Trigonometry reduction table. Fundamental trigonometric identity

In this article we will take a comprehensive look. Basic trigonometric identities are equalities that establish a connection between the sine, cosine, tangent and cotangent of one angle, and allow one to find any of these trigonometric functions through a known other.

Let us immediately list the main trigonometric identities that we will analyze in this article. Let's write them down in a table, and below we'll give the output of these formulas and provide the necessary explanations.

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Relationship between sine and cosine of one angle

Sometimes they do not talk about the main trigonometric identities listed in the table above, but about one single basic trigonometric identity type . The explanation for this fact is quite simple: the equalities are obtained from the basic trigonometric identity after dividing both its parts by and respectively, and the equality And follow from the definitions of sine, cosine, tangent and cotangent. We'll talk about this in more detail in the following paragraphs.

That is, it is the equality that is of particular interest, which was given the name of the main trigonometric identity.

Before proving the main trigonometric identity, we give its formulation: the sum of the squares of the sine and cosine of one angle is identically equal to one. Now let's prove it.

The basic trigonometric identity is very often used when converting trigonometric expressions. It allows the sum of the squares of the sine and cosine of one angle to be replaced by one. No less often, the basic trigonometric identity is used in the reverse order: unit is replaced by the sum of the squares of the sine and cosine of any angle.

Tangent and cotangent through sine and cosine

Identities connecting tangent and cotangent with sine and cosine of one angle of view and follow immediately from the definitions of sine, cosine, tangent and cotangent. Indeed, by definition, sine is the ordinate of y, cosine is the abscissa of x, tangent is the ratio of the ordinate to the abscissa, that is, , and the cotangent is the ratio of the abscissa to the ordinate, that is, .

Thanks to such obviousness of the identities and Tangent and cotangent are often defined not through the ratio of abscissa and ordinate, but through the ratio of sine and cosine. So the tangent of an angle is the ratio of the sine to the cosine of this angle, and the cotangent is the ratio of the cosine to the sine.

In conclusion of this paragraph, it should be noted that the identities and take place for all angles at which the trigonometric functions included in them make sense. So the formula is valid for any , other than (otherwise the denominator will have zero, and we did not define division by zero), and the formula - for all , different from , where z is any .

Relationship between tangent and cotangent

An even more obvious trigonometric identity than the previous two is the identity connecting the tangent and cotangent of one angle of the form . It is clear that it holds for any angles other than , otherwise either the tangent or the cotangent are not defined.

Proof of the formula very simple. By definition and from where . The proof could have been carried out a little differently. Since , That .

So, the tangent and cotangent of the same angle at which they make sense are .

This is the last and most main lesson, necessary to solve problems B11. We already know how to convert angles from a radian measure to a degree measure (see the lesson “Radian and degree measure of an angle”), and we also know how to determine the sign of a trigonometric function, focusing on the coordinate quarters (see the lesson “Signs of trigonometric functions”).

The only thing left to do is calculate the value of the function itself - the very number that is written in the answer. This is where the basic trigonometric identity comes to the rescue.

Basic trigonometric identity. For any angle α the following statement is true:

sin 2 α + cos 2 α = 1.

This formula relates the sine and cosine of one angle. Now, knowing the sine, we can easily find the cosine - and vice versa. It is enough to take the square root:

Note the "±" sign in front of the roots. The fact is that from the basic trigonometric identity it is not clear what the original sine and cosine were: positive or negative. After all, squaring - even function, which “burns” all the disadvantages (if there were any).

That is why in all problems B11, which are found in the Unified State Examination in mathematics, there are necessarily additional conditions that help get rid of uncertainty with signs. Usually this is an indication of the coordinate quarter, by which the sign can be determined.

An attentive reader will probably ask: “What about tangent and cotangent?” It is impossible to directly calculate these functions from the above formulas. However, there are important consequences from the basic trigonometric identity, which already contains tangents and cotangents. Namely:

An important corollary: for any angle α, the basic trigonometric identity can be rewritten as follows:

These equations are easily derived from the main identity - it is enough to divide both sides by cos 2 α (to obtain the tangent) or by sin 2 α (to obtain the cotangent).

Let's look at all this at specific examples. Below are the real B11 problems which are taken from the mock ones Unified State Exam options in mathematics 2012.

We know the cosine, but we don't know the sine. The main trigonometric identity (in its “pure” form) connects just these functions, so we will work with it. We have:

sin 2 α + cos 2 α = 1 ⇒ sin 2 α + 99/100 = 1 ⇒ sin 2 α = 1/100 ⇒ sin α = ±1/10 = ±0.1.

To solve the problem, it remains to find the sign of the sine. Since the angle α ∈ (π /2; π ), then in degree measure it is written as follows: α ∈ (90°; 180°).

Consequently, angle α lies in the II coordinate quarter - all sines there are positive. Therefore sin α = 0.1.

So, we know the sine, but we need to find the cosine. Both of these functions are in the basic trigonometric identity. Let's substitute:

sin 2 α + cos 2 α = 1 ⇒ 3/4 + cos 2 α = 1 ⇒ cos 2 α = 1/4 ⇒ cos α = ±1/2 = ±0.5.

It remains to deal with the sign in front of the fraction. What to choose: plus or minus? By condition, angle α belongs to the interval (π 3π /2). Let's convert the angles from radian measures to degrees - we get: α ∈ (180°; 270°).

Obviously, this is the III coordinate quarter, where all cosines are negative. Therefore cos α = −0.5.

Task. Find tan α if the following is known:

Tangent and cosine are related by the equation following from the basic trigonometric identity:

We get: tan α = ±3. The sign of the tangent is determined by the angle α. It is known that α ∈ (3π /2; 2π ). Let's convert the angles from radian measures to degrees - we get α ∈ (270°; 360°).

Obviously, this is the IV coordinate quarter, where all tangents are negative. Therefore tan α = −3.

Task. Find cos α if the following is known:

Again the sine is known and the cosine is unknown. Let us write down the main trigonometric identity:

sin 2 α + cos 2 α = 1 ⇒ 0.64 + cos 2 α = 1 ⇒ cos 2 α = 0.36 ⇒ cos α = ±0.6.

The sign is determined by the angle. We have: α ∈ (3π /2; 2π ). Let's convert the angles from degrees to radians: α ∈ (270°; 360°) is the IV coordinate quarter, the cosines there are positive. Therefore, cos α = 0.6.

Task. Find sin α if the following is known:

Let us write down a formula that follows from the basic trigonometric identity and directly connects sine and cotangent:

From here we get that sin 2 α = 1/25, i.e. sin α = ±1/5 = ±0.2. It is known that angle α ∈ (0; π /2). In degree measure, this is written as follows: α ∈ (0°; 90°) - I coordinate quarter.

So, the angle is in the I coordinate quadrant - all trigonometric functions there are positive, so sin α = 0.2.

Definition. Reduction formulas are formulas that allow you to move from trigonometric functions of the form to functions of argument. With their help, the sine, cosine, tangent and cotangent of an arbitrary angle can be reduced to the sine, cosine, tangent and cotangent of an angle from the interval from 0 to 90 degrees (from 0 to radians). Thus, reduction formulas allow us to move on to working with angles within 90 degrees, which is undoubtedly very convenient.

Reduction formulas:

There are two rules for using reduction formulas.

1. If the angle can be represented as (π/2 ±a) or (3*π/2 ±a), then function name changes sin to cos, cos to sin, tg to ctg, ctg to tg. If the angle can be represented in the form (π ±a) or (2*π ±a), then The function name remains unchanged.

Look at the picture below, it shows schematically when to change the sign and when not

2. Sign of the reduced function remains the same. If the original function had a plus sign, then the reduced function also has a plus sign. If the original function had a minus sign, then the reduced function also has a minus sign.

The figure below shows the signs of the basic trigonometric functions depending on the quarter.

Example:

Calculate

Let's use the reduction formulas:

Sin(150˚) is in the second quarter; from the figure we see that the sin sign in this quarter is equal to “+”. This means that the given function will also have a “+” sign. We applied the second rule.

Now 150˚ = 90˚ +60˚. 90˚ is π/2. That is, we are dealing with the case π/2+60, therefore, according to the first rule, we change the function from sin to cos. As a result, we get Sin(150˚) = cos(60˚) = ½.

Reduction formulas are relationships that allow you to go from sine, cosine, tangent and cotangent with angles `\frac (\pi)2 \pm \alpha`, `\pi \pm \alpha`, `\frac (3\pi) 2 \pm \alpha`, `2\pi \pm \alpha` to the same functions of the angle `\alpha`, which is located in the first quarter of the unit circle. Thus, the reduction formulas “lead” us to working with angles in the range from 0 to 90 degrees, which is very convenient.

All together there are 32 reduction formulas. They will undoubtedly come in handy during the Unified State Exam, exams, and tests. But let us immediately warn you that there is no need to memorize them! You need to spend a little time and understand the algorithm for their application, then it will not be difficult for you to right moment derive the necessary equality.

First, let's write down all the reduction formulas:

For angle (`\frac (\pi)2 \pm \alpha`) or (`90^\circ \pm \alpha`):

`sin(\frac (\pi)2 - \alpha)=cos \ \alpha;` ` sin(\frac (\pi)2 + \alpha)=cos \ \alpha`
`cos(\frac (\pi)2 - \alpha)=sin \ \alpha;` ` cos(\frac (\pi)2 + \alpha)=-sin \ \alpha`
`tg(\frac (\pi)2 — \alpha)=ctg \ \alpha;` ` tg(\frac (\pi)2 + \alpha)=-ctg \ \alpha`
`ctg(\frac (\pi)2 — \alpha)=tg \ \alpha;` ` ctg(\frac (\pi)2 + \alpha)=-tg \ \alpha`

For angle (`\pi \pm \alpha`) or (`180^\circ \pm \alpha`):

`sin(\pi - \alpha)=sin \ \alpha;` ` sin(\pi + \alpha)=-sin \ \alpha`
`cos(\pi - \alpha)=-cos \ \alpha;` ` cos(\pi + \alpha)=-cos \ \alpha`
`tg(\pi - \alpha)=-tg \ \alpha;` ` tg(\pi + \alpha)=tg \ \alpha`
`ctg(\pi - \alpha)=-ctg \ \alpha;` ` ctg(\pi + \alpha)=ctg \ \alpha`

For angle (`\frac (3\pi)2 \pm \alpha`) or (`270^\circ \pm \alpha`):

`sin(\frac (3\pi)2 — \alpha)=-cos \ \alpha;` ` sin(\frac (3\pi)2 + \alpha)=-cos \ \alpha`
`cos(\frac (3\pi)2 — \alpha)=-sin \ \alpha;` ` cos(\frac (3\pi)2 + \alpha)=sin \ \alpha`
`tg(\frac (3\pi)2 — \alpha)=ctg \ \alpha;` ` tg(\frac (3\pi)2 + \alpha)=-ctg \ \alpha`
`ctg(\frac (3\pi)2 — \alpha)=tg \ \alpha;` ` ctg(\frac (3\pi)2 + \alpha)=-tg \ \alpha`

For angle (`2\pi \pm \alpha`) or (`360^\circ \pm \alpha`):

`sin(2\pi - \alpha)=-sin \ \alpha;` ` sin(2\pi + \alpha)=sin \ \alpha`
`cos(2\pi - \alpha)=cos \ \alpha;` ` cos(2\pi + \alpha)=cos \ \alpha`
`tg(2\pi - \alpha)=-tg \ \alpha;` ` tg(2\pi + \alpha)=tg \ \alpha`
`ctg(2\pi - \alpha)=-ctg \ \alpha;` ` ctg(2\pi + \alpha)=ctg \ \alpha`

You can often find reduction formulas in the form of a table where angles are written in radians:

To use it, we need to select the row with the function we need and the column with the desired argument. For example, to find out using a table what ` sin(\pi + \alpha)` will be equal to, it is enough to find the answer at the intersection of the row ` sin \beta` and the column ` \pi + \alpha`. We get ` sin(\pi + \alpha)=-sin \ \alpha`.

And the second, similar table, where angles are written in degrees:

Mnemonic rule for reduction formulas or how to remember them

As we already mentioned, there is no need to memorize all the above relationships. If you looked at them carefully, you probably noticed some patterns. They allow us to formulate a mnemonic rule (mnemonic - remember), with the help of which we can easily obtain any reduction formula.

Let us immediately note that to apply this rule you need to be good at identifying (or remembering) the signs of trigonometric functions in different quarters of the unit circle.
The vaccine itself contains 3 stages:

1. 1. The function argument must be represented as `\frac (\pi)2 \pm \alpha`, `\pi \pm \alpha`, `\frac (3\pi)2 \pm \alpha`, `2\pi \ pm \alpha`, and `\alpha` is required sharp corner(from 0 to 90 degrees).
   2. For arguments `\frac (\pi)2 \pm \alpha`, `\frac (3\pi)2 \pm \alpha` trigonometric function the expression being converted changes to a cofunction, that is, the opposite (sine to cosine, tangent to cotangent and vice versa). For arguments `\pi \pm \alpha`, `2\pi \pm \alpha` the function does not change.
   3. The sign of the original function is determined. The resulting function on the right side will have the same sign.

To see how this rule can be applied in practice, let’s transform several expressions:

1. `cos(\pi + \alpha)`.

The function is not reversed. The angle `\pi + \alpha` is in the third quarter, the cosine in this quarter has a “-” sign, so the transformed function will also have a “-” sign.

Answer: ` cos(\pi + \alpha)= - cos \alpha`

2. `sin(\frac (3\pi)2 - \alpha)`.

According to the mnemonic rule, the function will be reversed. The angle `\frac (3\pi)2 - \alpha` is in the third quarter, the sine here has a “-” sign, so the result will also have a “-” sign.

Answer: `sin(\frac (3\pi)2 - \alpha)= - cos \alpha`

3. `cos(\frac (7\pi)2 - \alpha)`.

`cos(\frac (7\pi)2 - \alpha)=cos(\frac (6\pi)2+\frac (\pi)2-\alpha)=cos (3\pi+(\frac(\pi )2-\alpha))`. Let's represent `3\pi` as `2\pi+\pi`. `2\pi` is the period of the function.

Important: The functions `cos \alpha` and `sin \alpha` have a period of `2\pi` or `360^\circ`, their values ​​will not change if the argument is increased or decreased by these values.

Based on this, our expression can be written as follows: `cos (\pi+(\frac(\pi)2-\alpha)`. Applying the mnemonic rule twice, we get: `cos (\pi+(\frac(\pi) 2-\alpha)= - cos (\frac(\pi)2-\alpha)= - sin \alpha`.

Answer: `cos(\frac (7\pi)2 - \alpha)=- sin \alpha`.

Horse rule

The second point of the mnemonic rule described above is also called the horse rule of reduction formulas. I wonder why horses?

So, we have functions with arguments `\frac (\pi)2 \pm \alpha`, `\pi \pm \alpha`, `\frac (3\pi)2 \pm \alpha`, `2\pi \ pm \alpha`, points `\frac (\pi)2`, `\pi`, `\frac (3\pi)2`, `2\pi` are key, they are located on the coordinate axes. `\pi` and `2\pi` are on the horizontal x-axis, and `\frac (\pi)2` and `\frac (3\pi)2` are on vertical axis ordinate

We ask ourselves the question: “Does a function change into a cofunction?” To answer this question, you need to move your head along the axis on which the key point is located.

That is, for arguments with key points located on the horizontal axis, we answer “no” by shaking our heads to the sides. And for corners with key points located on the vertical axis, we answer “yes” by nodding our heads from top to bottom, like a horse :)

We recommend watching a video tutorial in which the author explains in detail how to remember reduction formulas without memorizing them.

Practical examples of using reduction formulas

The use of reduction formulas begins in grades 9 and 10. Many problems using them were submitted to the Unified State Exam. Here are some of the problems where you will have to apply these formulas:

• problems to solve a right triangle;
• transformation of numeric and alphabetic trigonometric expressions, calculation of their values;
• stereometric tasks.

Example 1. Calculate using reduction formulas a) `sin 600^\circ`, b) `tg 480^\circ`, c) `cos 330^\circ`, d) `sin 240^\circ`.

Solution: a) `sin 600^\circ=sin (2 \cdot 270^\circ+60^\circ)=-cos 60^\circ=-\frac 1 2`;

b) `tg 480^\circ=tg (2 \cdot 270^\circ-60^\circ)=ctg 60^\circ=\frac(\sqrt 3)3`;

c) `cos 330^\circ=cos (360^\circ-30^\circ)=cos 30^\circ=\frac(\sqrt 3)2`;

d) `sin 240^\circ=sin (270^\circ-30^\circ)=-cos 30^\circ=-\frac(\sqrt 3)2`.

Example 2. Having expressed cosine through sine using reduction formulas, compare the numbers: 1) `sin \frac (9\pi)8` and `cos \frac (9\pi)8`; 2) `sin \frac (\pi)8` and `cos \frac (3\pi)10`.

Solution: 1)`sin \frac (9\pi)8=sin (\pi+\frac (\pi)8)=-sin \frac (\pi)8`

`cos \frac (9\pi)8=cos (\pi+\frac (\pi)8)=-cos \frac (\pi)8=-sin \frac (3\pi)8`

`-sin \frac (\pi)8> -sin \frac (3\pi)8`

`sin \frac (9\pi)8>cos \frac (9\pi)8`.

2) `cos \frac (3\pi)10=cos (\frac (\pi)2-\frac (\pi)5)=sin \frac (\pi)5`

`sin \frac (\pi)8

`sin \frac (\pi)8

Let us first prove two formulas for the sine and cosine of the argument `\frac (\pi)2 + \alpha`: ` sin(\frac (\pi)2 + \alpha)=cos \ \alpha` and ` cos(\frac (\ pi)2 + \alpha)=-sin \\alpha`. The rest are derived from them.

Let's take a unit circle and point A on it with coordinates (1,0). Let after turning to angle `\alpha` it will go to point `A_1(x, y)`, and after turning by angle `\frac (\pi)2 + \alpha` to point `A_2(-y, x)`. Dropping the perpendiculars from these points to the line OX, we see that the triangles `OA_1H_1` and `OA_2H_2` are equal, since their hypotenuses and adjacent angles are equal. Then, based on the definitions of sine and cosine, we can write `sin \alpha=y`, `cos \alpha=x`, `sin(\frac (\pi)2 + \alpha)=x`, `cos(\frac (\ pi)2 + \alpha)=-y`. Where can we write that ` sin(\frac (\pi)2 + \alpha)=cos \alpha` and ` cos(\frac (\pi)2 + \alpha)=-sin \alpha`, which proves the reduction formulas for sine and cosine angles `\frac (\pi)2 + \alpha`.

Coming from the definition of tangent and cotangent, we obtain ` tan(\frac (\pi)2 + \alpha)=\frac (sin(\frac (\pi)2 + \alpha))(cos(\frac (\pi)2 + \alpha))=\frac (cos \alpha)(-sin \alpha)=-ctg \alpha` and ` сtg(\frac (\pi)2 + \alpha)=\frac (cos(\frac (\ pi)2 + \alpha))(sin(\frac (\pi)2 + \alpha))=\frac (-sin \alpha)(cos \alpha)=-tg \alpha`, which proves the reduction formulas for tangent and the cotangent of the angle `\frac (\pi)2 + \alpha`.

To prove formulas with the argument `\frac (\pi)2 - \alpha`, it is enough to represent it as `\frac (\pi)2 + (-\alpha)` and follow the same path as above. For example, `cos(\frac (\pi)2 - \alpha)=cos(\frac (\pi)2 + (-\alpha))=-sin(-\alpha)=sin(\alpha)`.

The angles `\pi + \alpha` and `\pi - \alpha` can be represented as `\frac (\pi)2 +(\frac (\pi)2+\alpha)` and `\frac (\pi) 2 +(\frac (\pi)2-\alpha)` respectively.

And `\frac (3\pi)2 + \alpha` and `\frac (3\pi)2 - \alpha` as `\pi +(\frac (\pi)2+\alpha)` and `\pi +(\frac (\pi)2-\alpha)`.