Reduction formulas with pi. Formulas for reducing trigonometric functions

How to remember formulas for reducing trigonometric functions? It's easy if you use an association. This association was not invented by me. As already mentioned, a good association should “catch”, that is, evoke vivid emotions. I cannot call the emotions caused by this association positive. But it gives a result - it allows you to remember reduction formulas, which means it has the right to exist. After all, if you don't like it, you don't have to use it, right?

The reduction formulas have the form: sin(πn/2±α), cos(πn/2±α), tg(πn/2±α), ctg(πn/2±α). Remember that +α gives counterclockwise movement, - α gives clockwise movement.

To work with reduction formulas, you need two points:

1) put the sign that the initial function has (in textbooks they write: reducible. But in order not to get confused, it is better to call it initial), if we consider α to be the angle of the first quarter, that is, small.

2) Horizontal diameter - π±α, 2π±α, 3π±α... - in general, when there is no fraction, the name of the function does not change. Vertical π/2±α, 3π/2±α, 5π/2±α... - when there is a fraction, the name of the function changes: sine - to cosine, cosine - to sine, tangent - to cotangent and cotangent - to tangent.

Now, actually, the association:

vertical diameter (there is a fraction) -

standing drunk. What will happen to him early?

or is it too late? That's right, it will fall.

The function name will change.

If the diameter is horizontal, the drunk is already lying down. He's probably asleep. Nothing will happen to him, he has already accepted horizontal position. Accordingly, the name of the function does not change.

That is, sin(π/2±α), sin(3π/2±α), sin(5π/2±α), etc. give ±cosα,

and sin(π±α), sin(2π±α), sin(3π±α), … - ±sinα.

We already know how.

How it works? Let's look at examples.

1) cos(π/2+α)=?

We become π/2. Since +α means we go forward, counterclockwise. We find ourselves in the second quarter, where the cosine has a “-“ sign. The name of the function changes (“a drunk person is standing”, which means he will fall). So,

cos(π/2+α)=-sin α.

Let's get to 2π. Since -α - we go backwards, that is, clockwise. We find ourselves in the IV quarter, where the tangent has a “-“ sign. The name of the function does not change (the diameter is horizontal, “the drunk is already lying down”). Thus, tan(2π-α)=- tanα.

3) ctg²(3π/2-α)=?

Examples in which a function is raised to an even power are even simpler to solve. The even degree “-” removes it, that is, you just need to find out whether the name of the function changes or remains. The diameter is vertical (there is a fraction, “standing drunk”, it will fall), the name of the function changes. We get: ctg²(3π/2-α)= tan²α.

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 derive the necessary equality at the right time.

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 necessarily an acute angle (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;
• numeric and alphabetic conversions 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)`.

They belong to the trigonometry section of mathematics. Their essence is to reduce trigonometric functions of angles to a “simple” form. Much can be written about the importance of knowing them. There are already 32 of these formulas!

Don’t be alarmed, you don’t need to learn them, like many other formulas in a math course. There is no need to fill your head with unnecessary information, you need to remember the “keys” or laws, and remembering or deriving the required formula will not be a problem. By the way, when I write in articles “... you need to learn!!!” - this means that it really needs to be learned.

If you are not familiar with reduction formulas, then the simplicity of their derivation will pleasantly surprise you - there is a “law” with the help of which this can be easily done. And you can write any of the 32 formulas in 5 seconds.

I will list only some of the problems that will appear on the Unified State Exam in mathematics, where without knowledge of these formulas there is a high probability of failing in solving them. For example:

– problems for solving a right triangle, where we are talking about the external angle, and problems for internal angles, some of these formulas are also necessary.

– tasks on calculating the values ​​of trigonometric expressions; converting numerical trigonometric expressions; converting literal trigonometric expressions.

– problems on the tangent and the geometric meaning of the tangent; a reduction formula for the tangent is required, as well as other problems.

– stereometric problems, in the course of solving it is often necessary to determine the sine or cosine of an angle that lies in the range from 90 to 180 degrees.

And these are just those points that relate to the Unified State Exam. And in the algebra course itself there are many problems, the solution of which simply cannot be done without knowledge of reduction formulas.

So what does this lead to and how do the specified formulas make it easier for us to solve problems?

For example, you need to determine the sine, cosine, tangent, or cotangent of any angle from 0 to 450 degrees:

the alpha angle ranges from 0 to 90 degrees

* * *

So, it is necessary to understand the “law” that works here:

1. Determine the sign of the function in the corresponding quadrant.

Let me remind you:

2. Remember the following:

function changes to cofunction

function does not change to cofunction

What does the concept mean - a function changes to a cofunction?

Answer: sine changes to cosine or vice versa, tangent to cotangent or vice versa.

That's all!

Now, according to the presented law, we will write down several reduction formulas ourselves:

This angle lies in the third quarter, the cosine in the third quarter is negative. We don’t change the function to a cofunction, since we have 180 degrees, which means:

The angle lies in the first quarter, the sine in the first quarter is positive. We do not change the function to a cofunction, since we have 360 ​​degrees, which means:

Here is another additional confirmation that the sines of adjacent angles are equal:

The angle lies in the second quarter, the sine in the second quarter is positive. We do not change the function to a cofunction, since we have 180 degrees, which means:

Work through each formula mentally or in writing, and you will be convinced that there is nothing complicated.

***

In the article on the solution, the following fact was noted - the sine of one acute angle in a right triangle is equal to the cosine of another acute angle in it.