The ratios between the main trigonometric functions - sine, cosine, tangent and cotangent - are given trigonometric formulas. And since there are quite a lot of connections between trigonometric functions, this also explains the abundance of trigonometric formulas. Some formulas connect the trigonometric functions of the same angle, others - the functions of a multiple angle, others - allow you to lower the degree, the fourth - to express all functions through the tangent of a half angle, etc.
In this article, we will list in order all the main trigonometric formulas, which are sufficient to solve the vast majority of trigonometry problems. For ease of memorization and use, we will group them according to their purpose, and enter them into tables.
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Basic trigonometric identities
Basic trigonometric identities set the relationship between the sine, cosine, tangent and cotangent of one angle. They follow from the definition of sine, cosine, tangent and cotangent, as well as the concept of the unit circle. They allow one to express trigonometric function through any other.
For a detailed description of these trigonometry formulas, their derivation and application examples, see the article.
Cast formulas
Cast formulas follow from the properties of sine, cosine, tangent and cotangent, that is, they reflect the property of periodicity of trigonometric functions, the property of symmetry, and also the property of shift by a given angle. These trigonometric formulas allow you to move from working with arbitrary angles to working with angles ranging from zero to 90 degrees.
The rationale for these formulas, a mnemonic rule for memorizing them, and examples of their application can be studied in the article.
Addition Formulas
Trigonometric addition formulas show how the trigonometric functions of the sum or difference of two angles are expressed in terms of the trigonometric functions of these angles. These formulas serve as the basis for the derivation of the following trigonometric formulas.
Formulas for double, triple, etc. corner
Formulas for double, triple, etc. angle (they are also called multiple angle formulas) show how the trigonometric functions of double, triple, etc. angles () are expressed in terms of trigonometric functions of a single angle. Their derivation is based on addition formulas.
More detailed information is collected in the article formulas for double, triple, etc. angle .
Half Angle Formulas
Half Angle Formulas show how the trigonometric functions of a half angle are expressed in terms of the cosine of an integer angle. These trigonometric formulas follow from the double angle formulas.
Their conclusion and examples of application can be found in the article.
Reduction Formulas
Trigonometric formulas for decreasing degrees designed to facilitate the transition from natural degrees trigonometric functions to sines and cosines to the first degree, but multiple angles. In other words, they allow one to reduce the powers of trigonometric functions to the first.
Formulas for the sum and difference of trigonometric functions
main destination sum and difference formulas for trigonometric functions is to pass to the product of functions, which is very useful when simplifying trigonometric expressions. These formulas are also widely used in solving trigonometric equations, since they allow factoring the sum and difference of sines and cosines.
Formulas for the product of sines, cosines and sine by cosine
The transition from the product of trigonometric functions to the sum or difference is carried out through the formulas for the product of sines, cosines and sine by cosine.
Universal trigonometric substitution
We complete the review of the basic formulas of trigonometry with formulas expressing trigonometric functions in terms of the tangent of a half angle. This replacement is called universal trigonometric substitution. Its convenience lies in the fact that all trigonometric functions are expressed in terms of the tangent of a half angle rationally without roots.
Bibliography.
- Algebra: Proc. for 9 cells. avg. school / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova; Ed. S. A. Telyakovsky.- M.: Enlightenment, 1990.- 272 p.: Ill.- ISBN 5-09-002727-7
- Bashmakov M.I. Algebra and the beginning of analysis: Proc. for 10-11 cells. avg. school - 3rd ed. - M.: Enlightenment, 1993. - 351 p.: ill. - ISBN 5-09-004617-4.
- Algebra and the beginning of the analysis: Proc. for 10-11 cells. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorova.- 14th ed.- M.: Enlightenment, 2004.- 384 p.: ill.- ISBN 5-09-013651-3.
- Gusev V. A., Mordkovich A. G. Mathematics (a manual for applicants to technical schools): Proc. allowance.- M.; Higher school, 1984.-351 p., ill.
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Examples:
\(2\sin(x) = \sqrt(3)\)
tg\((3x)=-\) \(\frac(1)(\sqrt(3))\)
\(4\cos^2x+4\sinx-1=0\)
\(\cos4x+3\cos2x=1\)
How to solve trigonometric equations:
Any trigonometric equation should be reduced to one of the following types:
\(\sint=a\), \(\cost=a\), tg\(t=a\), ctg\(t=a\)
where \(t\) is an expression with x, \(a\) is a number. Such trigonometric equations are called protozoa. They are easy to solve using () or special formulas:
See infographics on solving simple trigonometric equations here: , and .
Example . Solve the trigonometric equation \(\sinx=-\)\(\frac(1)(2)\).Solution:
Answer: \(\left[ \begin(gathered)x=-\frac(π)(6)+2πk, \\ x=-\frac(5π)(6)+2πn, \end(gathered)\right.\) \(k,n∈Z\)
What does each symbol mean in the formula for the roots of trigonometric equations, see.
Attention! The equations \(\sinx=a\) and \(\cosx=a\) have no solutions if \(a ϵ (-∞;-1)∪(1;∞)\). Because the sine and cosine for any x is greater than or equal to \(-1\) and less than or equal to \(1\):
\(-1≤\sin x≤1\) \(-1≤\cosx≤1\)
Example
. Solve the equation \(\cosx=-1,1\).
Solution:
\(-1,1<-1\), а значение косинуса не может быть меньше \(-1\). Значит у уравнения нет решения.
Answer
: no solutions.
Example . Solve the trigonometric equation tg\(x=1\).
Solution:
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Solve the equation using a number circle. For this: |
Example
. Solve the trigonometric equation \(\cos(3x+\frac(π)(4))=0\).
Solution:
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Let's use the number circle again. \(3x+\)\(\frac(π)(4)\) \(=±\)\(\frac(π)(2)\) \(+2πk\), \(k∈Z\) \(3x+\)\(\frac(π)(4)\) \(=\)\(\frac(π)(2)\) \(+2πk\) \(3x+\)\(\frac( π)(4)\) \(=-\)\(\frac(π)(2)\) \(+2πk\) 8) As usual, we will express \(x\) in equations. \(3x=-\)\(\frac(π)(4)\) \(+\)\(\frac(π)(2)\) \(+2πk\) \(3x=-\)\ (\frac(π)(4)\) \(+\)\(\frac(π)(2)\) \(+2πk\) |
Reducing trigonometric equations to the simplest ones is a creative task, here you need to use both, and special methods for solving equations:
- Method (the most popular in the exam).
- Method.
- Method of auxiliary arguments.
Consider an example of solving a square-trigonometric equation
Example . Solve the trigonometric equation \(2\cos^2x-5\cosx+2=0\)Solution:
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\(2\cos^2x-5\cosx+2=0\) |
Let's make the change \(t=\cosx\). |
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Our equation has become typical. You can solve it with . |
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\(D=25-4 \cdot 2 \cdot 2=25-16=9\) |
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\(t_1=\)\(\frac(5-3)(4)\) \(=\)\(\frac(1)(2)\) ; \(t_2=\)\(\frac(5+3)(4)\) \(=2\) |
We make a replacement. |
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\(\cosx=\)\(\frac(1)(2)\); \(\cosx=2\) |
We solve the first equation using a number circle. |
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Let us write down all the numbers lying on at these points. |
An example of solving a trigonometric equation with the study of ODZ:
Example (USE) . Solve the trigonometric equation \(=0\)|
\(\frac(2\cos^2x-\sin(2x))(ctg x)\)\(=0\) |
There is a fraction and there is a cotangent - so you need to write down. Let me remind you that the cotangent is actually a fraction: ctg\(x=\)\(\frac(\cosx)(\sinx)\) Therefore, the DPV for ctg\(x\): \(\sinx≠0\). |
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ODZ: ctg\(x ≠0\); \(\sinx≠0\)
\(x≠±\)\(\frac(π)(2)\) \(+2πk\); \(x≠πn\); \(k,n∈Z\) |
Note the "non-solutions" on the number circle. |
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\(\frac(2\cos^2x-\sin(2x))(ctg x)\)\(=0\) |
Let's get rid of the denominator in the equation by multiplying it by ctg\(x\). We can do this because we wrote above that ctg\(x ≠0\). |
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\(2\cos^2x-\sin(2x)=0\) |
Apply the double angle formula for the sine: \(\sin(2x)=2\sinx\cosx\). |
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\(2\cos^2x-2\sinx\cosx=0\) |
If your hands reached out to divide by cosine - pull them back! You can divide by an expression with a variable if it is definitely not equal to zero (for example, such: \(x^2+1,5^x\)). Instead, we take \(\cosx\) out of brackets. |
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\(\cosx (2\cosx-2\sinx)=0\) |
Let's split the equation into two. |
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\(\cosx=0\); \(2\cosx-2\sinx=0\) |
We solve the first equation with using a number circle. Divide the second equation by \(2\) and move \(\sinx\) to the right side. |
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\(x=±\)\(\frac(π)(2)\) \(+2πk\), \(k∈Z\). \(\cosx=\sinx\) |
The roots that turned out are not included in the ODZ. Therefore, we will not write them down in response. |
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Again we use a circle. |
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These roots are not excluded by the ODZ, so they can be written as a response. |
Trigonometric equations are not the easiest topic. Painfully they are diverse.) For example, these:
sin2x + cos3x = ctg5x
sin(5x+π /4) = ctg(2x-π /3)
sinx + cos2x + tg3x = ctg4x
Etc...
But these (and all other) trigonometric monsters have two common and obligatory features. First - you won't believe it - there are trigonometric functions in the equations.) Second: all expressions with x are within these same functions. And only there! If x appears somewhere outside, for example, sin2x + 3x = 3, this will be a mixed type equation. Such equations require an individual approach. Here we will not consider them.
We will not solve evil equations in this lesson either.) Here we will deal with the simplest trigonometric equations. Why? Yes, because the decision any trigonometric equations consists of two stages. At the first stage, the evil equation is reduced to a simple one by various transformations. On the second - this simplest equation is solved. No other way.
So, if you have problems in the second stage, the first stage does not make much sense.)
What do elementary trigonometric equations look like?
sinx = a
cosx = a
tgx = a
ctgx = a
Here but stands for any number. Any.
By the way, inside the function there may be not a pure x, but some kind of expression, such as:
cos(3x+π /3) = 1/2
etc. This complicates life, but does not affect the method of solving the trigonometric equation.
How to solve trigonometric equations?
Trigonometric equations can be solved in two ways. The first way: using logic and a trigonometric circle. We will explore this path here. The second way - using memory and formulas - will be considered in the next lesson.
The first way is clear, reliable, and hard to forget.) It is good for solving trigonometric equations, inequalities, and all sorts of tricky non-standard examples. Logic is stronger than memory!
We solve equations using a trigonometric circle.
We include elementary logic and the ability to use a trigonometric circle. Can't you!? However... It will be difficult for you in trigonometry...) But it doesn't matter. Take a look at the lessons "Trigonometric circle ...... What is it?" and "Counting angles on a trigonometric circle." Everything is simple there. Unlike textbooks...)
Ah, you know!? And even mastered "Practical work with a trigonometric circle"!? Accept congratulations. This topic will be close and understandable to you.) What is especially pleasing is that the trigonometric circle does not care which equation you solve. Sine, cosine, tangent, cotangent - everything is the same for him. The solution principle is the same.
So we take any elementary trigonometric equation. At least this:
cosx = 0.5
I need to find X. Speaking in human language, you need find the angle (x) whose cosine is 0.5.
How did we use the circle before? We drew a corner on it. In degrees or radians. And immediately seen trigonometric functions of this angle. Now let's do the opposite. Draw a cosine equal to 0.5 on the circle and immediately we'll see injection. It remains only to write down the answer.) Yes, yes!
We draw a circle and mark the cosine equal to 0.5. On the cosine axis, of course. Like this:
Now let's draw the angle that this cosine gives us. Hover your mouse over the picture (or touch the picture on a tablet), and see this same corner X.
Which angle has a cosine of 0.5?
x \u003d π / 3
cos 60°= cos( π /3) = 0,5
Some people will grunt skeptically, yes... They say, was it worth it to fence the circle, when everything is clear anyway... You can, of course, grunt...) But the fact is that this is an erroneous answer. Or rather, inadequate. Connoisseurs of the circle understand that there are still a whole bunch of angles that also give a cosine equal to 0.5.
If you turn the movable side OA for a full turn, point A will return to its original position. With the same cosine equal to 0.5. Those. the angle will change 360° or 2π radians, and cosine is not. The new angle 60° + 360° = 420° will also be a solution to our equation, because
There are an infinite number of such full rotations... And all these new angles will be solutions to our trigonometric equation. And they all need to be written down somehow. Everything. Otherwise, the decision is not considered, yes ...)
Mathematics can do this simply and elegantly. In one short answer, write down infinite set solutions. Here's what it looks like for our equation:
x = π /3 + 2π n, n ∈ Z
I will decipher. Still write meaningfully nicer than stupidly drawing some mysterious letters, right?)
π /3 is the same angle that we saw on the circle and determined according to the table of cosines.
2π is one full turn in radians.
n - this is the number of complete, i.e. whole revolutions. It is clear that n can be 0, ±1, ±2, ±3.... and so on. As indicated by the short entry:
n ∈ Z
n belongs ( ∈ ) to the set of integers ( Z ). By the way, instead of the letter n letters can be used k, m, t etc.
This notation means that you can take any integer n . At least -3, at least 0, at least +55. What do you want. If you plug that number into your answer, you get a specific angle, which is sure to be the solution to our harsh equation.)
Or, in other words, x \u003d π / 3 is the only root of an infinite set. To get all the other roots, it is enough to add any number of full turns to π / 3 ( n ) in radians. Those. 2πn radian.
Everything? No. I specifically stretch the pleasure. To remember better.) We received only a part of the answers to our equation. I will write this first part of the solution as follows:
x 1 = π /3 + 2π n, n ∈ Z
x 1 - not one root, it is a whole series of roots, written in short form.
But there are other angles that also give a cosine equal to 0.5!
Let's return to our picture, according to which we wrote down the answer. Here she is:
Move the mouse over the image and see another corner that also gives a cosine of 0.5. What do you think it equals? The triangles are the same... Yes! It is equal to the angle X , only plotted in the negative direction. This is the corner -X. But we have already calculated x. π /3 or 60°. Therefore, we can safely write:
x 2 \u003d - π / 3
And, of course, we add all the angles that are obtained through full turns:
x 2 = - π /3 + 2π n, n ∈ Z
That's all now.) In a trigonometric circle, we saw(who understands, of course)) all angles that give a cosine equal to 0.5. And they wrote down these angles in a short mathematical form. The answer is two infinite series of roots:
x 1 = π /3 + 2π n, n ∈ Z
x 2 = - π /3 + 2π n, n ∈ Z
This is the correct answer.
Hope, general principle for solving trigonometric equations with the help of a circle is understandable. We mark the cosine (sine, tangent, cotangent) from the given equation on the circle, draw the corresponding angles and write down the answer. Of course, you need to figure out what kind of corners we are saw on the circle. Sometimes it's not so obvious. Well, as I said, logic is required here.)
For example, let's analyze another trigonometric equation:
Please note that the number 0.5 is not the only possible number in the equations!) It's just more convenient for me to write it than roots and fractions.
We work according to the general principle. We draw a circle, mark (on the sine axis, of course!) 0.5. We draw at once all the angles corresponding to this sine. We get this picture:
Let's deal with the angle first. X in the first quarter. We recall the table of sines and determine the value of this angle. The matter is simple:
x \u003d π / 6
We recall full turns and, with a clear conscience, write down the first series of answers:
x 1 = π /6 + 2π n, n ∈ Z
Half the job is done. Now we need to define second corner... This is trickier than in cosines, yes ... But logic will save us! How to determine the second angle through x? Yes Easy! The triangles in the picture are the same, and the red corner X equal to the angle X . Only it is counted from the angle π in the negative direction. That's why it's red.) And for our answer, we need an angle measured correctly from the positive semiaxis OX, i.e. from an angle of 0 degrees.
Hover the cursor over the picture and see everything. I removed the first corner so as not to complicate the picture. The angle of interest to us (drawn in green) will be equal to:
π - x
x we know it π /6 . So the second angle will be:
π - π /6 = 5π /6
Again, we recall the addition of full revolutions and write down the second series of answers:
x 2 = 5π /6 + 2π n, n ∈ Z
That's all. A complete answer consists of two series of roots:
x 1 = π /6 + 2π n, n ∈ Z
x 2 = 5π /6 + 2π n, n ∈ Z
Equations with tangent and cotangent can be easily solved using the same general principle for solving trigonometric equations. Unless, of course, you know how to draw the tangent and cotangent on a trigonometric circle.
In the examples above, I used the tabular value of sine and cosine: 0.5. Those. one of those meanings that the student knows must. Now let's expand our capabilities to all other values. Decide, so decide!)
So, let's say we need to solve the following trigonometric equation:
There is no such value of the cosine in the short tables. We coolly ignore this terrible fact. We draw a circle, mark 2/3 on the cosine axis and draw the corresponding angles. We get this picture.
We understand, for starters, with an angle in the first quarter. To know what x is equal to, they would immediately write down the answer! We don't know... Failure!? Calm! Mathematics does not leave its own in trouble! She invented arc cosines for this case. Do not know? In vain. Find out. It's a lot easier than you think. According to this link, there is not a single tricky spell about "inverse trigonometric functions" ... It's superfluous in this topic.
If you're in the know, just say to yourself, "X is an angle whose cosine is 2/3." And immediately, purely by definition of the arccosine, we can write:
We remember about additional revolutions and calmly write down the first series of roots of our trigonometric equation:
x 1 = arccos 2/3 + 2π n, n ∈ Z
The second series of roots is also written almost automatically, for the second angle. Everything is the same, only x (arccos 2/3) will be with a minus:
x 2 = - arccos 2/3 + 2π n, n ∈ Z
And all things! This is the correct answer. Even easier than with tabular values. You don’t need to remember anything.) By the way, the most attentive will notice that this picture with the solution through the arc cosine essentially no different from the picture for cos equations x = 0.5.
Exactly! The general principle on that and the general! I specifically drew two almost identical pictures. The circle shows us the angle X by its cosine. It is a tabular cosine, or not - the circle does not know. What kind of angle is this, π / 3, or what kind of arc cosine is up to us to decide.
With a sine the same song. For example:
Again we draw a circle, mark the sine equal to 1/3, draw the corners. It turns out this picture:
And again the picture is almost the same as for the equation sinx = 0.5. Again we start from the corner in the first quarter. What is x equal to if its sine is 1/3? No problem!
So the first pack of roots is ready:
x 1 = arcsin 1/3 + 2π n, n ∈ Z
Let's take a look at the second angle. In the example with a table value of 0.5, it was equal to:
π - x
So here it will be exactly the same! Only x is different, arcsin 1/3. So what!? You can safely write the second pack of roots:
x 2 = π - arcsin 1/3 + 2π n, n ∈ Z
This is a completely correct answer. Although it does not look very familiar. But it's understandable, I hope.)
This is how trigonometric equations are solved using a circle. This path is clear and understandable. It is he who saves in trigonometric equations with the selection of roots on a given interval, in trigonometric inequalities - they are generally solved almost always in a circle. In short, in any tasks that are a little more complicated than standard ones.
Putting knowledge into practice?
Solve trigonometric equations:
At first it is simpler, directly on this lesson.
Now it's more difficult.
Hint: here you have to think about the circle. Personally.)
And now outwardly unpretentious ... They are also called special cases.
sinx = 0
sinx = 1
cosx = 0
cosx = -1
Hint: here you need to figure out in a circle where there are two series of answers, and where there is one ... And how to write down one instead of two series of answers. Yes, so that not a single root from an infinite number is lost!)
Well, quite simple):
sinx = 0,3
cosx = π
tgx = 1,2
ctgx = 3,7
Hint: here you need to know what is the arcsine, arccosine? What is arc tangent, arc tangent? The simplest definitions. But you don’t need to remember any tabular values!)
The answers are, of course, in disarray):
x 1= arcsin0,3 + 2πn, n ∈ Z
x 2= π - arcsin0.3 + 2
Not everything works out? It happens. Read the lesson again. Only thoughtfully(there is such an obsolete word...) And follow the links. The main links are about the circle. Without it in trigonometry - how to cross the road blindfolded. Sometimes it works.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
The simplest trigonometric equations are the equations
Cos(x)=a, sin(x)=a, tg(x)=a, ctg(x)=a
Equation cos(x) = a
Explanation and rationale
- Roots cox equations= a. When | a | > 1 the equation has no roots because | cosx |< 1 для любого x (прямая y = а при а >1 or at a< -1 не пересекает график функцииy = cosx).
Let | a |< 1. Тогда прямая у = а пересекает график функции
y = cos x. On the interval, the function y = cos x decreases from 1 to -1. But a decreasing function takes on each of its values only at one point of its domain of definition, therefore the equation cos x = a has only one root on this interval, which, by definition of the arc cosine, is: x 1 = arccos a (and for this root cos x = but).
Cosine - even function, so on the interval [-p; 0] the equation cos x = and also has only one root - the number opposite to x 1, that is
x 2 = -arccos a.
Thus, on the interval [-n; n] (length 2n) the equation cos x = a for | a |< 1 имеет только корни x = ±arccos а.
The function y = cos x is periodic with a period of 2n, so all other roots differ from those found by 2np (n € Z). We get the following formula for the roots of the equation cos x = a when
x = ± arccos a + 2n, n £ Z.
- Particular cases of solving the equation cosx = a.
It is useful to remember the special notation for the roots of the equation cos x = a when
a \u003d 0, a \u003d -1, a \u003d 1, which can be easily obtained using the unit circle as a guide.
Since the cosine is equal to the abscissa of the corresponding point unit circle, we obtain that cos x = 0 if and only if the corresponding point of the unit circle is point A or point B.
Similarly, cos x = 1 if and only if the corresponding point of the unit circle is the point C, therefore,
x = 2πp, k € Z.
Also cos x \u003d -1 if and only if the corresponding point of the unit circle is the point D, thus x \u003d n + 2n,
Equation sin(x) = a
Explanation and rationale
- Roots sinx equations= a. When | a | > 1 the equation has no roots because | sinx |< 1 для любого x (прямая y = а на рисунке при а >1 or at a< -1 не пересекает график функции y = sinx).
The concept of solving trigonometric equations.
- To solve a trigonometric equation, convert it to one or more basic trigonometric equations. Solving the trigonometric equation ultimately comes down to solving the four basic trigonometric equations.
Solution of basic trigonometric equations.
- There are 4 types of basic trigonometric equations:
- sin x = a; cos x = a
- tan x = a; ctg x = a
- Solving basic trigonometric equations involves looking at the different x positions on the unit circle, as well as using a conversion table (or calculator).
- Example 1. sin x = 0.866. Using a conversion table (or calculator), you get the answer: x = π/3. The unit circle gives another answer: 2π/3. Remember: all trigonometric functions are periodic, that is, their values are repeated. For example, the periodicity of sin x and cos x is 2πn, and the periodicity of tg x and ctg x is πn. So the answer is written like this:
- x1 = π/3 + 2πn; x2 = 2π/3 + 2πn.
- Example 2 cos x = -1/2. Using a conversion table (or calculator), you get the answer: x = 2π/3. The unit circle gives another answer: -2π/3.
- x1 = 2π/3 + 2π; x2 = -2π/3 + 2π.
- Example 3. tg (x - π/4) = 0.
- Answer: x \u003d π / 4 + πn.
- Example 4. ctg 2x = 1.732.
- Answer: x \u003d π / 12 + πn.
Transformations used in solving trigonometric equations.
- To transform trigonometric equations, algebraic transformations are used (factoring, reduction homogeneous members etc.) and trigonometric identities.
- Example 5. Using trigonometric identities, the equation sin x + sin 2x + sin 3x = 0 is converted to the equation 4cos x*sin (3x/2)*cos (x/2) = 0. Thus, the following basic trigonometric equations need to be solved: cos x = 0; sin(3x/2) = 0; cos(x/2) = 0.
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Finding angles from known values of functions.
- Before learning how to solve trigonometric equations, you need to learn how to find angles from known values of functions. This can be done using a conversion table or calculator.
- Example: cos x = 0.732. The calculator will give the answer x = 42.95 degrees. The unit circle will give additional angles, the cosine of which is also equal to 0.732.
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Set aside the solution on the unit circle.
- You can put solutions to the trigonometric equation on the unit circle. The solutions of the trigonometric equation on the unit circle are the vertices of a regular polygon.
- Example: The solutions x = π/3 + πn/2 on the unit circle are the vertices of the square.
- Example: The solutions x = π/4 + πn/3 on the unit circle are the vertices of a regular hexagon.
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Methods for solving trigonometric equations.
- If the given trigonometric equation contains only one trigonometric function, solve this equation as a basic trigonometric equation. If this equation includes two or more trigonometric functions, then there are 2 methods for solving such an equation (depending on the possibility of its transformation).
- Method 1
- Transform this equation into an equation of the form: f(x)*g(x)*h(x) = 0, where f(x), g(x), h(x) are the basic trigonometric equations.
- Example 6. 2cos x + sin 2x = 0. (0< x < 2π)
- Solution. Using the double angle formula sin 2x = 2*sin x*cos x, replace sin 2x.
- 2cos x + 2*sin x*cos x = 2cos x*(sin x + 1) = 0. Now solve two basic trigonometric equations: cos x = 0 and (sin x + 1) = 0.
- Example 7 cos x + cos 2x + cos 3x = 0. (0< x < 2π)
- Solution: Using trigonometric identities, transform this equation into an equation of the form: cos 2x(2cos x + 1) = 0. Now solve two basic trigonometric equations: cos 2x = 0 and (2cos x + 1) = 0.
- Example 8. sin x - sin 3x \u003d cos 2x. (0< x < 2π)
- Solution: Using trigonometric identities, transform this equation into an equation of the form: -cos 2x*(2sin x + 1) = 0. Now solve two basic trigonometric equations: cos 2x = 0 and (2sin x + 1) = 0.
- Method 2
- Convert the given trigonometric equation to an equation containing only one trigonometric function. Then replace this trigonometric function with some unknown, for example, t (sin x = t; cos x = t; cos 2x = t, tg x = t; tg (x/2) = t, etc.).
- Example 9. 3sin^2 x - 2cos^2 x = 4sin x + 7 (0< x < 2π).
- Solution. IN given equation replace (cos^2 x) with (1 - sin^2 x) (according to the identity). The transformed equation looks like:
- 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Replace sin x with t. Now the equation is: 5t^2 - 4t - 9 = 0. This is quadratic equation, which has two roots: t1 = -1 and t2 = 9/5. The second root t2 does not satisfy the range of the function (-1< sin x < 1). Теперь решите: t = sin х = -1; х = 3π/2.
- Example 10. tg x + 2 tg^2 x = ctg x + 2
- Solution. Replace tg x with t. Rewrite the original equation as follows: (2t + 1)(t^2 - 1) = 0. Now find t and then find x for t = tg x.
- If the given trigonometric equation contains only one trigonometric function, solve this equation as a basic trigonometric equation. If this equation includes two or more trigonometric functions, then there are 2 methods for solving such an equation (depending on the possibility of its transformation).
