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In a trigonometry course that covers a large number of lessons in 10th grade, the main four are studied trigonometric functions: sine, cosine, tangent and cotangent. Schoolchildren must be able to handle these functions, build their graphs, analyze each of the functions, build graphs of transformed functions, and be able to work with tables trigonometric values etc.
Also, they should be able to handle and reproduce some basic trigonometry formulas and use them when solving practical examples. This was all covered in previous video tutorials. Students can review and refresh the material in their heads.
So, this type This lesson is devoted to the study of formulas for the tangent of the sum and difference of arguments. Previously, we studied formulas for the sine of the sum and difference of arguments, as well as the cosine.
They are demonstrated by the announcer and displayed on the screen, circled in red frames, in order to emphasize the importance of remembering these formulas.
As for the tangent, we know how to write this concept, that is, expressed through sine and cosine. The tangent of the sum of arguments can be written as the sine of the sum of arguments divided by the cosine of the sum of arguments. We have a fraction where the numerator and denominator can be written using formulas studied in advance. We get a ready-made new formula that can be slightly simplified and transformed. The speaker suggests dividing each term of the polynomial by the product of the cosine of one argument and the sine of another. By dividing, some members will contract and the expression will decrease as a whole.
We get a simplified new formula that is worth remembering. If you understand the principle of its receipt, then no problems will arise in further understanding and memorization.
It is further stated that the arguments cannot take values that lie on the asymptotes of the graph of the tangent function. Exceptions are also printed for argument sums. The teacher should definitely consider this point with the class.
In the first example, which is displayed in the video tutorial, it is proposed to calculate some fairly large fractional expression, which contains the sum of tangents in both the denominator and the numerator. Since the tangent arguments are not tabular values, it is recommended to present them as a sum of more convenient degrees. Having done this procedure You can use the learned formula to further solve and obtain the answer.
The second example suggests simplifying an expression that is the sum of two fractions. WITH right side All callouts that are used to solve the problem are given. The announcer explains everything step by step in a calm and clear voice. Not a single moment was missed.
The third example is more complex. Here it is proposed to calculate the tangent of a certain value if some data is known. When solving, they also use previously studied formulas that appear in the callouts on the right side.
The solution is quite long. The answer is finally displayed. After this example, the video discusses another example equation. Since when solving it we use trigonometric table values, it is displayed on the screen for clarity and simplicity. In this way, students can see where certain values are taken from and understand better.
Similar examples can be given to students to complete at home. If they have problems solving it, they can refer to this video and watch it again.
This electronic resource can be used for demonstration at school during a lesson. The teacher will be able to “revive” the lesson with the help of such materials. It will become more memorable and interesting. If students have questions, the teacher or tutor viewing the lesson with them will be able to comment in more detail and explain. Smart students will be able to independently understand the material and master it without additional help.
TEXT DECODING:
Tangent of the sum and difference of arguments
We have already become acquainted with formulas that express the sine and cosine of the sum and difference of arguments. Show formulas
Let's consider how we can express the tangent of the sum and difference of the arguments. Recall that tangent is the ratio of the sine of a number to the cosine of this number
Then we express the tangent of the sum of two angles through the sine and cosine of the sum of two angles, using the formulas sine of the sum and cosine of the sum:
sin(x + y) = sin x cos y + cos x sin y,
cos (x + y) = cos x cos y - sin x sin y.
(after all, if there are tangents of the angles x and y, the product of the cosines of these angles is different from zero), after dividing the numerator and denominator by cos x cos y we get the sum in the numerator and this is equal to tgx and and this is equal to tgy.
We reduce the denominator and get one
as in the numerator and this is equal to tgx and and this is equal to tgy.
Therefore, tan(x+y) =.
(Tangent of the sum of two arguments equal to the sum tangents of these arguments divided by one minus the product of the tangents of these arguments.)
The formula for the tangent of the difference of arguments is proved in a similar way:
tg(x-y) =. (Tangent of the difference between two arguments equal to the difference tangents of these arguments divided by one plus the product of the tangents of these arguments.)
Of course, all tangents make sense, i.e. x+ πn, y + πn,
x + y + πn (for the tangent of the sum of two arguments), x - y + πn (for the tangent of the difference of two arguments).
Let's look at examples.
EXAMPLE 1. Calculate.
Solution. This expression represents the right side of the tan sum formula for the arguments 16° and 44°. Therefore, we reduce the expression to the form of the left side and find that the tangent is equal to 60 0, therefore equal. (Show table of values)
Tg(16°+44°) = tg 60° = .
EXAMPLE 2. Simplify the expression + (the quotient of the sum of the tangents of the arguments x and y by the tangent of the sum of these arguments plus the quotient of the difference of the tangents of the arguments x and y by the tangent of the difference of these arguments).
Solution. In the denominator of the first and second fractions, we apply the formulas for the tangent of the sum and the difference of the arguments, make reductions and get 1 - tgxtgy + 1 + tgxtgy, - tgxtgy and tgxtgy as a result gives zero, then the answer is 2.
1 - tgxtgy + 1 + tgxtgy = 2.
EXAMPLE 3. Calculate tan(+y) (tangent pi times four plus y) if it is known that cozy = , π<у< (игрек больше пи, но меньше трех пи на два).
Solution. Applying the formulas for the tangent of the sum of arguments, we obtain
Let's find tgy (knowing cozy = , π<у<), воспользовавшись формулой. Получим tg 2 у = - 1 подставим значение косинуса в формулу, тогда получим - 1 = .
tg 2 y = - 1 - 1 = .
tan 2 y =. Let's extract the square root tg y = and tg y =
By condition, the argument y (y) belongs to the third quarter, and there the tangent is positive. This means tg y = . Now let’s return to the original formula and substitute the found value:
Answer: = 7.
EXAMPLE 4. Solve the equation = -1 (The difference between the tangents of three x and x, divided by the sum of one and the product of the tangents of three x and x is equal to minus one).
Solution. Note on the left side of the equation the formula for the tangent of the difference between the arguments three x and x. We have
tg(3x-x) =- 1, from which we get 2x, which means
2x = arctan (-1) + πn, (two x equals arctangent minus one plus pi).
Since arctg (-1)= -arctg 1, then tg (-1) = Show table
We substitute the data into the expression and get:
2x =-+ πn, (two x equals minus pi times four plus pi en)
x = -+, (x equals minus pi times eight plus pi divided by two)
Most Frequently Asked Questions
Is it possible to make a stamp on a document according to the sample provided? Answer Yes, it's possible. Send a scanned copy or a good quality photo to our email address, and we will make the necessary duplicate.
What types of payment do you accept?
Answer You can pay for the document upon receipt by the courier, after checking the correctness of completion and quality of execution of the diploma. This can also be done at the office of postal companies offering cash on delivery services.
All terms of delivery and payment for documents are described in the “Payment and Delivery” section. We are also ready to listen to your suggestions regarding the terms of delivery and payment for the document.
Can I be sure that after placing an order you will not disappear with my money? Answer We have quite a long experience in the field of diploma production. We have several websites that are constantly updated. Our specialists work in different parts of the country, producing over 10 documents a day. Over the years, our documents have helped many people solve employment problems or move to higher-paying jobs. We have earned trust and recognition among clients, so there is absolutely no reason for us to do this. Moreover, this is simply impossible to do physically: you pay for your order the moment you receive it in your hands, there is no prepayment.
Can I order a diploma from any university? Answer In general, yes. We have been working in this field for almost 12 years. During this time, an almost complete database of documents issued by almost all universities in the country and for different years of issue was formed. All you need is to select a university, specialty, document, and fill out the order form.
What to do if you find typos and errors in a document?
Answer When receiving a document from our courier or postal company, we recommend that you carefully check all the details. If a typo, error or inaccuracy is discovered, you have the right not to pick up the diploma, but you must indicate the detected defects personally to the courier or in writing by sending an email.
We will correct the document as soon as possible and resend it to the specified address. Of course, shipping will be paid by our company.
To avoid such misunderstandings, before filling out the original form, we email the customer a mock-up of the future document for checking and approval of the final version. Before sending the document by courier or mail, we also take additional photos and videos (including in ultraviolet light) so that you have a clear idea of what you will receive in the end.
What should I do to order a diploma from your company?
Answer To order a document (certificate, diploma, academic certificate, etc.), you must fill out the online order form on our website or provide your email so that we can send you an application form, which you need to fill out and send back to us.
If you do not know what to indicate in any field of the order form/questionnaire, leave them blank. Therefore, we will clarify all the missing information over the phone.
Latest reviews
Alexei:
I needed to acquire a diploma to get a job as a manager. And the most important thing is that I have both experience and skills, but I can’t get a job without a document. Once I came across your site, I finally decided to buy a diploma. The diploma was completed in 2 days!! Now I have a job that I never dreamed of before!! Thank you!
Reference data for tangent (tg x) and cotangent (ctg x). Geometric definition, properties, graphs, formulas. Table of tangents and cotangents, derivatives, integrals, series expansions. Expressions through complex variables. Connection with hyperbolic functions.
Geometric definition
|BD| - length of the arc of a circle with center at point A.
α is the angle expressed in radians.
Tangent ( tan α) is a trigonometric function depending on the angle α between the hypotenuse and the leg of a right triangle, equal to the ratio of the length of the opposite leg |BC| to the length of the adjacent leg |AB| .
Cotangent ( ctg α) is a trigonometric function depending on the angle α between the hypotenuse and the leg of a right triangle, equal to the ratio of the length of the adjacent leg |AB| to the length of the opposite leg |BC| .
Tangent
Where n- whole.
In Western literature, tangent is denoted as follows:
.
;
;
.
Graph of the tangent function, y = tan x
Cotangent
Where n- whole.
In Western literature, cotangent is denoted as follows:
.
The following notations are also accepted:
;
;
.
Graph of the cotangent function, y = ctg x
Properties of tangent and cotangent
Periodicity
Functions y = tg x and y = ctg x are periodic with period π.
Parity
The tangent and cotangent functions are odd.
Areas of definition and values, increasing, decreasing
The tangent and cotangent functions are continuous in their domain of definition (see proof of continuity). The main properties of tangent and cotangent are presented in the table ( n- whole).
| y = tg x | y = ctg x | |
| Scope and continuity | ||
| Range of values | -∞ < y < +∞ | -∞ < y < +∞ |
| Increasing | - | |
| Descending | - | |
| Extremes | - | - |
| Zeros, y = 0 | ||
| Intercept points with the ordinate axis, x = 0 | y = 0 | - |
Formulas
Expressions using sine and cosine
;
;
;
;
;
Formulas for tangent and cotangent from sum and difference
The remaining formulas are easy to obtain, for example
Product of tangents
Formula for the sum and difference of tangents
This table presents the values of tangents and cotangents for certain values of the argument. 
Expressions using complex numbers
Expressions through hyperbolic functions
;
;
Derivatives
; .
.
Derivative of the nth order with respect to the variable x of the function:
.
Deriving formulas for tangent > > > ; for cotangent > > >
Integrals
Series expansions
To obtain the expansion of the tangent in powers of x, you need to take several terms of the expansion in a power series for the functions sin x And cos x and divide these polynomials by each other, . This produces the following formulas.
At .
at .
Where Bn- Bernoulli numbers. They are determined either from the recurrence relation:
;
;
Where .
Or according to Laplace's formula: 
Inverse functions
The inverse functions of tangent and cotangent are arctangent and arccotangent, respectively.
Arctangent, arctg
, Where n- whole.
Arccotangent, arcctg
, Where n- whole.
References:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
G. Korn, Handbook of Mathematics for Scientists and Engineers, 2012.
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