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 the right angle (in our example this is the side \(AC\)); legs are the two remaining sides \(AB\) and \(BC\) (those adjacent to right angle), and, if we consider the legs relative to the angle \(BC\), then the leg \(AB\) is the adjacent leg, and the leg \(BC\) is the 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, this circle is constructed in Cartesian system coordinates The radius of the circle is equal to one, and the center of the circle lies at the origin, starting position 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 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, you will also get an angle of a certain value, 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 in 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 of 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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We will begin our study of trigonometry with the right triangle. Let's define what sine and cosine are, as well as tangent and cotangent of an acute angle. This is the basics of trigonometry.
Let us remind you that right angle is an angle equal to 90 degrees. In other words, half a turned angle.
Sharp corner- less than 90 degrees.
Obtuse angle- greater than 90 degrees. In relation to such an angle, “obtuse” is not an insult, but a mathematical term :-)
Let's draw a right triangle. A right angle is usually denoted by . Please note that the side opposite the corner is indicated by the same letter, only small. Thus, the side opposite angle A is designated .
The angle is indicated by the corresponding Greek letter.
Hypotenuse of a right triangle is the side opposite the right angle.
Legs- sides lying opposite acute angles.
The leg lying opposite the angle is called opposite(relative to angle). The other leg, which lies on one of the sides of the angle, is called adjacent.
Sinus acute angle in right triangle- this is an attitude opposite leg to the hypotenuse:
Cosine acute angle in a right triangle - ratio adjacent leg to the hypotenuse:
Tangent acute angle in a right triangle - the ratio of the opposite side to the adjacent:
Another (equivalent) definition: the tangent of an acute angle is the ratio of the sine of the angle to its cosine:
Cotangent acute angle in a right triangle - the ratio of the adjacent side to the opposite (or, which is the same, the ratio of cosine to sine):
Note the basic relationships for sine, cosine, tangent, and cotangent below. They will be useful to us when solving problems.
Let's prove some of them.
Okay, we have given definitions and written down formulas. But why do we still need sine, cosine, tangent and cotangent?
We know that the sum of the angles of any triangle is equal to.
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. Knowing the two sides of a right triangle, you can find the third. This means that the angles have their own ratio, and the sides have their own. But what should you do if in a right triangle you know one angle (except the right angle) and one side, but you need to find the other sides?
This is what people in the past encountered when making maps of the area and the starry sky. After all, it is not always possible to directly measure all sides of a triangle.
Sine, cosine and tangent - they are also called trigonometric angle functions- give relationships 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 the values of sine, cosine, tangent and cotangent for “good” angles from to.
Please note the two red dashes in the table. At appropriate angle values, tangent and cotangent do not exist.
Let's look at several trigonometry problems from the FIPI Task Bank.
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 it using the Pythagorean theorem.
The problem is solved.
Often in problems there are triangles with angles and or with angles and. Remember 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 looked at problems solving right triangles - that is, finding unknown sides or angles. But that's not all! IN Unified State Exam options in mathematics there are many problems where the sine, cosine, tangent or cotangent of the external angle of a triangle appears. More on this in the next article.
Lecture: Sine, cosine, tangent, cotangent of an arbitrary angle
Sine, cosine of an arbitrary angle
To understand what trigonometric functions are, let's look at a circle with unit radius. Given circle has center at the origin at coordinate plane. For determining specified functions we will use the radius vector OR, which starts at the center of the circle, and the point R is a point on the circle. This radius vector forms an angle alpha with the axis OH. Since a circle has a radius, equal to one, That OR = R = 1.
If from the point R lower the perpendicular to the axis OH, then we get a right triangle with a hypotenuse equal to one.
If the radius vector moves clockwise, then this direction called negative, if it moves counterclockwise - positive.
Sine of the angle OR, is the ordinate of the point R vector on a circle.
That is, to obtain the sine value given angle alpha it is necessary to determine the coordinate U on surface.
How given value was received? Since we know that the sine of an arbitrary angle in a right triangle is the ratio of the opposite leg to the hypotenuse, we get that
And since R=1, That sin(α) = y 0 .
In a unit circle, the ordinate value cannot be less than -1 and greater than 1, which means
Sinus accepts positive value in the first and second quarters of the unit circle, and in the third and fourth - negative.
Cosine of the angle given circle formed by the radius vector OR, is the abscissa of the point R vector on a circle.
That is, to obtain the cosine value of a given angle alpha, it is necessary to determine the coordinate X on surface.
The cosine of an arbitrary angle in a right triangle is the ratio of the adjacent leg to the hypotenuse, we get that
And since R=1, That cos(α) = x 0 .
In the unit circle, the abscissa value cannot be less than -1 and greater than 1, which means
The cosine takes a positive value in the first and fourth quarters of the unit circle, and negative in the second and third.
Tangentarbitrary angle The ratio of sine to cosine is calculated.
If we consider a right triangle, then this is the ratio of the opposite side to the adjacent side. If we're talking about about the unit circle, then this is the ratio of the ordinate to the abscissa.
Judging by these relationships, it can be understood that the tangent cannot exist if the abscissa value is zero, that is, at an angle of 90 degrees. The tangent can take all other values.
The tangent is positive in the first and third quarters of the unit circle, and negative in the second and fourth.
Sine is one of the basic trigonometric functions, the use of which is not limited to geometry alone. Tables for calculating trigonometric functions, like engineering calculators, are not always at hand, and calculating the sine is sometimes necessary to solve various tasks. In general, calculating the sine will help consolidate drawing skills and knowledge of trigonometric identities.
Games with ruler and pencil
A simple task: how to find the sine of an angle drawn on paper? To solve, you will need a regular ruler, a triangle (or compass) and a pencil. The simplest way to calculate the sine of an angle is by dividing the far leg of a triangle with a right angle by the long side - the hypotenuse. Thus, you first need to complete the acute angle to the shape of a right triangle by drawing a line perpendicular to one of the rays at an arbitrary distance from the vertex of the angle. We will need to maintain an angle of exactly 90°, for which we need a clerical triangle.
Using a compass is a little more accurate, but will take more time. On one of the rays you need to mark 2 points at a certain distance, adjust the radius on the compass, approximately equal to distance between points, and draw semicircles with centers at these points until the intersections of these lines are obtained. By connecting the intersection points of our circles with each other, we get a strict perpendicular to the ray of our angle; all that remains is to extend the line until it intersects with another ray.
In the resulting triangle, you need to use a ruler to measure the side opposite the corner and the long side on one of the rays. The ratio of the first dimension to the second will be the desired value of the sine of the acute angle.
Find the sine for an angle greater than 90°
For obtuse angle the task is not much more difficult. You need to draw a ray from the vertex to the opposite side using a ruler to form a straight line with one of the rays of the angle we are interested in. With the received acute angle should proceed as described above, sinuses adjacent corners, forming together a reverse angle of 180°, are equal.
Calculating sine using other trigonometric functions
Also, calculating the sine is possible if the values of other trigonometric functions of the angle or at least the lengths of the sides of the triangle are known. Trigonometric identities will help us with this. Let's look at common examples.
How to find the sine with a known cosine of an angle? The first trigonometric identity, based on the Pythagorean theorem, states that the sum of the squares of the sine and cosine of the same angle is equal to one.
How to find the sine with a known tangent of an angle? The tangent is obtained by dividing the far side by the near side or dividing the sine by the cosine. Thus, the sine will be the product of the cosine and the tangent, and the square of the sine will be the square of this product. We replace the squared cosine with the difference between one and the square sine according to the first trigonometric identity and through simple manipulations we reduce the equation to the calculation of the square sine through the tangent; accordingly, to calculate the sine, you will have to extract the root of the result obtained.
How to find the sine with a known cotangent of an angle? The value of the cotangent can be calculated by dividing the length of the leg closest to the angle by the length of the far one, and also by dividing the cosine by the sine, that is, the cotangent is a function, reciprocal of tangent relative to the number 1. To calculate the sine, you can calculate the tangent using the formula tg α = 1 / ctg α and use the formula in the second option. You can also derive a direct formula by analogy with the tangent, which will look like in the following way.
How to find the sine of three sides of a triangle
There is a formula for finding the length unknown side any triangle, not just right-angled, in two known parties using the trigonometric function of the cosine of the opposite angle. She looks like this.
I think you deserve more than this. Here is my key to trigonometry:
- Draw the dome, wall and ceiling
- Trigonometric functions are nothing more than percentage these three forms.
Metaphor for sine and cosine: dome
Instead of just looking at the triangles themselves, imagine them in action by finding some special example from life.
Imagine you are in the middle of a dome and want to hang a movie projector screen. You point your finger at the dome at a certain angle “x”, and the screen should be suspended from this point.
The angle you point to determines:
- sine(x) = sin(x) = screen height (from 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 large as possible? Hang it directly 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 in this position and will hang furthest away, as you asked.
Height and distance from the screen are inversely proportional: the closer the screen hangs, the greater its height.
Sine and cosine are percentages
No one during my years of study, alas, explained to me that the trigonometric functions sine and cosine are nothing more than 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 fleeced.
Absolute height doesn't mean anything. But if the sine value is 0.95, then I understand that the TV is hanging almost on the top of your dome. Very soon he will reach maximum height in the center of the dome, and then begins to decline again.
How can we calculate this percentage? It's 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 side / hypotenuse.” It's all about getting interest! It is best to define sine as “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 toward the 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 unit, enlarged or reduced in scale to the right size. So determine the unit circle connections and apply the results to your specific circle size.
Experiment: take any corner and see what percentage of height to width it displays:
The graph of the growth of the sine value is not just a straight line. The first 45 degrees cover 70% of the height, but the last 10 degrees (from 80° to 90°) cover only 2%.
This will make it clearer to you: if you walk 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 next to each other to your dome. Cried your view from the window and a good price for resale!
But is it possible to somehow win in this situation?
Of course yes. What if we hung a movie screen right on our neighbor's wall? You target the angle (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 is not moving anywhere from you, right?)
- secant(x) = sec(x) = “length of the ladder” from you standing in the center of the dome to the top of the suspended screen
Let's clarify a couple of points regarding 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 create 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 a larger version of sine! And while the increase in 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, from you to the wall) and starts to rise from there
- The secant is always longer than the tangent. The slanted ladder you use to hang your screen should be longer than the screen itself, right? (With 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 an angle of 50 degrees, tan(50)=1.19. Your screen is 19% larger than the distance to the wall (dome radius).
(Enter x=0 and check your intuition - tan(0) = 0 and sec(0) = 1.)
Cotangent and cosecant. Ceiling
Incredibly, your neighbor has now decided to build a roof over your dome. (What's wrong with him? Apparently he doesn't want you to spy on him while he's walking around the yard naked...)
Well, it's time to build an exit to the roof and talk to your neighbor. You choose the angle of inclination and begin construction:
- the vertical distance between the roof outlet and the floor is always 1 (the radius of the dome)
- cotangent(x) = cot(x) = distance between the top of the dome and the exit point
- cosecant(x) = csc(x) = length of your path to the roof
Tangent and secant describe the wall, and COtangent and COsecant describe the ceiling.
Our intuitive conclusions this time are similar to the previous ones:
- If you take the angle equal to 0°, your exit to the roof will last forever, since it will never reach the ceiling. Problem.
- The shortest “ladder” 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-ceiling combination, the result will be the following:
Well, it’s still the same triangle, increased in size to reach the wall and the ceiling. We have vertical sides (sine, tangent), horizontal sides (cosine, cotangent) and “hypotenuses” (secant, cosecant). (By the arrows you can see where 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, the “height to width” ratios should also be the same for all triangles. (Just step back from the very big triangle to less. Yes, the size has changed, but the aspect ratios will remain the same).
Knowing which side in each triangle is equal to 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 clearly see these dependencies and understand where they come from. This technique is much better than memorization dry formulas.
Don't forget about other angles
Psst... Don't get stuck 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 relative sizes may be different.
(You may have noticed that the sine and cosine ratios are always the smallest because they are contained within the dome).
To summarize: what do we need to remember?
For most of us, I'd say 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 trigonometric functions result in percentages, which we apply to our script.
You don't need to memorize formulas like 1 2 + cot 2 = csc 2 . They are only suitable for stupid tests, in which knowledge of a fact is passed off as understanding it. Take a minute to draw a semicircle in the form of a dome, a wall and a roof, label the elements, and all the formulas will come to 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 an angle of 30 degrees takes up 50% of the maximum height.
The inverse trigonometric function is written as sin -1 or arcsin. It is also often written asin in various languages programming.
If our height is 25% of the dome's height, what is our angle?
In our table of proportions you can find a ratio where the secant is divided by 1. For example, the secant by 1 (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 radius of a unit circle. What angle of inclination to the wall does this value correspond to?
Appendix: Some examples
Example: Find the sine of angle x.
A 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's nothing wrong with that. The triangle also has a height, it is indicated in green in the figure.
What is the hypotenuse equal to? According to the Pythagorean theorem, we know that:
3 2 + 4 2 = hypotenuse 2 25 = hypotenuse 2 5 = hypotenuse
Fine! Sine is the percentage of the height of the triangle's longest side, or hypotenuse. In our example, the sine is 3/5 or 0.60.
Of course, we can go several ways. Now we know that the sine is 0.60, we can simply find the arcsine:
Asin(0.6)=36.9
Here's another approach. Note that the triangle is “facing the wall,” so we can use the tangent instead of the sine. The height is 3, the distance to the wall is 4, so the tangent is ¾ or 75%. We can use the arctangent to go from a percentage value back to an angle:
Tan = 3/4 = 0.75 atan(0.75) = 36.9 Example: Will you swim to the shore?
You are in a boat and you have enough fuel to travel 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 problem statement: we only have a table of arc cosine values.
What we have? coastline can be represented as a “wall” in our famous triangle, and the “length of a ladder” attached to the wall is the maximum possible distance to be covered by boat to the shore (2 km). A secant appears.
First, you need to go to percentages. We have 2 / 0.25 = 8, that is, we can swim a distance that is 8 times the straight distance to the shore (or to the wall).
The question arises: “What is the secant of 8?” But we cannot answer it, since we only have arc cosines.
We use our previously derived dependencies to relate the secant to the cosine: “sec/1 = 1/cos”
Sekans 8 equal to cosine⅛. An angle whose cosine is ⅛ is equal to acos(1/8) = 82.8. And this is the largest angle we can afford on a boat with the specified amount of fuel.
Not bad, right? Without the dome-wall-ceiling analogy, I would have gotten lost in a bunch of formulas and calculations. Visualizing the problem greatly simplifies the search for a solution, and it is also interesting to see which trigonometric function will ultimately help.
For each problem, think like this: Am I interested in the dome (sin/cos), wall (tan/sec), or ceiling (cot/csc)?
And trigonometry will become much more enjoyable. Easy calculations for you!