EntranceTrigonometric functions of a numeric argument. Properties and graphs of trigonometric functions
We looked at the most basic trigonometric functions(don’t be fooled, in addition to sine, cosine, tangent and cotangent, there are a whole host of other functions, but more on them later), but for now let’s look at some basic properties of the functions already studied.
Trigonometric functions of numeric argument
Whatever real number t is taken, it can be associated with a uniquely defined number sin(t) . True, the matching rule is quite complex and consists of the following.
To find the value of sin(t) from the number t, you need:
- place the number circle on coordinate plane so that the center of the circle coincides with the origin of coordinates, and starting point And the circle hits the point (1; 0);
- find a point on the circle corresponding to the number t;
- find the ordinate of this point.
- this ordinate is the desired sin(t) .
In fact, we are talking about the function s = sin(t) , where t is any real number. We can calculate some values of this function (for example, sin(0) = 0, \(sin \frac (\pi)(6) = \frac(1)(2) \) etc.), we know some of its properties.
In the same way, we can consider that we have already received some ideas about three more functions: s = cos(t) s = tan(t) s = ctg(t) All these functions are called trigonometric functions of the numerical argument t.
Relationship between trigonometric functions
As you, I hope, can guess, all trigonometric functions are interconnected and even without knowing the meaning of one, it can be found through another.
For example, the most important formula in all trigonometry is basic trigonometric identity :
\[ sin^(2) t + cos^(2) t = 1 \]
As you can see, knowing the value of the sine, you can find the value of the cosine, and also vice versa. Also very common formulas connecting sine and cosine with tangent and cotangent:
\[ \boxed (\tan\; t=\frac(\sin\; t)(\cos\; t), \qquad t \neq \frac(\pi)(2)+ \pi k) \]
\[ \boxed (\cot\; t=\frac(\cos\; )(\sin\; ), \qquad t \neq \pi k) \]
From the last two formulas one can derive another trigometric identity, this time connecting tangent and cotangent:
\[ \boxed (\tan \; t \cdot \cot \; t = 1, \qquad t \neq \frac(\pi k)(2)) \]
Now let's see how these formulas work in practice.
EXAMPLE 1. Simplify the expression: a) \(1+ \tan^2 \; t \), b) \(1+ \cot^2 \; t \)
a) First of all, let’s write the tangent, keeping the square:
\[ 1+ \tan^2 \; t = 1 + \frac(\sin^2 \; t)(\cos^2 \; t) \]
\[ 1 + \frac(\sin^2 \; t)(\cos^2 \; t)= \sin^2\; t + \cos^2\; t + \frac(\sin^2 \; t)(\cos^2 \; t) \]
Now let’s put everything under a common denominator, and we get:
\[ \sin^2\; t + \cos^2\; t + \frac(\sin^2 \; t)(\cos^2 \; t) = \frac(\cos^2 \; t + \sin^2 \; t)(\cos^2 \; t ) \]
And finally, as we see, the numerator can be reduced to one by the main trigonometric identity, as a result we get: \[ 1+ \tan^2 \; = \frac(1)(\cos^2 \; t) \]
b) With the cotangent we perform all the same actions, only the denominator will no longer be a cosine, but a sine, and the answer will be like this:
\[ 1+ \cot^2 \; = \frac(1)(\sin^2 \; t) \]
Having completed this task, we derived two more very important formulas that connect our functions, which we also need to know like the back of our hands:
\[ \boxed (1+ \tan^2 \; = \frac(1)(\cos^2 \; t), \qquad t \neq \frac(\pi)(2)+ \pi k) \]
\[ \boxed (1+ \cot^2 \; = \frac(1)(\sin^2 \; t), \qquad t \neq \pi k) \]
You must know all the formulas presented by heart, otherwise further study of trigonometry without them is simply impossible. In the future there will be more formulas and there will be a lot of them and I assure you that you will definitely remember all of them for a long time, or maybe you won’t remember them, but EVERYONE should know these six things!
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Whatever real number t is taken, it can be associated with a uniquely defined number sin t. True, the matching rule is quite complex; as we saw above, it is as follows.
To find the value of sin t using the number t, you need:
1) position the number circle in 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 point (1; 0);
2) find a point on the circle corresponding to the number t;
3) find the ordinate of this point.
This ordinate is sin t.
In fact, we are talking about the function u = sin t, where t is any real number.
All these functions are called trigonometric functions of the numerical argument t.
There are a number of relations that connect the values of various trigonometric functions; we have already obtained some of these relations:
sin 2 t+cos 2 t = 1
From the last two formulas it is easy to obtain a relationship connecting tg t and ctg t:
All of these formulas are used in cases where, knowing the value of a trigonometric function, it is necessary to calculate the values of other trigonometric functions.
The terms “sine”, “cosine”, “tangent” and “cotangent” were actually familiar, however, they were still used in a slightly different interpretation: in geometry and physics they considered sine, cosine, tangent and cotangent at the head(but not
numbers, as was in the previous paragraphs).
From geometry it is known that the sine (cosine) of an acute angle is the ratio of the leg right triangle to its hypotenuse, and the tangent (cotangent) of the angle is the ratio of the legs of a right triangle. A different approach to the concepts of sine, cosine, tangent and cotangent was developed in the previous paragraphs. In fact, these approaches are interrelated.
Let's take an angle with degree measure b o and place it in the “numeric circle in a rectangular coordinate system” model as shown in Fig. 14
the apex of the angle is compatible with the center
circles (with the origin of the coordinate system),
and one side of the angle is compatible with
the positive ray of the x-axis. Full stop
intersection of the second side of the angle with
denote by the circle the letter M. Ordina-
Fig. 14 b o, and the abscissa of this point is the cosine of the angle b o.
To find the sine or cosine of an angle b o it is not at all necessary to do these very complex constructions every time.
It is enough to note that the arc AM is the same part of the length number circle, what angle b o is from the corner of 360°. If the length of the arc AM is denoted by the letter t, we get:
Thus,
For example,
It is believed that 30° is a degree measure of an angle, and a radian measure of the same angle: 30° = rad. At all:
In particular, I’m glad where, in turn, we get it from.
So what is 1 radian? There are various measures of length of segments: centimeters, meters, yards, etc. There are also various measures to indicate the magnitude of angles. We are looking at central angles unit circle. An angle of 1° is the central angle subtended by an arc that is part of a circle. An angle of 1 radian is the central angle subtended by an arc of length 1, i.e. on an arc whose length is equal to the radius of the circle. From the formula, we find that 1 rad = 57.3°.
When considering the function u = sin t (or any other trigonometric function), we can consider the independent variable t to be a numerical argument, as was the case in the previous paragraphs, but we can also consider this variable to be a measure of the angle, i.e. corner argument. Therefore, when talking about a trigonometric function, in a certain sense it makes no difference to consider it a function of a numerical or angular argument.
Lesson and presentation on the topic: "Trigonometric function of a numerical argument, definition, identities"
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What we will study:
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's define the trigonometric function numeric element, as: $y= sin(t)$, $y= cos(t)$, $y= tg(t)$, $y= ctg(t)$.
Let's remember the basic formulas:
$sin^2(t)+cos^2(t)=1$. By the way, what is the name of this formula?
$tg(t)=\frac(sin(t))(cos(t))$, with $t≠\frac(π)(2)+πk$.
$ctg(t)=\frac(cos(t))(sin(t))$, for $t≠πk$.
Let's derive new formulas.
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 $t≠\frac(π)(2)+πk$.
Now let's divide both sides of the identity by $sin^2(t)$.
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 $t≠πk$.
We managed to obtain two new formulas. Remember them.
These formulas are used if for some reason known value A trigonometric function needs to calculate the value of another function.
Solving examples on trigonometric functions of a numerical argument
Example 1.$cos(t) =\frac(5)(7)$, find $sin(t)$; $tg(t)$; $ctg(t)$ for all t.
Solution:
$sin^2(t)+cos^2(t)=1$.
Then $sin^2(t)=1-cos^2(t)$.
$sin^2(t)=1-(\frac(5)(7))^2=1-\frac(25)(49)=\frac(49-25)(49)=\frac(24) (49)$.
$sin(t)=±\frac(\sqrt(24))(7)=±\frac(2\sqrt(6))(7)$.
$tg(t)=±\sqrt(\frac(1)(cos^2(t))-1)=±\sqrt(\frac(1)(\frac(25)(49))-1)= ±\sqrt(\frac(49)(25)-1)=±\sqrt(\frac(24)(25))=±\frac(\sqrt(24))(5)$.
$ctg(t)=±\sqrt(\frac(1)(sin^2(t))-1)=±\sqrt(\frac(1)(\frac(24)(49))-1)= ±\sqrt(\frac(49)(24)-1)=±\sqrt(\frac(25)(24))=±\frac(5)(\sqrt(24))$.
Example 2.
$tg(t) = \frac(5)(12)$, find $sin(t)$; $cos(t)$; $ctg(t)$, for all $0 Solution: 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.
$tg^2(t)+1=\frac(1)(cos^2(t))$.
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 $t$.
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 just memorizing them: what you learn by heart is quickly forgotten, but what you understand 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