Sin x sin x equals what. Solving simple trigonometric equations

– there will certainly be tasks on trigonometry. Trigonometry is often disliked for the need to cram a huge number of difficult formulas, teeming with sines, cosines, tangents and cotangents. The site already once gave advice on how to remember a forgotten formula, using the example of the Euler and Peel formulas.

And in this article we will try to show that it is enough to firmly know only five simple trigonometric formulas, and know about the rest general idea and bring them out as you go. It’s like with DNA: the molecule does not store the complete blueprints of a finished living being. Rather, it contains instructions for assembling it from available amino acids. So in trigonometry, knowing some general principles, we will get everything necessary formulas from small set those that must be kept in mind.

We will rely on following formulas:

From the formulas for sine and cosine sums, knowing about the parity of the cosine function and the oddness of the sine function, substituting -b instead of b, we obtain formulas for differences:

1. Sine of the difference: sin(a-b) = sinacos(-b)+cosasin(-b) = sinacosb-cosasinb
2. Cosine of the difference: cos(a-b) = cosacos(-b)-sinasin(-b) = cosacosb+sinasinb

Putting a = b into the same formulas, we obtain the formulas for sine and cosine of double angles:

1. Sinus double angle : sin2a = sin(a+a) = sinacosa+cosasina = 2sinacosa
2. Cosine of double angle: cos2a = cos(a+a) = cosacosa-sinasina = cos2 a-sin2 a

The formulas for other multiple angles are obtained similarly:

1. Sine of a triple angle: sin3a = sin(2a+a) = sin2acosa+cos2asina = (2sinacosa)cosa+(cos2 a-sin2 a)sina = 2sinacos2 a+sinacos2 a-sin 3 a = 3 sinacos2 a-sin 3 a = 3 sina(1-sin2 a)-sin 3 a = 3 sina-4sin 3a
2. Cosine of triple angle: cos3a = cos(2a+a) = cos2acosa-sin2asina = (cos2 a-sin2 a)cosa-(2sinacosa)sina = cos 3 a- sin2 acosa-2sin2 acosa = cos 3 a-3 sin2 acosa = cos 3 a-3(1- cos2 a)cosa = 4cos 3 a-3 cosa

Before we move on, let's look at one problem.
Given: the angle is acute.
Find its cosine if
Solution given by one student:
Because , That sina= 3,a cosa = 4.
(From math humor)

So, the definition of tangent relates this function to both sine and cosine. But you can get a formula that relates the tangent only to the cosine. To derive it, we take the main trigonometric identity: sin 2 a+cos 2 a= 1 and divide it by cos 2 a. We get:

So the solution to this problem would be:

(Since the angle is acute, when extracting the root, the + sign is taken)

The formula for the tangent of a sum is another one that is difficult to remember. Let's output it like this:

Immediately displayed and

From the cosine formula for a double angle, you can obtain the sine and cosine formulas for half angles. To do this, to the left side of the double angle cosine formula:
cos2 a = cos 2 a-sin 2 a
we add one, and to the right - a trigonometric unit, i.e. the sum of the squares of sine and cosine.
cos2a+1 = cos2 a-sin2 a+cos2 a+sin2 a
2cos 2 a = cos2 a+1
Expressing cosa through cos2 a and performing a change of variables, we get:

The sign is taken depending on the quadrant.

Similarly, subtracting one from the left side of the equality and the sum of the squares of the sine and cosine from the right, we get:
cos2a-1 = cos2 a-sin2 a-cos2 a-sin2 a
2sin 2 a = 1-cos2 a

And finally, to convert the sum of trigonometric functions into a product, we use the following technique. Let's say we need to represent the sum of sines as a product sina+sinb. Let's introduce variables x and y such that a = x+y, b+x-y. Then
sina+sinb = sin(x+y)+ sin(x-y) = sin x cos y+ cos x sin y+ sin x cos y- cos x sin y=2 sin x cos y. Let us now express x and y in terms of a and b.

Since a = x+y, b = x-y, then . That's why

You can withdraw immediately

1. Formula for partitioning products of sine and cosine V amount: sinacosb = 0.5(sin(a+b)+sin(a-b))

We recommend that you practice and derive formulas on your own for converting the difference of sines and the sum and difference of cosines into the product, as well as for dividing the products of sines and cosines into the sum. Having completed these exercises, you will thoroughly master the skill of deriving trigonometric formulas and will not get lost even in the most difficult test, olympiad or testing.

Let's deal with simple concepts: sine and cosine and calculation cosine squared and sine squared.

Sine and cosine are studied in trigonometry (the study of right-angle triangles).

Therefore, first, let’s remember the basic concepts of a right triangle:

Hypotenuse- the side that always lies opposite right angle(90 degree angle). The hypotenuse is the longest side of a right angle triangle.

The remaining two sides in a right triangle are called legs.

You should also remember that three angles in a triangle always add up to 180°.

Now let's move on to cosine and sine of the angle alpha (∠α)(this can be called any indirect angle in a triangle or used as a designation x - "x", which does not change the essence).

Sine of angle alpha (sin ∠α)- this is an attitude opposite leg (the side opposite the corresponding angle) to the hypotenuse. If you look at the figure, then sin ∠ABC = AC / BC

Cosine of angle alpha (cos ∠α)- attitude adjacent to the angle of the leg to the hypotenuse. Looking again at the figure above, cos ∠ABC = AB / BC

And just to remind you: cosine and sine will never be more than one, since any roll is shorter than the hypotenuse (and the hypotenuse is the longest side of any triangle, because the longest side is located opposite the largest angle in the triangle).

Cosine squared, sine squared

Now let's move on to the main ones trigonometric formulas: Calculate cosine squared and sine squared.

To calculate them, you should remember the basic trigonometric identity:

sin 2 α + cos 2 α = 1(sine square plus cosine square of one angle always equals one).

From trigonometric identity we draw conclusions about the sine:

sin 2 α = 1 - cos 2 α

sine square alpha equal to one minus the cosine of the double angle alpha and divide it all by two.

sin 2 α = (1 – cos(2α)) / 2

​​​​​​​From the trigonometric identity we draw conclusions about the cosine:

cos 2 α = 1 - sin 2 α

or more difficult option formulas: cosine square alpha is equal to one plus the cosine of the double angle alpha and also divide everything by two.

cos 2 α = (1 + cos(2α)) / 2

These two are more complex formulas sine squared and cosine squared are also called “reducing the degree for squares of trigonometric functions.” Those. there was a second degree, they lowered it to the first and the calculations became more convenient.

The simplest solution trigonometric equations.

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

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 unit circle, corresponding to 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:

Let's carry out 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 by radians, go around full circle, then we will arrive at a point corresponding to the rotation angle 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're in this let's take the notes(that is, even), then we 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:

Let's carry out 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 line of cotangents:

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:

1.

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:

2.

Cosine equal to zero, if the cosine argument is equal to

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 tutorial “Selecting roots in a trigonometric equation using trigonometric circle"

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

Sine and cosine originally arose from the need to calculate quantities in right triangles. It was noticed that if the degree measure of the angles in a right triangle is not changed, then the aspect ratio, no matter how much these sides change in length, always remains the same.

This is how the concepts of sine and cosine were introduced. Sinus acute angle in a right triangle is the ratio opposite side to the hypotenuse, and the cosine is adjacent to the hypotenuse.

Theorems of cosines and sines

But cosines and sines can be used for more than just right triangles. To find the value of an obtuse or acute angle or side of any triangle, it is enough to apply the theorem of cosines and sines.

The cosine theorem is quite simple: “The square of the side of a triangle equal to the sum the squares of the other two sides minus twice the product of these sides by the cosine of the angle between them.”

There are two interpretations of the sine theorem: small and extended. According to the small one: “In a triangle, the angles are proportional opposing parties». This theorem often expanded due to the property of the circumscribed circle of a triangle: “In a triangle, the angles are proportional to the opposite sides, and their ratio is equal to the diameter of the circumscribed circle.”

Derivatives

The derivative is a mathematical tool that shows how quickly a function changes relative to a change in its argument. Derivatives are used in geometry, and in a number of technical disciplines.

When solving problems, you need to know the tabular values ​​of the derivatives of trigonometric functions: sine and cosine. The derivative of a sine is a cosine, and a cosine is a sine, but with a minus sign.

Application in mathematics

Sines and cosines are especially often used when solving right triangles and tasks associated with them.

The convenience of sines and cosines is also reflected in technology. It was easy to evaluate angles and sides using the theorems of cosines and sines, breaking down complex figures and objects into “simple” triangles. Engineers often deal with aspect ratio calculations and degree measures, spent a lot of time and effort to calculate the cosines and sines of non-tabular angles.

Then Bradis tables came to the rescue, containing thousands of values ​​of sines, cosines, tangents and cotangents different angles. IN Soviet time some teachers forced their students to memorize pages of Bradis tables.

Radian - angular magnitude arcs, length equal to the radius or 57.295779513° degrees.

Degree (in geometry) - 1/360th part of a circle or 1/90th part of a right angle.

π = 3.141592653589793238462… ( approximate value Pi numbers).

Cosine table for angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360°.

Angle x (in degrees)	0°	30°	45°	60°	90°	120°	135°	150°	180°	210°	225°	240°	270°	300°	315°	330°	360°
Angle x (in radians)	0	π/6	π/4	π/3	π/2	2 x π/3	3 x π/4	5 x π/6	π	7 x π/6	5 x π/4	4 x π/3	3 x π/2	5 x π/3	7 x π/4	11 x π/6	2 x π
cos x	1	√3/2 (0,8660)	√2/2 (0,7071)	1/2 (0,5)	0	-1/2 (-0,5)	-√2/2 (-0,7071)	-√3/2 (-0,8660)	-1	-√3/2 (-0,8660)	-√2/2 (-0,7071)	-1/2 (-0,5)	0	1/2 (0,5)	√2/2 (0,7071)	√3/2 (0,8660)	1