In algebra classes, teachers tell us that there is a small (in fact, very large) class trigonometric equations that are not resolved using standard methods- neither through factorization, nor through a change of variable, nor even through homogeneous terms. In this case, a fundamentally different approach comes into play - the method auxiliary angle.

What is this method and how to apply it? First, let’s remember the formulas for the sine of the sum/difference and the cosine of the sum/difference:

\[\begin(align)& \sin \left(\alpha \pm \beta \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\& \cos \left(\ alpha \pm \beta \right)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\\end(align)\]

I think these formulas are well known to you - from them the double argument formulas are derived, without which there is absolutely nowhere in trigonometry. But let's now look at a simple equation:

Divide both sides by 5:

Note that $((\left(\frac(3)(5) \right))^(2))+((\left(\frac(4)(5) \right))^(2))= 1$, which means that there is sure to be an angle $\alpha $ for which these numbers are cosine and sine, respectively. Therefore, our equation will be rewritten as follows:

\[\begin(align)& \cos \alpha \sin x+\sin \alpha \cos x=1 \\& \sin \left(\alpha +x \right)=1 \\\end(align)\]

And this is already easily solved, after which all that remains is to find out why equal to the angle$\alpha$. How to find out, and also how to choose the correct number to divide both sides of the equation (in this simple example we divided by 5) - about this in today's video lesson:

Today we will analyze the solution of trigonometric equations, or, more precisely, a single technique called the “auxiliary angle method.” Why this method? Simply because over the past two or three days, when I was teaching students to whom I told about solving trigonometric equations, and we were examining, among other things, the auxiliary angle method, and all the students, as one, made the same mistake. But the method is generally simple and, moreover, it is one of the main techniques in trigonometry. Therefore many trigonometric problems they cannot be solved at all except by the auxiliary angle method.

Therefore, now, first, we will look at a couple of simple tasks, and then we will move on to more serious tasks. However, all of these one way or another will require us to use the auxiliary angle method, the essence of which I will tell in the first design.

Solving simple trigonometric problems

Example #1

\[\cos 2x=\sqrt(3)\sin 2x-1\]

Let's transform our expression a little:

\[\cos 2x-\sqrt(3)\sin 2x=-1\left| \left(-1 \right) \right.\]

\[\sqrt(3)\cdot \sin 2x-\cos 2x=1\]

How will we solve it? The standard trick is to solve $\sin 2x$ and $\cos 2x$ using the formulas double angle, and then rewrite the unit as $((\sin )^(2))x((\cos )^(2))x$, get homogeneous equation, bring it to tangents and solve. However, this is a long and tedious path that requires a large amount of calculations.

I suggest you think about this. We have $\sin$ and $\cos$. Let us recall the formula for cosine and sine of sum and difference:

\[\sin \left(\alpha \pm \beta \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \]

\[\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \]

\[\cos \left(\alpha -\beta \right)=\cos a\cos \beta +\sin \alpha \sin \beta \]

Let's return to our example. Let's reduce everything to the sine of the difference. But first, the equation needs to be transformed a little. Let's find the coefficient:

$\sqrt(l)$ is the very coefficient by which it is necessary to divide both sides of the equation so that before the sine and cosine there appear numbers that are themselves sines and cosines. Let's divide:

\[\frac(\sqrt(3))(2)\cdot \sin 2x-\frac(1)(2)\cdot \cos 2x=\frac(1)(2)\]

Let's look at what we got on the left: does there exist a $\sin $ and $\cos $ such that $\cos \alpha =\frac(\sqrt(3))(2)$ and $\sin \alpha =\frac(1)(2)$? Obviously there is: $\alpha =\frac(\text( )\!\!\pi\!\!\text( ))(6)$. Therefore we can rewrite our expression as follows:

\[\cos \frac(\text( )\!\!\pi\!\!\text( ))(\text(6))\cdot \sin 2x-\sin \frac(\text( )\! \!\pi\!\!\text( ))(\text(6))\cdot \cos 2x=\frac(1)(2)\]

\[\sin 2x\cdot \cos \frac(\text( )\!\!\pi\!\!\text( ))(\text(6))-\cos 2x\cdot \sin \frac(\ text( )\!\!\pi\!\!\text( ))(\text(6))=\frac(1)(2)\]

Now we have the formula for the sine of the difference. We can write like this:

\[\sin \left(2x-\frac(\text( )\!\!\pi\!\!\text( ))(\text(6)) \right)=\frac(1)(2) \]

Here we have the simplest classical trigonometric construction. Let me remind you:

We will write this down for our specific expression:

\[\left[ \begin(align)& 2x-\frac(\text( )\!\!\pi\!\!\text( ))(6)=\frac(\text( )\!\! \pi\!\!\text( ))(6)=2\text( )\!\!\pi\!\!\text( )n \\& 2x-\frac(\text( )\!\ !\pi\!\!\text( ))(\text(6))=\text( )\!\!\pi\!\!\text( )-\frac(\text( )\!\! \pi\!\!\text( ))(\text(6))+2\text( )\!\!\pi\!\!\text( )n \\\end(align) \right.\ ]

\[\left[ \begin(align)& 2x=\frac(\text( )\!\!\pi\!\!\text( ))(3)+2\text( )\!\!\pi \!\!\text( )n \\& 2x=\text( )\!\!\pi\!\!\text( )+2\text( )\!\!\pi\!\!\text ( )n \\\end(align) \right.\]

\[\left[ \begin(align)& x=\frac(\text( )\!\!\pi\!\!\text( ))(6)+\text( )\!\!\pi\ !\!\text( )n \\& x=\frac(\text( )\!\!\pi\!\!\text( ))(2)+\text( )\!\!\pi\ !\!\text( )n \\\end(align) \right.\]

Nuances of the solution

So, what should you do if you come across a similar example:

1. Modify the design if necessary.
2. Find the correction factor, take the root from it and divide both sides of the example by it.
3. Let's see what sine and cosine values ​​the numbers get.
4. We expand the equation using the sine or cosine difference or sum formulas.
5. We solve the simplest trigonometric equation.

In this regard, attentive students will probably have two questions.

What prevents us from writing down $\sin $ and $\cos $ at the stage of finding the correction factor? — The basic trigonometric identity is preventing us. The fact is that the resulting $\sin $ and $\cos $, like any others with the same argument, should, when squared, give exactly “one” in total. During the decision process, you need to be very careful and not lose the “2” before the “X”.

The auxiliary angle method is a tool that helps reduce an “ugly” equation to a completely adequate and “beautiful” one.

Example No. 2

\[\sqrt(3)\sin 2x+2((\sin )^(2))x-1=2\cos x\]

We see that we have $((\sin )^(2))x$, so let's use the power reduction calculations. However, before we use them, let's take them out. To do this, remember how to find the cosine of a double angle:

\[\cos 2x=((\cos )^(2))x-((\sin )^(2))x=2((\cos )^(2))x-1=1-2(( \sin )^(2))x\]

If we write $\cos 2x$ in the third option, we get:

\[\cos 2x=1-2((\sin )^(2))x\]

\[((\sin )^(2))x=\frac(1-((\cos )^(2))x)(x)\]

I'll write it out separately:

\[((\sin )^(2))x=\frac(1-\cos 2x)(2)\]

The same can be done for $((\cos )^(2))x$:

\[((\cos )^(2))x=\frac(1+\cos 2x)(2)\]

We only need the first calculations. Let's start working on the task:

\[\sqrt(3)\cdot \sin 2x+2\cdot \frac(1-\cos 2x)(2)-1=2\cos x\]

\[\sqrt(3)\cdot \sin 2x+1-\cos 2x-1=2\cos x\]

\[\sqrt(3)\cdot \sin 2x-\cos 2x=2\cos x\]

Now let's use the calculations of the cosine of the difference. But first, let’s calculate the $l$ correction:

Let's rewrite taking into account this fact:

\[\frac(\sqrt(3))(2)\cdot \sin 2x-\frac(1)(2)\cdot \cos 2x=\cos x\]

In this case, we can write that $\frac(\sqrt(3))(2)=\frac(\text( )\!\!\pi\!\!\text( ))(3)$, and $\frac(1)(2)=\cos \frac(\text( )\!\!\pi\!\!\text( ))(3)$. Let's rewrite:

\[\sin \frac(\text( )\!\!\pi\!\!\text( ))(\text(3))\cdot \sin 2x-\cos \frac(\text( )\! \!\pi\!\!\text( ))(\text(3))\cdot \cos 2x=\cos x\]

\[-\cos \left(\frac(\text( )\!\!\pi\!\!\text( ))(\text(3))+2x \right)=\cos x\]

Let’s add a “minus” into the bracket in a clever way. To do this, note the following:

\[\cos \left(\frac(\text( )\!\!\pi\!\!\text( ))(\text(3))+2x \right)=\cos \left(\text( )\!\!\pi\!\!\text( )-\text( )\!\!\pi\!\!\text( +)\frac(\text( )\!\!\pi\! \!\text( ))(\text(3))+2x \right)=\]

\[=\cos \left(\text( )\!\!\pi\!\!\text( )-\frac(2\text( )\!\!\pi\!\!\text( )) (3)+2x \right)=\cos \left(\text( )\!\!\pi\!\!\text( )+\varphi \right)=-\cos \varphi \]

Let's return to our expression and remember that in the role of $\varphi $ we have the expression $-\frac(2\text( )\!\!\pi\!\!\text( ))(3)+2x$. Therefore, let's write:

\[-\left(-\cos \left(-\frac(2\text( )\!\!\pi\!\!\text( ))(3)+2x \right) \right)=\cos x\]

\[\cos \left(2x-\frac(2\text( )\!\!\pi\!\!\text( ))(3) \right)=\cos x\]

To decide similar task, you need to remember this:

\[\cos \alpha =\cos \beta \]

\[\left[ \begin(align)& \alpha =\beta +2\text( )\!\!\pi\!\!\text( )n \\& \alpha =-\beta +2\text ( )\!\!\pi\!\!\text( )n \\\end(align) \right.\]

Let's look at our example:

\[\left[ \begin(align)& 2x-\frac(2\text( )\!\!\pi\!\!\text( ))(3)=x+2\text( )\!\ !\pi\!\!\text( )n \\& 2x-\frac(2\text( )\!\!\pi\!\!\text( ))(3)=-x+2\text ( )\!\!\pi\!\!\text( )n \\\end(align) \right.\]

Let's calculate each of these equations:

And the second:

Let's write down the final answer:

\[\left[ \begin(align)& x=\frac(2\text( )\!\!\pi\!\!\text( ))(3)+2\text( )\!\!\ pi\!\!\text( )n \\& x=\frac(2\text( )\!\!\pi\!\!\text( ))(9)+\frac(2\text( ) \!\!\pi\!\!\text( )n)(3) \\\end(align) \right.\]

Nuances of the solution

In fact, this expression can be solved in many different ways, but it is the auxiliary angle method that is in this case optimal. In addition, using this design as an example, I would like to draw your attention to several more interesting techniques and facts:

• Formulas for reducing degrees. These formulas do not need to be memorized, but you need to know how to derive them, which is what I told you about today.
• Solving equations of the form $\cos \alpha =\cos \beta $.
• Adding a "zero".

But that's not all. Until now, $\sin $ and $\cos $, which we derived as an additional argument, we believed that they must be positive. Therefore, now we will solve more complex problems.

Analysis of more complex problems

Example #1

\[\sin 3x+4((\sin )^(3))x+4\cos x=5\]

Let's transform the first term:

\[\sin 3x=\sin \left(2x+x \right)=\sin 2x\cdot \cos x+\cos 2x\cdot \sin x\]

\[=2\left(1-\cos 2x \right)\cdot \sin x\]

Now let’s substitute all this into our original construction:

\[\sin 2x\cos x+\cos 2x\sin x+2\sin x-2\cos x\sin x+4\cos x=5\]

\[\sin 2x\cos x-\operatorname(cosx)-cos2\sin x+2\sin x+4\cos x=5\]

\[\sin \left(2x-x \right)+2\sin x+4\cos x=5\]

Let's introduce our amendment:

We write down:

\[\frac(3)(5)\sin x+\frac(4)(5)\cos x=1\]

Such $\alpha $ for which $\sin $ or $\cos $ would be equal to $\frac(3)(5)$ and $\frac(4)(5)$ in trigonometric table No. So let’s just write it like this and reduce the expression to the sine of the sum:

\[\sin x\cdot \cos \varphi +\cos x\cdot \sin \varphi =1\]

\[\sin \left(x+\varphi \right)=1\]

This special case, the simplest trigonometric construction:

It remains to find what $\varphi $ is equal to. This is where many students go wrong. The fact is that $\varphi $ is subject to two requirements:

\[\left\( \begin(align)& \cos \varphi =\frac(3)(5) \\& \sin \varphi =\frac(4)(5) \\\end(align) \right .\]

Let's draw a radar and see where such values ​​occur:

Returning to our expression, we write the following:

But this entry can be optimized a little. Because we know the following:

\[\alpha:\arcsin \alpha +\arccos \alpha =\frac(\text( )\!\!\pi\!\!\text( ))(\text(2)),\]

then in our case we can write it like this:

Example No. 2

This will require an even deeper understanding of solution techniques. standard tasks no trigonometry. But to solve this example we also use the auxiliary angle method.\[\]

The first thing that catches your eye is that there are no degrees higher than the first and therefore nothing can be expanded according to the formulas for decomposition of degrees. Use reverse calculations:

Why did I shell out $5$. Look here:

One unit according to basic trigonometric identity we can write it as $((\sin )^(2))x+((\cos )^(2))x$:

What does such a record give us? The fact is that the first bracket contains an exact square. Let's collapse it and get:

I suggest introducing a new variable:

\[\sin x+\cos x=t\]

In this case we will get the expression:

\[((t)_(1))=\frac(5+1)(4)=\frac(3)(2)\]

\[((t)_(2))=\frac(5-1)(4)=1\]

In total we get:

\[\left[ \begin(align)& \sin x+\cos x=\frac(3)(2) \\& \sin x+\cos x=1 \\\end(align) \right.\]

Of course knowledgeable students Now they will say that such constructions are easily solved by reducing them to a homogeneous structure. However, we will solve each equation using the auxiliary angle method. To do this, we first calculate the $l$ correction:

\[\sqrt(l)=\sqrt(2)\]

Let's divide everything by $\sqrt(2)$:

\[\left[ \begin(align)& \frac(\sqrt(2))(2)\sin x+\frac(\sqrt(2))(2)\cos x=\frac(3)(2\ sqrt(2)) \\& \frac(\sqrt(2))(2)\sin x+\frac(\sqrt(2))(2)\cos x=\frac(\sqrt(2))(2 ) \\\end(align) \right.\]

Let's reduce everything to $\cos $:

\[\cos x\cdot \cos \frac(\text( )\!\!\pi\!\!\text( ))(4)+\sin x\sin \frac(\text( )\!\ !\pi\!\!\text( ))(\text(4))\]

\[\left[ \begin(align)& \cos \left(x-\frac(\text( )\!\!\pi\!\!\text( ))(\text(4)) \right) =\frac(3)(2\sqrt(2)) \\& \cos \left(x-\frac(\text( )\!\!\pi\!\!\text( ))(4) \ right)=\frac(\sqrt(2))(2) \\\end(align) \right.\]

Let's look at each of these expressions.

The first equation has no roots, and to prove this fact, irrationality in the denominator will help us. Let's note the following:

\[\sqrt(2)<1,5\]

\[\frac(3)(2\sqrt(2))>\frac(3)(3\cdot 1.5)=\frac(3)(3)=1\]

In total, we have clearly proven that it is required that $\cos \left(x-\frac(\text( )\!\!\pi\!\!\text( ))(4) \right)$ be equal to the number, which is greater than “one” and, therefore, this construction has no roots.

Let's deal with the second one:

Let's solve this construction:

In principle, you can leave the answer like this, or you can write it down:

Important points

In conclusion, I would like to once again draw your attention to working with “ugly” arguments, i.e. when $\sin $ and $\cos $ are not table values. The problem is that if we say that in our equation $\frac(3)(5)$ is $\cos $ and $\frac(4)(5)$ is $\sin $, then in the end, after we decide on the design, we need to take both of these requirements into account. We get a system of two equations. If we do not take this into account, we will get the following situation. In this case, we will get two points and in place of $\varphi $ we will have two numbers: $\arcsin \frac(4)(5)$ and $-\arcsin \frac(4)(5)$, but the latter is us not satisfied in any way. The same will happen with the point $\frac(3)(5)$.

This problem only arises when we're talking about about “ugly” arguments. When we have table values, then there is nothing like that.

I hope today's lesson helped you understand what the auxiliary angle method is and how to apply it with examples different levels complexity. But this is not the only lesson devoted to solving problems using the auxiliary angle method. So stay tuned!

Formula for an additional (auxiliary) argument

Consider an expression of the form

in which the numbers and are not equal to zero at the same time. Let's multiply and divide each of the terms by and take out common multiplier outside of brackets:

It is easy to check that

which means, by Theorem 2, there is a real angle such that

Thus, using the sine of the sum formula, we get

where the angle such as and is called the auxiliary argument formula and is used when solving inhomogeneous problems linear equations and inequalities.

Inverse trigonometric functions

Definitions

So far we have solved the problem of determining trigonometric functions given angles. What if it's worth it? inverse problem: knowing any trigonometric function, determine the corresponding angle.

arcsine

Consider the expression where is the well-known real number. By definition, the sine is the ordinate of the point of intersection of the ray forming an angle with the abscissa axis and the trigonometric circle. Thus, to solve the equation, you need to find the intersection points of a straight line and a trigonometric circle.

Obviously, when a straight line and a circle do not have common points, which means the equation has no solutions. That is, it is impossible to find an angle whose sine would be greater than 1 in absolute value.

When, a straight line and a circle have points of intersection, for example, and (see figure). Thus, the given sine will have, and all angles differing from them by an integer amount full revolutions, i.e. , - an infinite number of angles. How to choose one angle among this infinite variety?

To uniquely determine the angle corresponding to the number, it is necessary to require the fulfillment of an additional condition: this angle must belong to the segment. This angle is called the arcsine of the number. angle trigonometric function identity

Arcsine real number is a real number whose sine is equal to. This number is designated.

arc cosine

Let us now consider an equation of the form. To solve it, it is necessary to find all points on the trigonometric circle that have an abscissa, i.e. points of intersection with a line. As in the previous case, the equation under consideration has no solutions. And if there are points of intersection of a straight line and a circle, corresponding infinite number corners, .

To uniquely determine the angle corresponding to given cosine, enter additional condition: this angle must belong to the segment; such an angle is called the arc cosine of the number.

arc cosine real number is a real number whose cosine is equal to. This number is designated.

Arctangent and arccotangent

Let's look at the expression. To solve it, you need to find on the circle all the points of intersection with the line, slope which equal to tangent the angle of inclination of the straight line to the positive direction of the x-axis. So direct in front of everyone real values crosses trigonometric circle at two points. These points are symmetrical about the origin and correspond to the angles, .

For unambiguous definition angle with a given tangent, it is selected from the interval.

Arctangent An arbitrary real number is a real number whose tangent is equal to. This number is designated.

To determine the arc tangent of an angle, similar reasoning is used, with the only difference being that the intersection of a circle with a straight line is considered and the angle is selected from the interval.

Arccotangent An arbitrary real number is a real number whose cotangent is equal to. This number is designated.

Properties of inverse trigonometric functions

Domain and Domain

Even/odd

Converting inverse trigonometric functions

To transform expressions containing inverse trigonometric functions, properties following from the definition of these functions are often used:

For any real number it holds

and vice versa:

Similarly for any real number it holds

and vice versa:

Graphs of trigonometric and inverse trigonometric functions

Graphs of trigonometric functions

Let's start by plotting a graph of a function on a segment. To do this, we will use the definition of sine on a trigonometric circle. Let's divide the trigonometric circle into (in this case 16) equal parts and place a coordinate system nearby, where the segment on the axis is also divided into equal parts. By drawing straight lines parallel to the axis through the dividing points of the circle, at the intersection of these lines with the perpendiculars restored from the corresponding dividing points on the axis, we obtain points whose coordinates, by definition, are equal to the sines of the corresponding angles. Drawing a smooth curve through these points, we obtain a graph of the function for. To obtain a graph of a function on the entire number line, use the periodicity of the sine: , .

To obtain the graph of the function, we will use the reduction formula. Thus, the graph of a function is obtained from the graph of a function by parallel transfer to the left by a length segment.

Using graphs of trigonometric functions provides another simple way to obtain reduction formulas. Let's look at a few examples.

Let's simplify the expression. On the axis we denote the angle and denote its sine and cosine as and respectively. Let's find the angle on the axis and restore the perpendicular to the intersection with the sine graph. It is obvious from the figure that.

Task: simplify the expression.

Let's move on to constructing a graph of the function. First, remember that for an angle, the tangent is the length of the segment AB. By analogy with constructing a sine graph, dividing the right semicircle into equal parts and plotting the resulting tangent values, we obtain the graph shown in the figure. For other values, the graph is obtained using the tangent periodicity property, .

The dotted lines on the graph represent asymptotes. Asymptote a curve is a straight line to which the curve approaches as close as desired when moving to infinity, but does not intersect it.

For a tangent, the asymptotes are straight lines, the appearance of which is associated with the conversion to zero at these points.

Using similar reasoning, a graph of the function is obtained. For it, the asymptotes are straight lines, . This graph can also be obtained using the reduction formula, i.e. transformation of symmetry about the axis and shift to the right.

Properties of trigonometric functions

Graphs of inverse trigonometric functions

First we introduce the concept of an inverse function.

If a function monotonically increases or decreases, then for it there exists inverse function. To construct a graph of the inverse function, the graph should be subjected to a symmetry transformation with respect to the straight line. The figures show an example of obtaining a graph of the inverse function.

Since the arcsine, arccosine, arctangent and arccotangent functions are the inverses of the sine, cosine, tangent and cotangent functions, respectively, their graphs are obtained by the transformation described above. The graphs of the original functions in the figures are shaded.

From the above figures, one of the main properties of inverse trigonometric functions is obvious: the sum of co-functions of the same number gives.

Lemma. If the sum of the squares of two real numbers is equal to one, then one of these numbers can be considered as a cosine, and the other as the sine of some angle.

In other words, if A 2 + b 2 = 1 , then there is an angle φ , such that

A = cosφ; b= sinφ.

Before proving this lemma, let us explain it in following example:

$$ (\frac(\sqrt3)(2))^2 + (\frac(1)(2)) = \frac(3)(4) + \frac(1)(4) = 1 $$

Therefore there is an angle φ , such that \(\frac(\sqrt3)(2) \) = cos φ ; 1 / 2 = sin φ .

As φ in this case, you can select any of the angles 30°, 30° ± 360°, 30° ± 2 360°, etc.

Proof of the lemma:

Consider a vector \(\vec(0A)\) with coordinates ( a, b ). Since A 2 + b 2 = 1 , the length of this vector is 1. But in this case its coordinates must be equal cos φ And sinφ, Where φ - the angle that forms given vector with the abscissa axis.

So, A = cosφ; b=sinφ, which was what needed to be proven.

The proven lemma allows us to transform the expression a sin x + b cos x to a form more convenient for study.

First of all, let’s take the expression \(\sqrt(a^2 + b^2)\) out of brackets

$$ a sinx + b cosx = \sqrt(a^2 + b^2)(\frac(a)(\sqrt(a^2 + b^2))sinx + \frac(b)(\sqrt(a ^2 + b^2))cosx) $$

Since

$$ (\frac(a)(\sqrt(a^2 + b^2)))^2 + (\frac(b)(\sqrt(a^2 + b^2)))^2 = 1 $ $

the first of the numbers \(\frac(a)(\sqrt(a^2 + b^2)) \) and \(\frac(b)(\sqrt(a^2 + b^2)) \) can be considered as the cosine of some angle φ , and the second - as the sine of the same angle φ :

$$ \frac(a)(\sqrt(a^2 + b^2)) = cos\phi, \;\; \frac(b)(\sqrt(a^2 + b^2)) = sin\phi $$

But in that case

a sin x + b cos x = \(\sqrt(a^2 + b^2)\)(cos φ sin x + sin φ cos x) = \(\sqrt(a^2 + b^2)\) sin (x + φ )

a sin x + b cos x = \(\sqrt(a^2 + b^2)\) sin (x + φ), where the angle φ is determined from the conditions

$$ sin\phi = \frac(b)(\sqrt(a^2 + b^2)) \;\; cos\phi = \frac(a)(\sqrt(a^2 + b^2)) $$

Examples.

1) \(sin x + cos x = \sqrt2 (\frac(1)(\sqrt2) sin x + \frac(1)(\sqrt2)cos x) = \sqrt2 (cos\frac(\pi)(4 )sin x + sin\frac(\pi)(4)cos x) =\\= \sqrt2(sinx + \frac(\pi)(4)) \)

The resulting formula sin x+cos x= \(\sqrt2(sinx + \frac(\pi)(4))\) useful to remember.

2) If one of the numbers A And b positive and the other negative, then the expression
a sin x + b cos x It is more convenient to convert not to the sine of the sum, but to the sine of the difference of two angles. So,

$$ 3sinx - 4cosx = \sqrt(9+16)(\frac(3)(\sqrt(9+16))sinx - \frac(4)(\sqrt(9+16))cosx) =\\= 5(sinx\cdot\frac(3)(5) - cosx\cdot\frac(4)(5)) = 5sin(x - \phi), $$

where under φ we can mean any angle that satisfies the following conditions:

cos φ = 3 / 5, sin φ = 4 / 5

In particular, one can put φ = arctan 4 / 3 . Then we get:

3 sin x - 4 cos x = 5 sin (x - arctan 4 / 3).

Elementary trigonometric equations are equations of the form, where --- one of the trigonometric functions: , .

Elementary trigonometric equations have an infinite number of roots. For example, the equation is satisfied following values: , etc. General formula along which all the roots of the equation are found, where is:

Here it can take any integer values, each of them corresponds to a specific root of the equation; in this formula (as well as in other formulas by which elementary trigonometric equations are solved) are called parameter. They are usually written to emphasize that the parameter can accept any integer values.

The solutions to the equation where are found by the formula

The equation is solved using the formula

and the equation is by the formula

Let us especially note some special cases of elementary trigonometric equations, when the solution can be written without using general formulas:

When solving trigonometric equations important role plays the period of trigonometric functions. Therefore, we present two useful theorems:

Theorem If --- basic period of the function, then the number is the main period of the function.

The periods of the functions and are said to be commensurable if there exist natural numbers So what.

Theorem If periodic functions and, have commensurate and, then they have general period, which is the period of the functions, .

The theorem states what is the period of a function, and is not necessarily the main period. For example, the main period of functions and --- , and the main period of their product --- .

Introducing an auxiliary argument

The standard way to transform expressions of the form is the following: let --- corner, given by the equalities, . For any, such an angle exists. Thus. If, or, in other cases.

Scheme for solving trigonometric equations

The basic scheme that we will follow when solving trigonometric equations is as follows:

solution given equation comes down to a decision elementary equations. Solutions --- conversions, factorization, replacement of unknowns. The guiding principle is not to lose your roots. This means that when going to to the following equation(equations) we are not afraid of the appearance of extra (extraneous) roots, but only care that each subsequent equation of our “chain” (or a set of equations in the case of branching) is a consequence of the previous one. One of possible methods root selection is a check. Let us immediately note that in the case of trigonometric equations, the difficulties associated with selecting roots and checking, as a rule, increase sharply compared to algebraic equations. After all, we have to check series consisting of infinite number members.

Special mention should be made of the replacement of unknowns when solving trigonometric equations. In most cases, after the necessary replacement, it turns out algebraic equation. Moreover, equations are not so rare that, although they are trigonometric in appearance, essentially they are not, since after the first step --- replacements variables --- turn into algebraic ones, and a return to trigonometry occurs only at the stage of solving elementary trigonometric equations.

Let us remind you once again: the replacement of the unknown should be done at the first opportunity; the resulting equation after the replacement must be solved to the end, including the stage of selecting the roots, and only then returned to the original unknown.

One of the features of trigonometric equations is that the answer in many cases can be written in various ways. Even to solve the equation, the answer can be written as follows:

1) in the form of two series: , ;

2) in standard form, which is a combination of the above series: , ;

3) since, the answer can be written in the form, . (In the future, the presence of a parameter, or in a response record, automatically means that this parameter accepts all possible integer values. Exceptions will be specified.)

Obviously, the three listed cases do not exhaust all the possibilities for writing the answer to the equation under consideration (there are infinitely many of them).

For example, when the equality is true. Therefore, in the first two cases, if, we can replace by.

Usually the answer is written based on point 2. It is useful to remember the following recommendation: if the work does not end with solving the equation, it is still necessary to conduct research and select roots, then the most convenient form of recording is indicated in point 1. (A similar recommendation should be given for the equation.)

Let's consider an example illustrating what has been said.

Example Solve the equation.

Solution. The most obvious way is the following. This equation splits into two: i. Solving each of them and combining the answers obtained, we will find.

Another way. Since, then, replacing and using the formulas for reducing the degree. After some minor transformations we get where.

At first glance, the second formula does not have any special advantages over the first. However, if we take, for example, it turns out that, i.e. the equation has a solution, while the first method leads us to the answer. “Seeing” and proving equality is not so easy.