The simplest equations with sine. Trigonometric equations

Solving simple trigonometric equations.

Solving trigonometric equations of any level of complexity ultimately comes down to solving the simplest trigonometric equations. And in this the trigonometric circle again turns out to be the best assistant.

Let's recall the definitions of cosine and sine.

The cosine of an angle is the abscissa (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.

The sine of an angle is the ordinate (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.

The positive direction of movement on the trigonometric circle is counterclockwise. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1;0)

We use these definitions to solve simple trigonometric equations.

1. Solve the equation

This equation is satisfied by all values of the rotation angle that correspond to points on the circle whose ordinate is equal to .

Let's mark a point with ordinate on the ordinate axis:

Draw a horizontal line parallel to the x-axis until it intersects with the circle. We get two points lying on the circle and having an ordinate. These points correspond to rotation angles in and radians:

If we, leaving the point corresponding to the angle of rotation per radian, go around a full circle, then we will arrive at a point corresponding to the angle of rotation per radian and having the same ordinate. That is, this rotation angle also satisfies our equation. We can make as many “idle” revolutions as we like, returning to the same point, and all these angle values will satisfy our equation. The number of “idle” revolutions will be denoted by the letter (or). Since we can make these revolutions in both positive and negative directions, (or) can take on any integer values.

That is, the first series of solutions to the original equation has the form:

, , - set of integers (1)

Similarly, the second series of solutions has the form:

, Where , . (2)

As you might have guessed, this series of solutions is based on the point on the circle corresponding to the angle of rotation by .

These two series of solutions can be combined into one entry:

If we take (that is, even) in this entry, then we will get the first series of solutions.

If we take (that is, odd) in this entry, then we get the second series of solutions.

2. Now let's solve the equation

Since this is the abscissa of a point on the unit circle obtained by rotating through an angle, we mark the point with the abscissa on the axis:

Draw a vertical line parallel to the axis until it intersects with the circle. We will get two points lying on the circle and having an abscissa. These points correspond to rotation angles in and radians. Recall that when moving clockwise we get a negative rotation angle:

Let us write down two series of solutions:

(We get to the desired point by going from the main full circle, that is.

Let's combine these two series into one entry:

3. Solve the equation

The tangent line passes through the point with coordinates (1,0) of the unit circle parallel to the OY axis

Let's mark a point on it with an ordinate equal to 1 (we are looking for the tangent of which angles is equal to 1):

Let's connect this point to the origin of coordinates with a straight line and mark the points of intersection of the line with the unit circle. The intersection points of the straight line and the circle correspond to the angles of rotation on and:

Since the points corresponding to the rotation angles that satisfy our equation lie at a distance of radians from each other, we can write the solution this way:

4. Solve the equation

The line of cotangents passes through the point with the coordinates of the unit circle parallel to the axis.

Let's mark a point with abscissa -1 on the cotangent line:

Let's connect this point to the origin of the straight line and continue it until it intersects with the circle. This straight line will intersect the circle at points corresponding to the angles of rotation in and radians:

Since these points are separated from each other by a distance equal to , we can write the general solution of this equation as follows:

In the given examples illustrating the solution of the simplest trigonometric equations, tabular values of trigonometric functions were used.

However, if the right side of the equation contains a non-tabular value, then we substitute the value into the general solution of the equation:

SPECIAL SOLUTIONS:

Let us mark the points on the circle whose ordinate is 0:

Let us mark a single point on the circle whose ordinate is 1:

Let us mark a single point on the circle whose ordinate is equal to -1:

Since it is customary to indicate values closest to zero, we write the solution as follows:

Let us mark the points on the circle whose abscissa is equal to 0:

5.
Let us mark a single point on the circle whose abscissa is equal to 1:

Let us mark a single point on the circle whose abscissa is equal to -1:

And slightly more complex examples:

The sine is equal to one if the argument is equal to

The argument of our sine is equal, so we get:

Let's divide both sides of the equality by 3:

Answer:

Cosine is zero if the argument of cosine is

The argument of our cosine is equal to , so we get:

Let's express , to do this we first move to the right with the opposite sign:

Let's simplify the right side:

Divide both sides by -2:

Note that the sign in front of the term does not change, since k can take any integer value.

Answer:

And finally, watch the video lesson “Selecting roots in a trigonometric equation using a trigonometric circle”

This concludes our conversation about solving simple trigonometric equations. Next time we will talk about how to decide.

Maintaining your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please review our privacy practices and let us know if you have any questions.

Collection and use of personal information

Personal information refers to data that can be used to identify or contact a specific person.

You may be asked to provide your personal information at any time when you contact us.

Below are some examples of the types of personal information we may collect and how we may use such information.

What personal information do we collect:

When you submit an application on the site, we may collect various information, including your name, phone number, email address, etc.

How we use your personal information:

The personal information we collect allows us to contact you with unique offers, promotions and other events and upcoming events.
From time to time, we may use your personal information to send important notices and communications.
We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
If you participate in a prize draw, contest or similar promotion, we may use the information you provide to administer such programs.

Disclosure of information to third parties

We do not disclose the information received from you to third parties.

Exceptions:

If necessary - in accordance with the law, judicial procedure, in legal proceedings, and/or on the basis of public requests or requests from government authorities in the territory of the Russian Federation - to disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public importance purposes.
In the event of a reorganization, merger, or sale, we may transfer the personal information we collect to the applicable successor third party.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as unauthorized access, disclosure, alteration and destruction.

Respecting your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security standards to our employees and strictly enforce privacy practices.

Lesson and presentation on the topic: "Solving simple trigonometric equations"

Additional materials
Dear users, do not forget to leave your comments, reviews, wishes! All materials have been checked by an anti-virus program.

Manuals and simulators in the Integral online store for grade 10 from 1C
We solve problems in geometry. Interactive tasks for building in space
Software environment "1C: Mathematical Constructor 6.1"

What we will study:
1. What are trigonometric equations?

3. Two main methods for solving trigonometric equations.
4. Homogeneous trigonometric equations.
5. Examples.

What are trigonometric equations?

Guys, we have already studied arcsine, arccosine, arctangent and arccotangent. Now let's look at trigonometric equations in general.

Trigonometric equations are equations in which a variable is contained under the sign of a trigonometric function.

Let us repeat the form of solving the simplest trigonometric equations:

1)If |a|≤ 1, then the equation cos(x) = a has a solution:

X= ± arccos(a) + 2πk

2) If |a|≤ 1, then the equation sin(x) = a has a solution:

3) If |a| > 1, then the equation sin(x) = a and cos(x) = a have no solutions 4) The equation tg(x)=a has a solution: x=arctg(a)+ πk

5) The equation ctg(x)=a has a solution: x=arcctg(a)+ πk

For all formulas k is an integer

The simplest trigonometric equations have the form: T(kx+m)=a, T is some trigonometric function.

Example.

Solve the equations: a) sin(3x)= √3/2

Solution:

A) Let us denote 3x=t, then we will rewrite our equation in the form:

The solution to this equation will be: t=((-1)^n)arcsin(√3 /2)+ πn.

From the table of values we get: t=((-1)^n)×π/3+ πn.

Let's return to our variable: 3x =((-1)^n)×π/3+ πn,

Then x= ((-1)^n)×π/9+ πn/3

Answer: x= ((-1)^n)×π/9+ πn/3, where n is an integer. (-1)^n – minus one to the power of n.

More examples of trigonometric equations.

Solve the equations: a) cos(x/5)=1 b)tg(3x- π/3)= √3

Solution:

A) This time let’s move directly to calculating the roots of the equation right away:

X/5= ± arccos(1) + 2πk. Then x/5= πk => x=5πk

Answer: x=5πk, where k is an integer.

B) We write it in the form: 3x- π/3=arctg(√3)+ πk. We know that: arctan(√3)= π/3

3x- π/3= π/3+ πk => 3x=2π/3 + πk => x=2π/9 + πk/3

Answer: x=2π/9 + πk/3, where k is an integer.

Solve the equations: cos(4x)= √2/2. And find all the roots on the segment.

Solution:

Let us solve our equation in general form: 4x= ± arccos(√2/2) + 2πk

4x= ± π/4 + 2πk;

X= ± π/16+ πk/2;

Now let's see what roots fall on our segment. At k At k=0, x= π/16, we are in the given segment.
With k=1, x= π/16+ π/2=9π/16, we hit again.
For k=2, x= π/16+ π=17π/16, but here we didn’t hit, which means that for large k we also obviously won’t hit.

Answer: x= π/16, x= 9π/16

Two main solution methods.

We looked at the simplest trigonometric equations, but there are also more complex ones. To solve them, the method of introducing a new variable and the method of factorization are used. Let's look at examples.

Let's solve the equation:

Solution:
To solve our equation, we will use the method of introducing a new variable, denoting: t=tg(x).

As a result of the replacement we get: t 2 + 2t -1 = 0

Let's find the roots of the quadratic equation: t=-1 and t=1/3

Then tg(x)=-1 and tg(x)=1/3, we get the simplest trigonometric equation, let’s find its roots.

X=arctg(-1) +πk= -π/4+πk; x=arctg(1/3) + πk.

Answer: x= -π/4+πk; x=arctg(1/3) + πk.

An example of solving an equation

Solve equations: 2sin 2 (x) + 3 cos(x) = 0

Solution:

Let's use the identity: sin 2 (x) + cos 2 (x)=1

Our equation will take the form: 2-2cos 2 (x) + 3 cos (x) = 0

2 cos 2 (x) - 3 cos(x) -2 = 0

Let us introduce the replacement t=cos(x): 2t 2 -3t - 2 = 0

The solution to our quadratic equation is the roots: t=2 and t=-1/2

Then cos(x)=2 and cos(x)=-1/2.

Because cosine cannot take values greater than one, then cos(x)=2 has no roots.

For cos(x)=-1/2: x= ± arccos(-1/2) + 2πk; x= ±2π/3 + 2πk

Answer: x= ±2π/3 + 2πk

Homogeneous trigonometric equations.

Definition: Equations of the form a sin(x)+b cos(x) are called homogeneous trigonometric equations of the first degree.

Equations of the form

homogeneous trigonometric equations of the second degree.

To solve a homogeneous trigonometric equation of the first degree, divide it by cos(x): You cannot divide by the cosine if it is equal to zero, let's make sure that this is not the case:
Let cos(x)=0, then asin(x)+0=0 => sin(x)=0, but sine and cosine are not equal to zero at the same time, we get a contradiction, so we can safely divide by zero.

Solve the equation:
Example: cos 2 (x) + sin(x) cos(x) = 0

Solution:

Let's take out the common factor: cos(x)(c0s(x) + sin (x)) = 0

Then we need to solve two equations:

Cos(x)=0 and cos(x)+sin(x)=0

Cos(x)=0 at x= π/2 + πk;

Consider the equation cos(x)+sin(x)=0 Divide our equation by cos(x):

1+tg(x)=0 => tg(x)=-1 => x=arctg(-1) +πk= -π/4+πk

Answer: x= π/2 + πk and x= -π/4+πk

How to solve homogeneous trigonometric equations of the second degree?
Guys, always follow these rules!

1. See what the coefficient a is equal to, if a=0 then our equation will take the form cos(x)(bsin(x)+ccos(x)), an example of the solution of which is on the previous slide

2. If a≠0, then you need to divide both sides of the equation by the cosine squared, we get:

We change the variable t=tg(x) and get the equation:

Solve example No.:3

Solve the equation:
Solution:

Let's divide both sides of the equation by the cosine square:

We change the variable t=tg(x): t 2 + 2 t - 3 = 0

Let's find the roots of the quadratic equation: t=-3 and t=1

Then: tg(x)=-3 => x=arctg(-3) + πk=-arctg(3) + πk

Tg(x)=1 => x= π/4+ πk

Answer: x=-arctg(3) + πk and x= π/4+ πk

Solve example No.:4

Solve the equation:

Solution:
Let's transform our expression:

We can solve such equations: x= - π/4 + 2πk and x=5π/4 + 2πk

Answer: x= - π/4 + 2πk and x=5π/4 + 2πk

Solve example no.:5

Solve the equation:

Solution:
Let's transform our expression:

Let us introduce the replacement tg(2x)=t:2 2 - 5t + 2 = 0

The solution to our quadratic equation will be the roots: t=-2 and t=1/2

Then we get: tg(2x)=-2 and tg(2x)=1/2
2x=-arctg(2)+ πk => x=-arctg(2)/2 + πk/2

2x= arctg(1/2) + πk => x=arctg(1/2)/2+ πk/2

Answer: x=-arctg(2)/2 + πk/2 and x=arctg(1/2)/2+ πk/2

Problems for independent solution.

1) Solve the equation

A) sin(7x)= 1/2 b) cos(3x)= √3/2 c) cos(-x) = -1 d) tg(4x) = √3 d) ctg(0.5x) = -1.7

2) Solve the equations: sin(3x)= √3/2. And find all the roots on the segment [π/2; π].

3) Solve the equation: cot 2 (x) + 2 cot (x) + 1 =0

4) Solve the equation: 3 sin 2 (x) + √3sin (x) cos(x) = 0

5) Solve the equation: 3sin 2 (3x) + 10 sin(3x)cos(3x) + 3 cos 2 (3x) =0

6) Solve the equation: cos 2 (2x) -1 - cos(x) =√3/2 -sin 2 (2x)

The video course “Get an A” includes all the topics necessary to successfully pass the Unified State Exam in mathematics with 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.

All the necessary theory. Quick solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. A basis for solving complex problems of Part 2 of the Unified State Exam.

You can order a detailed solution to your problem!!!

An equality containing an unknown under the sign of a trigonometric function (`sin x, cos x, tan x` or `ctg x`) is called a trigonometric equation, and it is their formulas that we will consider further.

The simplest equations are `sin x=a, cos x=a, tg x=a, ctg x=a`, where `x` is the angle to be found, `a` is any number. Let us write down the root formulas for each of them.

1. Equation `sin x=a`.

For `|a|>1` it has no solutions.

When `|a| \leq 1` has an infinite number of solutions.

Root formula: `x=(-1)^n arcsin a + \pi n, n \in Z`

2. Equation `cos x=a`

For `|a|>1` - as in the case of sine, it has no solutions among real numbers.

When `|a| \leq 1` has an infinite number of solutions.

Root formula: `x=\pm arccos a + 2\pi n, n \in Z`

Special cases for sine and cosine in graphs.

3. Equation `tg x=a`

Has an infinite number of solutions for any values of `a`.

Root formula: `x=arctg a + \pi n, n \in Z`

4. Equation `ctg x=a`

Also has an infinite number of solutions for any values of `a`.

Root formula: `x=arcctg a + \pi n, n \in Z`

Formulas for the roots of trigonometric equations in the table

For sine:
For cosine:
For tangent and cotangent:
Formulas for solving equations containing inverse trigonometric functions:

Methods for solving trigonometric equations

Solving any trigonometric equation consists of two stages:

with the help of transforming it to the simplest;
solve the simplest equation obtained using the root formulas and tables written above.

Let's look at the main solution methods using examples.

Algebraic method.

This method involves replacing a variable and substituting it into an equality.

Example. Solve the equation: `2cos^2(x+\frac \pi 6)-3sin(\frac \pi 3 - x)+1=0`

`2cos^2(x+\frac \pi 6)-3cos(x+\frac \pi 6)+1=0`,

make a replacement: `cos(x+\frac \pi 6)=y`, then `2y^2-3y+1=0`,

we find the roots: `y_1=1, y_2=1/2`, from which two cases follow:

1. `cos(x+\frac \pi 6)=1`, `x+\frac \pi 6=2\pi n`, `x_1=-\frac \pi 6+2\pi n`.

2. `cos(x+\frac \pi 6)=1/2`, `x+\frac \pi 6=\pm arccos 1/2+2\pi n`, `x_2=\pm \frac \pi 3- \frac \pi 6+2\pi n`.

Answer: `x_1=-\frac \pi 6+2\pi n`, `x_2=\pm \frac \pi 3-\frac \pi 6+2\pi n`.

Factorization.

Example. Solve the equation: `sin x+cos x=1`.

Solution. Let's move all the terms of the equality to the left: `sin x+cos x-1=0`. Using , we transform and factorize the left-hand side:

`sin x — 2sin^2 x/2=0`,

`2sin x/2 cos x/2-2sin^2 x/2=0`,

`2sin x/2 (cos x/2-sin x/2)=0`,

`sin x/2 =0`, `x/2 =\pi n`, `x_1=2\pi n`.
`cos x/2-sin x/2=0`, `tg x/2=1`, `x/2=arctg 1+ \pi n`, `x/2=\pi/4+ \pi n` , `x_2=\pi/2+ 2\pi n`.

Answer: `x_1=2\pi n`, `x_2=\pi/2+ 2\pi n`.

Reduction to a homogeneous equation

First, you need to reduce this trigonometric equation to one of two forms:

`a sin x+b cos x=0` (homogeneous equation of the first degree) or `a sin^2 x + b sin x cos x +c cos^2 x=0` (homogeneous equation of the second degree).

Then divide both parts by `cos x \ne 0` - for the first case, and by `cos^2 x \ne 0` - for the second. We obtain equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which need to be solved using known methods.

Example. Solve the equation: `2 sin^2 x+sin x cos x - cos^2 x=1`.

Solution. Let's write the right side as `1=sin^2 x+cos^2 x`:

`2 sin^2 x+sin x cos x — cos^2 x=` `sin^2 x+cos^2 x`,

`2 sin^2 x+sin x cos x — cos^2 x -` ` sin^2 x — cos^2 x=0`

`sin^2 x+sin x cos x — 2 cos^2 x=0`.

This is a homogeneous trigonometric equation of the second degree, we divide its left and right sides by `cos^2 x \ne 0`, we get:

`\frac (sin^2 x)(cos^2 x)+\frac(sin x cos x)(cos^2 x) — \frac(2 cos^2 x)(cos^2 x)=0`

`tg^2 x+tg x — 2=0`. Let's introduce the replacement `tg x=t`, resulting in `t^2 + t - 2=0`. The roots of this equation are `t_1=-2` and `t_2=1`. Then:

`tg x=-2`, `x_1=arctg (-2)+\pi n`, `n \in Z`
`tg x=1`, `x=arctg 1+\pi n`, `x_2=\pi/4+\pi n`, ` n \in Z`.

Answer. `x_1=arctg (-2)+\pi n`, `n \in Z`, `x_2=\pi/4+\pi n`, `n \in Z`.

Go to half corner

Example. Solve the equation: `11 sin x - 2 cos x = 10`.

Solution. Let's apply the double angle formulas, resulting in: `22 sin (x/2) cos (x/2) -` `2 cos^2 x/2 + 2 sin^2 x/2=` `10 sin^2 x/2 +10 cos^2 x/2`

`4 tg^2 x/2 — 11 tg x/2 +6=0`

Applying the algebraic method described above, we obtain:

`tg x/2=2`, `x_1=2 arctg 2+2\pi n`, `n \in Z`,
`tg x/2=3/4`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Answer. `x_1=2 arctg 2+2\pi n, n \in Z`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Introduction of auxiliary angle

In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, divide both sides by `sqrt (a^2+b^2)`:

`\frac a(sqrt (a^2+b^2)) sin x +` `\frac b(sqrt (a^2+b^2)) cos x =` `\frac c(sqrt (a^2) +b^2))`.

The coefficients on the left side have the properties of sine and cosine, namely the sum of their squares is equal to 1 and their modules are not greater than 1. Let us denote them as follows: `\frac a(sqrt (a^2+b^2))=cos \varphi` , ` \frac b(sqrt (a^2+b^2)) =sin \varphi`, `\frac c(sqrt (a^2+b^2))=C`, then:

`cos \varphi sin x + sin \varphi cos x =C`.

Let's take a closer look at the following example:

Example. Solve the equation: `3 sin x+4 cos x=2`.

Solution. Divide both sides of the equality by `sqrt (3^2+4^2)`, we get:

`\frac (3 sin x) (sqrt (3^2+4^2))+` `\frac(4 cos x)(sqrt (3^2+4^2))=` `\frac 2(sqrt (3^2+4^2))`

`3/5 sin x+4/5 cos x=2/5`.

Let's denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, then we take `\varphi=arcsin 4/5` as an auxiliary angle. Then we write our equality in the form:

`cos \varphi sin x+sin \varphi cos x=2/5`

Applying the formula for the sum of angles for the sine, we write our equality in the following form:

`sin (x+\varphi)=2/5`,

`x+\varphi=(-1)^n arcsin 2/5+ \pi n`, `n \in Z`,

`x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Answer. `x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Fractional rational trigonometric equations

These are equalities with fractions whose numerators and denominators contain trigonometric functions.

Example. Solve the equation. `\frac (sin x)(1+cos x)=1-cos x`.

Solution. Multiply and divide the right side of the equality by `(1+cos x)`. As a result we get:

`\frac (sin x)(1+cos x)=` `\frac ((1-cos x)(1+cos x))(1+cos x)`

`\frac (sin x)(1+cos x)=` `\frac (1-cos^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)=` `\frac (sin^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)-` `\frac (sin^2 x)(1+cos x)=0`

`\frac (sin x-sin^2 x)(1+cos x)=0`

Considering that the denominator cannot be equal to zero, we get `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.

Let's equate the numerator of the fraction to zero: `sin x-sin^2 x=0`, `sin x(1-sin x)=0`. Then `sin x=0` or `1-sin x=0`.

`sin x=0`, `x=\pi n`, `n \in Z`
`1-sin x=0`, `sin x=-1`, `x=\pi /2+2\pi n, n \in Z`.

Given that ` x \ne \pi+2\pi n, n \in Z`, the solutions are `x=2\pi n, n \in Z` and `x=\pi /2+2\pi n` , `n \in Z`.

Answer. `x=2\pi n`, `n \in Z`, `x=\pi /2+2\pi n`, `n \in Z`.

Trigonometry, and trigonometric equations in particular, are used in almost all areas of geometry, physics, and engineering. Studying begins in the 10th grade, there are always tasks for the Unified State Exam, so try to remember all the formulas of trigonometric equations - they will definitely be useful to you!

However, you don’t even need to memorize them, the main thing is to understand the essence and be able to derive it. It's not as difficult as it seems. See for yourself by watching the video.