The video lesson “Trigonometric functions of a numerical argument” provides visual material to provide clarity when explaining the topic in class. During the demonstration, the principle of forming the value of trigonometric functions from a number is considered, a number of examples are described that teach how to calculate the values of trigonometric functions from a number. With the help of this manual, it is easier to develop skills in solving relevant problems and to achieve memorization of the material. Using the manual increases the effectiveness of the lesson and helps quickly achieve learning goals.
At the beginning of the lesson, the title of the topic is shown. Then the task is to find the corresponding cosine to some numerical argument. It is noted that this problem can be solved simply and this can be clearly demonstrated. The screen displays a unit circle with its center at the origin. It is noted that the point of intersection of the circle with the positive semi-axis of the abscissa axis is located at point A(1;0). An example of point M is given, which represents the argument t=π/3. This point is marked on the unit circle, and a perpendicular to the abscissa axis descends from it. The found abscissa of the point is the cosine of cos t. In this case, the abscissa of the point will be x=1/2. Therefore cos t=1/2.
Summarizing the facts considered, it is noted that it makes sense to talk about the function s=cos t. It is noted that students already have some knowledge about this function. Some cosine values are calculated: cos 0=1, cos π/2=0, cos π/3=1/2. Also related to this function are the functions s=sin t, s=tg t, s=ctg t. It is noted that they have a common name for all - trigonometric functions.
Important relationships that are used in solving problems with trigonometric functions are demonstrated: the main identity sin 2 t+ cos 2 t=1, the expression of tangent and cotangent through sine and cosine tg t=sin t/cos t, where t≠π/2+πk for kϵZ, ctg t= cos t/sin t, where t≠πk for kϵZ, as well as the ratio of tangent to cotangent tg t·ctg t=1 where t≠πk/2 for kϵZ.
Next, we propose to consider the proof of the relation 1+ tg 2 t=1/ cos 2 t, with t≠π/2+πk for kϵZ. To prove the identity, it is necessary to represent tg 2 t in the form of a ratio of sine and cosine, and then bring the terms on the left side to a common denominator 1+ tg 2 t=1+sin 2 t/cos 2 t = (sin 2 t+cos 2 t )/ cos 2 t. Using the basic trigonometric identity, we obtain 1 in the numerator, that is, the final expression 1/ cos 2 t. Q.E.D.
The identity 1+ cot 2 t=1/ sin 2 t is proved in a similar way, for t≠πk for kϵZ. Just as in the previous proof, the cotangent is replaced by the corresponding ratio of cosine and sine, and both terms on the left side are reduced to a common denominator 1+ cot 2 t=1+ cos 2 t/sin 2 t= (sin 2 t+cos 2 t)/sin 2 t. After applying the basic trigonometric identity to the numerator we get 1/ sin 2 t. This is the expression we are looking for.
The solution of examples in which the acquired knowledge is applied is considered. In the first task, you need to find the values of cost, tgt, ctgt, if the sine of the number sint=4/5 is known, and t belongs to the interval π/2< t<π. Для нахождения косинуса в данном примере рекомендуется использовать тождество sin 2 t+ cos 2 t=1, из которого следует cos 2 t=1-sin 2 t. Зная значение синуса, можно найти косинус cos 2 t=1-(4/5) 2 =9/25. То есть значение косинуса cost=3/5 и cost=-3/5. В условии указано, что аргумент принадлежит второй четверти координатной плоскости. В этой четверти значение косинуса отрицательное. С учетом данного ограничения находим cost=-3/5. Для нахождения тангенса числа пользуемся его определением tgt= sint/cost. Подставив известные значения синуса и косинуса, получаем tgt=4/5:(-3/5)=-4/3. Чтобы найти значение котангенса, также используется определение котангенса ctgt= cost/sint. Подставив известные значения синуса и косинуса в отношение, получаем ctgt=(-3/5):4/5=-3/4.
Next, we consider the solution to a similar problem in which the tangent tgt = -8/15 is known, and the argument is limited to the values 3π/2 To find the value of the sine, we use the definition of tangent tgt= sint/cost. From it we find sint= tgt·cost=(-8/15)·(15/17)=-8/17. Knowing that cotangent is the inverse function of tangent, we find ctgt=1/(-8/15)=-15/8. The video lesson “Trigonometric functions of a numerical argument” is used to increase the effectiveness of a mathematics lesson at school. During distance learning, this material can be used as a visual aid for developing skills in solving problems that involve trigonometric functions of a number. To acquire these skills, the student may be advised to independently examine visual material. TEXT DECODING: The topic of the lesson is “Trigonometric functions of a numerical argument.” Any real number t can be associated with a uniquely defined number cos t. To do this you need to do the following: 1) position the number circle on the coordinate plane so that the center of the circle coincides with the origin of coordinates, and the starting point A of the circle falls at the point (1;0); 2) find a point on the circle that corresponds to the number t; 3) find the abscissa of this point. This is cos t. Therefore, we will talk about the function s = cos t (es equals cosine te), where t is any real number. We already got some idea of this function: All these functions are called trigonometric functions of the numerical argument t. From the definitions of sine, cosine, tangent and cotangent, some relationships follow: 1) sin 2 t + cos 2 t = 1 (sine square te plus cosine square te equals one) 2)tgt = for t ≠ + πk, kϵZ (tangent te is equal to the ratio of sine te to cosine te with te not equal to pi by two plus pi ka, ka belongs to zet) 3) ctgt = for t ≠ πk, kϵZ (cotangent te is equal to the ratio of cosine te to sine te when te is not equal to pi ka, ka belongs to zet). 4) tgt ∙ ctgt = 1 for t ≠ , kϵZ (the product of tangent te by cotangent te is equal to one when te is not equal to peak ka, divided by two, ka belongs to zet) Let us prove two more important formulas: One plus tangent squared te is equal to the ratio of one to cosine squared te when te is not equal to pi by two plus pi ka. Proof. Let us reduce the expression one plus tangent squared te to the common denominator cosine squared te. We get in the numerator the sum of the squares of the cosine te and sine te, which is equal to one. And the denominator remains the square of the cosine te. The sum of unity and the square of the cotangent te is equal to the ratio of unity to the square of the sine te when te is not equal to pi ka. Proof. The expression one plus cotangent squared te, similarly, we bring to a common denominator and apply the first relation. Let's look at examples. EXAMPLE 1. Find cost, tgt, ctgt if sint = and< t < π.(если синус тэ равен четырем пятым и тэ из промежутка от пи на два до пи) Solution. From the first relation we find the cosine squared te is equal to one minus sine squared te: cos 2 t = 1 - sin 2 t. This means that cos 2 t = 1 -() 2 = (cosine square te is equal to nine twenty-fifths), that is, cost = (cosine te is equal to three fifths) or cost = - (cosine te is equal to minus three fifths). By condition, the argument t belongs to the second quarter, and in it cos t< 0 (косинус тэ отрицательный). This means that the cosine te is equal to minus three-fifths, cost = - . Let's calculate tangent te: tgt = = ׃ (-)= - ;(tangent te is equal to the ratio of sine te to cosine te, and therefore four-fifths to minus three-fifths and equal to minus four-thirds) Accordingly, we calculate (the cotangent of the number te. since the cotangent te is equal to the ratio of the cosine of te to the sine of te,) ctgt = = - . (cotangent te is equal to minus three-fourths). Answer: cost = - , tgt= - ; ctgt = - . (we fill in the answer as we solve it) EXAMPLE 2. It is known that tgt = - and< t < 2π(тангенс тэ равен минус восемь пятнадцатых и тэ принадлежит промежутку от трех пи на два до двух пи). Найти значения cost, sint, ctgt. Solution. Let’s use this relationship and substitute the value into this formula to obtain: 1 + (-) 2 = (one per cosine square te is equal to the sum of one and the square minus eight fifteenths). From here we find cos 2 t = (cosine square te is equal to two hundred twenty-five two hundred eighty-ninths). This means cost = (cosine te is fifteen seventeenths) or cost = . By condition, the argument t belongs to the fourth quarter, where cost>0. Therefore cost = .(cosenus te is equal to fifteen seventeenths) Let's find the value of the argument sine te. Since from the relation (show the relation tgt = for t ≠ + πk, kϵZ) sine te is equal to the product of tangent te and cosine te, then substituting the value of the argument te..tangent te is equal to minus eight fifteenths.. by condition, and cosine te is equal to solved earlier, we get sint = tgt ∙ cost = (-) ∙ = - , (sine te is equal to minus eight seventeenths) ctgt = = - . (since cotangent te is the reciprocal of tangent, which means cotangent te is equal to minus fifteen eighteenths) Additional materials Teaching aids and simulators in the Integral online store for grade 10
What we will study: Let's remember the basic formulas: $tg(t)=\frac(sin(t))(cos(t))$, with $t≠\frac(π)(2)+πk$. Let's derive new formulas. Now let's divide both sides of the identity by $sin^2(t)$. We managed to obtain two new formulas. Remember them. $cos(t) =\frac(5)(7)$, find $sin(t)$; $tg(t)$; $ctg(t)$ for all t. Solution: $sin^2(t)+cos^2(t)=1$. Example 2. $tg(t) = \frac(5)(12)$, find $sin(t)$; $cos(t)$; $ctg(t)$, for all $0 Solution: Lesson objectives: Educational: Educational: Educational: Lesson type: lesson of generalization and systematization of knowledge. Teaching methods: partially search, (heuristic). Test checking the level of knowledge, solving cognitive generalization problems, self-test, system generalizations. Lesson plan. Lesson progress 1. Organizational moment. Homework: Paragraph 1, clause 1.4 The French writer Anatole France once remarked: “You can only learn through fun. To digest knowledge, you need to absorb it with appetite.” Let's follow this advice from the writer today in class, let's be active, attentive, and absorb knowledge with great desire. After all, they will be useful to you in the future. Today we have the final lesson on the topic: “Trigonometric functions of a numerical argument.” We repeat and generalize the material studied, methods and techniques for solving trigonometric expressions. 2. Self-check test. The work is carried out in two versions. Questions on the screen. Students mark incorrect steps. The number of correct answers is recorded on the knowledge sheet. 3. Message. Report on the history of the development of trigonometry (speaking by a trained student). 4. Systematization of theoretical material. Oral tasks. 1) What are we talking about? What's special? Determine the sign of the expression: a) cos (700°) tg 380°, 2) What does this block of formulas say? What's wrong? 3) Consider the table: Trigonometric transformations 4) Solving problems of each type of trigonometric transformations. Finding the meanings of trigonometric expressions. Finding the value of a trigonometric function from a known value of a given trigonometric function. Given: sin = ;< < Find cos2, ctg2. Answer: .< < 2 Find: cos2 , tg2 Third difficulty level: Given: sin = ;< < Find: sin2 ; sin(60° - ); tg (45° + ) Additional task. Prove the identity: 4 sin 4 - 4 sin 2 = cos 2 2 - 1 6. The result of independent work. Students check their work and record the results on their knowledge sheet. 7. The lesson is summarized. Trigonometric functions of a numeric argument.
Trigonometric functions of numeric argumentt are functions of the form y= cos t, Using these formulas, through the known value of one trigonometric function, you can find the unknown values of other trigonometric functions. Explanations. 1) Take the formula cos 2 t + sin 2 t = 1 and use it to derive a new formula. To do this, divide both sides of the formula by cos 2 t (for t ≠ 0, that is, t ≠ π/2 + π k). So: cos 2 t sin 2 t 1 The first term is equal to 1. We know that the ratio of sine to conis is tangent, which means the second term is equal to tg 2 t. As a result, we get a new (and already known to you) formula: 2) Now divide cos 2 t + sin 2 t = 1 by sin 2 t (for t ≠ π k): cos 2 t sin 2 t 1 The ratio of cosine to sine is the cotangent. Means: sin 2 t 1 sin 2 t cos 2 t + sin 2 t 1 In the same way, you can easily find the sum of one and the square of the cotangent, as well as many other identities. Trigonometric functions of angular argument.
In functionsat =
cost,
at =
sint,
at =
tgt,
at =
ctgt variablet can be more than just a numeric argument. It can also be considered a measure of the angle - that is, the angular argument. Using the number circle and coordinate system, you can easily find the sine, cosine, tangent, and cotangent of any angle. To do this, two important conditions must be met: 2) one of the sides of the angle must be a positive axis beam x. In this case, the ordinate of the point at which the circle and the second side of the angle intersect is the sine of this angle, and the abscissa of this point is the cosine of this angle. Explanation. Let's draw an angle, one side of which is the positive ray of the axis x, and the second side comes out from the origin of the coordinate axis (and from the center of the circle) at an angle of 30º (see figure). Then the point of intersection of the second side with the circle corresponds to π/6. We know the ordinate and abscissa of this point. They are also the cosine and sine of our angle: √3 1 And knowing the sine and cosine of an angle, you can easily find its tangent and cotangent. Thus, the number circle, located in a coordinate system, is a convenient way to find the sine, cosine, tangent, or cotangent of an angle. But there is an easier way. You don’t have to draw a circle and a coordinate system. You can use simple and convenient formulas: Example: find the sine and cosine of an angle equal to 60º. Solution : π 60 π √3 π 1 Explanation: we found out that the sine and cosine of an angle of 60º correspond to the values of a point on a circle π/3. Next, we simply find the values of this point in the table - and thus solve our example. The table of sines and cosines of the main points of the number circle is in the previous section and on the “Tables” page.
Lesson and presentation on the topic: "Trigonometric function of a numerical argument, definition, identities"
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Algebraic problems with parameters, grades 9–11
Software environment "1C: Mathematical Constructor 6.1"
1. Definition of a numeric argument.
2. Basic formulas.
3. Trigonometric identities.
4. Examples and tasks for independent solution.Definition of a trigonometric function of a numeric argument
Guys, we know what sine, cosine, tangent and cotangent are.
Let's see if it is possible to find the values of other trigonometric functions using the values of some trigonometric functions?
Let us define the trigonometric function of a numerical element as: $y= sin(t)$, $y= cos(t)$, $y= tg(t)$, $y= ctg(t)$.
$sin^2(t)+cos^2(t)=1$. By the way, what is the name of this formula?
$ctg(t)=\frac(cos(t))(sin(t))$, for $t≠πk$.Trigonometric identities
We know the basic trigonometric identity: $sin^2(t)+cos^2(t)=1$.
Guys, let's divide both sides of the identity by $cos^2(t)$.
We get: $\frac(sin^2(t))(cos^2(t))+\frac(cos^2(t))(cos^2(t))=\frac(1)(cos^2 (t))$.
Let's transform: $(\frac(sin(t))(cos(t)))^2+1=\frac(1)(cos^2(t)).$
We get the identity: $tg^2(t)+1=\frac(1)(cos^2(t))$, with $t≠\frac(π)(2)+πk$.
We get: $\frac(sin^2(t))(sin^2(t))+\frac(cos^2(t))(sin^2(t))=\frac(1)(sin^2 (t))$.
Let's transform: $1+(\frac(cos(t))(sin(t)))^2=\frac(1)(sin^2(t)).$
We get a new identity that is worth remembering:
$ctg^2(t)+1=\frac(1)(sin^2(t))$, for $t≠πk$.
These formulas are used if, from some known value of a trigonometric function, it is necessary to calculate the value of another function.Solving examples on trigonometric functions of a numerical argument
Example 1.
Then $sin^2(t)=1-cos^2(t)$.
$sin^2(t)=1-(\frac(5)(7))^2=1-\frac(25)(49)=\frac(49-25)(49)=\frac(24) (49)$.
$sin(t)=±\frac(\sqrt(24))(7)=±\frac(2\sqrt(6))(7)$.
$tg(t)=±\sqrt(\frac(1)(cos^2(t))-1)=±\sqrt(\frac(1)(\frac(25)(49))-1)= ±\sqrt(\frac(49)(25)-1)=±\sqrt(\frac(24)(25))=±\frac(\sqrt(24))(5)$.
$ctg(t)=±\sqrt(\frac(1)(sin^2(t))-1)=±\sqrt(\frac(1)(\frac(24)(49))-1)= ±\sqrt(\frac(49)(24)-1)=±\sqrt(\frac(25)(24))=±\frac(5)(\sqrt(24))$.
$tg^2(t)+1=\frac(1)(cos^2(t))$.
Then $\frac(1)(cos^2(t))=1+\frac(25)(144)=\frac(169)(144)$.
We get that $cos^2(t)=\frac(144)(169)$.
Then $cos^2(t)=±\frac(12)(13)$, but $0
We get: $sin(t)=tg(t)*cos(t)=\frac(5)(12)*\frac(12)(13)=\frac(5)(13)$.
$ctg(t)=\frac(1)(tg(t))=\frac(12)(5)$.Problems to solve independently
1. $tg(t) = -\frac(3)(4)$, find $sin(t)$; $cos(t)$; $ctg(t)$, for all $\frac(π)(2)
4. $cos(t) = \frac(12)(13)$, find $sin(t)$; $tg(t)$; $ctg(t)$ for all $t$.
- Test work (tasks were posted on the stand).№
1 option
Option 2
1
Define sine and cosine of an acute angle
Define tangent and cotangent of an acute angle
2
What numerical functions are called tangent and cotangent? Give a definition.
What numerical functions are called sine and cosine? Give a definition.
3
A point on the unit circle has coordinates . Find the values of sin, cos.
The unit circle point has coordinates (- 0.8; - 0.6). Find the value of tg, ctg.
4
Which of the basic trigonometric functions are odd? Write down the corresponding equalities.
Which of the basic trigonometric functions are even? Write down the corresponding equalities.
5
How do the values of sine and cosine change when the angle changes by an integer number of revolutions? Write down the corresponding equalities.
How do the values of tangent and cotangent change when the angle changes by an integer number of revolutions? What's special? Write down the corresponding equalities.
6
Find the values of sin cos, sin(- 630°), cos (- 630°).
Find the values of tg, ctg, tg 540°, ctg(-450°).
7
Which figure shows the graph of the function y = sin x?
Which figure shows the graph of the function y = tg x?
8
Write down the reduction formulas for angles ( - ), ( - ).
Write down the reduction formulas for angles (+), (+).
9
Write addition formulas.
Write the basic trigonometric identities.
10
Write formulas for reducing the degree.
Write double argument formulas.
b) cos (- 1) sin (- 2)
Finding the meaning of trigonometric expressions
Finding the value of a trigonometric function from a known value of a given trigonometric function
Simplifying trigonometric expressions
Identities
y= sin t, y= tg t, y= ctg t.
--- + --- = ---
cos 2 t cos 2 t cos 2 t
--- + --- = ---, where t ≠ π k + π k, k– integer
sin 2 t sin 2 t sin 2 t
Knowing the basic principles of mathematics and having learned the basic formulas of trigonometry, you can easily derive most of the other trigonometric identities on your own. And this is even better than simply memorizing them: what is learned by heart is quickly forgotten, but what is understood is remembered for a long time, if not forever. For example, it is not necessary to memorize what the sum of one and the square of the tangent is equal to. If you forgot, you can easily remember if you know the simplest thing: tangent is the ratio of sine to cosine. In addition, apply the simple rule of adding fractions with different denominators and get the result:
1 + tg 2 t = 1 + --- = - + --- = ------ = ---
cos 2 t 1 cos 2 t cos 2 t cos 2 t
1) the vertex of the angle must be the center of the circle, which is also the center of the coordinate axis;
--; --
2 2
sin 60º = sin --- = sin -- = --
180 3 2
cos 60º = cos -- = -
3 2
