), we can easily calculate that "sin/cos = tan/1".

I have always tried to remember these facts through simple visualization. In the picture you can clearly see these dependencies and understand where they come from. This technique is much better than memorizing dry formulas.

Don't Forget Other Angles

Shh… No need to get hung up on one graph, thinking that the tangent is always less than 1. If you increase the angle, you can reach the ceiling without reaching the wall:

Pythagorean connections always work, but the relative sizes can be different.

(You've probably noticed that the ratio of sine and cosine is always the smallest because they are enclosed within a dome.)

To summarize: what do we need to remember?

For most of us, I would say that this will be enough:

• trigonometry explains the anatomy of mathematical objects such as circles and repeating intervals
• the dome/wall/roof analogy shows the relationship between different trigonometric functions
• the result of the trigonometric functions are the percentages that we apply to our scenario.

You don't need to memorize formulas like 1 2 + cot 2 = csc 2 . They are suitable only for stupid tests in which knowledge of a fact is presented as understanding it. Take a minute to draw a semicircle in the form of a dome, a wall and a roof, sign the elements, and all the formulas will be asked for you on paper.

Application: Inverse Functions

Any trigonometric function takes an angle as input and returns the result as a percentage. sin(30) = 0.5. This means that a 30 degree angle takes up 50% of the maximum height.

The inverse trigonometric function is written as sin -1 or arcsin (“arxine”). It is also often written asin in various programming languages.

If our height is 25% of the dome's height, what is our angle?

In our table of proportions, you can find the ratio where the secant is divided by 1. For example, the secant by 1 (the hypotenuse to the horizontal) will be equal to 1 divided by the cosine:

Let's say our secant is 3.5, i.e. 350% of the unit circle radius. What angle of inclination to the wall does this value correspond to?

Appendix: Some examples

Boring task. Let's complicate the banal “find the sine” to “What is the height as a percentage of the maximum (hypotenuse)?”.

First, notice that the triangle is rotated. There is nothing wrong with this. The triangle also has a height, it is shown in green in the figure.

What is the hypotenuse equal to? By the Pythagorean theorem, we know that:

3 2 + 4 2 = hypotenuse 2 25 = hypotenuse 2 5 = hypotenuse

Well! The sine is the percentage of the height from the longest side of the triangle, or the hypotenuse. In our example, the sine is 3/5 or 0.60.

Of course, we can go in several ways. Now we know that the sine is 0.60 and we can simply find the arcsine:

Asin(0.6)=36.9

And here is another approach. Note that the triangle is "face to face with the wall", so we can use tangent instead of sine. The height is 3, the distance to the wall is 4, so the tangent is ¾ or 75%. We can use the arc tangent to go from percentage back to angle:

Tan = 3/4 = 0.75 atan(0.75) = 36.9 Example: Will you swim to shore?

You are in a boat and you have enough fuel to sail 2 km. You are now 0.25 km from the coast. At what maximum angle to the shore can you swim to it so that you have enough fuel? Addition to the condition of the problem: we only have a table of arc cosine values.

What we have? coastline can be imagined as a “wall” in our famous triangle, and the “length of a ladder” attached to the wall is the maximum possible surmountable distance by boat to the shore (2 km). A secant emerges.

First, you need to switch to percentages. We have 2 / 0.25 = 8, meaning we can swim 8 times the straight distance to the shore (or wall).

The question arises “What is the secant 8?”. But we cannot give an answer to it, since we only have arc cosines.

We use our previously derived dependencies to map the secant to the cosine: “sec/1 = 1/cos”

The secant of 8 is equal to the cosine of ⅛. An angle whose cosine is ⅛ is acos(1/8) = 82.8. And this is the largest angle that we can afford on a boat with the specified amount of fuel.

Not bad, right? Without the dome-wall-ceiling analogy, I would be confused in a bunch of formulas and calculations. Visualizing the problem greatly simplifies the search for a solution, besides, it is interesting to see which trigonometric function will eventually help.

For each task, think like this: Am I interested in a dome (sin/cos), a wall (tan/sec), or a ceiling (cot/csc)?

And trigonometry will become much more pleasant. Easy calculations for you!

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.

Decision .

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)