At what points does the sine change to cosine. Casting formulas: proof, examples, mnemonic rule

Definition. Reduction formulas are called formulas that allow you to go from trigonometric functions of the form to functions of the 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, the casting formulas allow us to switch to working with angles within 90 degrees, which is undoubtedly very convenient.

Casting formulas:

There are two rules for using cast 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 as (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. The 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 main trigonometric functions depending on the quarter.

Example:

Calculate

Let's use the casting formulas:

Sin (150˚) is in the second quarter, according to the picture we see that the sin sign in this quarter is "+". This means that the reduced 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 of 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.

Lesson topic

Change in sine, cosine and tangent with increasing angle.

Lesson objectives

Get acquainted with new definitions and recall some already studied.
Get acquainted with the regularity of changes in the values of the sine cosine and tangent with increasing angle.
Developing - to develop students' attention, perseverance, perseverance, logical thinking, mathematical speech.
Educational - through the lesson to bring up an attentive attitude to each other, instill the ability to listen to comrades, mutual assistance, independence.

Lesson Objectives

Check student knowledge.

Lesson plan

Repetition of previously studied material.
Repetition tasks.
Change in sine, cosine and tangent with increasing angle.
Practical use.

Repetition of previously learned material

Let's start from the very beginning and remember what will be useful to refresh your memory. What are sine, cosine and tangent and which section of geometry these concepts relate to?

Trigonometry is such a complex Greek word: trigon is a triangle, metro is to measure. So in Greek it means: measured by triangles.

Subjects> Mathematics> Grade 8 Mathematics

Trigonometry. Reduction formulas.

Casting formulas do not need to be taught; they need to be understood. Understand the algorithm for their output. It is very easy!

Take the unit circle and place all the degree measures (0 °; 90 °; 180 °; 270 °; 360 °) on it.

Let us analyze the functions sin (a) and cos (a) in each quarter.

Remember that we look at the sin (a) function along the Y axis, and the cos (a) function along the X axis.

In the first quarter, it is seen that the function sin (a)> 0
And the function cos (a)> 0
The first quarter can be described in terms of a degree measure, like (90-α) or (360 + α).

In the second quarter, it is seen that the function sin (a)> 0 because the y-axis is positive in that quarter.
And the function cos (a) because the x-axis is negative in that quarter.
The second quarter can be described in terms of a degree measure like (90 + α) or (180-α).

In the third quarter, it is seen that the functions sin (a) The third quarter can be described in terms of a degree measure like (180 + α) or (270-α).

In the fourth quarter, it can be seen that the function sin (a) because the y-axis is negative in that quarter.
And the function cos (a)> 0 because the x-axis is positive in that quarter.
The fourth quarter can be described in terms of a degree measure like (270 + α) or (360-α).

Now let's consider the reduction formulas themselves.

Let's remember a simple algorithm:
1. Fourth.(Always look in which quarter you are).
2. Sign.(For a quarter, see the positive or negative cosine or sine functions.)
3. If you have (90 ° or π / 2) and (270 ° or 3π / 2) in parentheses, then function changes.

And so we begin to disassemble this algorithm in quarters.

Find out what will be equal to the expression cos (90-α)
We argue according to the algorithm:
1. Quarter one.

Will be cos (90-α) = sin (α)

Find out what will be equal to the expression sin (90-α)
We argue according to the algorithm:
1. Quarter one.

Will be sin (90-α) = cos (α)

Find out what will be equal to the expression cos (360 + α)
We argue according to the algorithm:
1. Quarter one.
2. In the first quarter, the sign of the cosine function is positive.

Will be cos (360 + α) = cos (α)

Find out what will be equal to the expression sin (360 + α)
We argue according to the algorithm:
1. Quarter one.
2. In the first quarter, the sign of the sine function is positive.
3. There is no (90 ° or π / 2) and (270 ° or 3π / 2) in brackets, the function does not change.
Will be sin (360 + α) = sin (α)

Find out what will be equal to the expression cos (90 + α)
We argue according to the algorithm:
1. Quarter two.

3. In parentheses there is (90 ° or π / 2), then the function changes from cosine to sine.
Will be cos (90 + α) = -sin (α)

Find out what will be equal to the expression sin (90 + α)
We argue according to the algorithm:
1. Quarter two.

3. In parentheses there is (90 ° or π / 2), then the function changes from sine to cosine.
Will be sin (90 + α) = cos (α)

Find out what will be equal to the expression cos (180-α)
We argue according to the algorithm:
1. Quarter two.
2. In the second quarter, the sign of the cosine function is negative.
3. There is no (90 ° or π / 2) and (270 ° or 3π / 2) in brackets, the function does not change.
Will be cos (180-α) = cos (α)

Find out what will be equal to the expression sin (180-α)
We argue according to the algorithm:
1. Quarter two.
2. In the second quarter, the sign of the sine function is positive.
3. There is no (90 ° or π / 2) and (270 ° or 3π / 2) in brackets, the function does not change.
Will be sin (180-α) = sin (α)

I argue about the third and fourth quarters in a similar way, let's make a table:

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Casting formulas are ratios 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 in the first quarter of the unit circle. Thus, the reduction formulas "lead" us to work 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 for the exam, exams, tests. But let us warn you right away 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 deduce the necessary equality at the right time.

First, let's write down all the casting 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 the angles are written in radians:

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

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

Mnemonic rule of reduction formulas or how to remember them

As we mentioned, you don't need to memorize all of the above relationships. If you looked at them carefully, you probably noticed some patterns. They allow us to formulate a mnemonic rule (mnemonic - to remember), with the help of which you can easily get any of the reduction formulas.

Immediately, we note that to apply this rule, you need to be able to determine (or remember) the signs of trigonometric functions in different quarters of the unit circle.
The privil itself contains 3 stages:

1. The function argument must be in the form `\ 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 the arguments `\ frac (\ pi) 2 \ pm \ alpha`,` \ frac (3 \ pi) 2 \ pm \ alpha`, the trigonometric function of the expression being converted is changed to a cofunction, that is, the opposite (sine to cosine, tangent to cotangent, and vice versa). For the arguments `\ pi \ pm \ alpha`,` 2 \ pi \ pm \ alpha`, the function is unchanged.
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 III quarter, the cosine in this quarter has a“ - ”sign, so the transformed function will also be with 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 be with 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)) `. We represent `3 \ pi` as` 2 \ pi + \ pi`. `2 \ pi` - function period.

Important: The functions `cos \ alpha` and` sin \ alpha` have periods 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 above mnemonic rule is also called the equine rule of reduction formulas. I wonder why equine?

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` on the horizontal abscissa, and `\ frac (\ pi) 2` and` \ frac (3 \ pi) 2` on the vertical ordinate.

We ask ourselves the question: "Does a function change to a co-function?" 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 head to the sides. And for corners with key points located on the vertical axis, we answer "yes", nodding our head from top to bottom, like a horse 🙂

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

Practical examples of using cast formulas

Application of formulas of reduction begins in the 9th and 10th grade. A lot of tasks with their use were taken out on the exam. Here are some of the tasks where you will need to apply these formulas:

tasks for solving a right-angled triangle;
converting numeric and alphabetic trigonometric expressions, calculating their values;
stereometric tasks.

Example 1. Calculate using the 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. Expressing the cosine in terms of the sine using the 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

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.

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

Coming out of the definition of tangent and cotangent, we get `tg (\ frac (\ pi) 2 + \ alpha) = \ frac (sin (\ frac (\ pi) 2 + \ alpha)) (cos (\ frac (\ pi) 2 + \ alpha)) = \ frac (cos \ alpha) (- sin \ alpha) = - ctg \ alpha` and `ctg (\ 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 the tangent and the cotangent of the angle `\ frac (\ pi) 2 + \ alpha`.

To prove formulas with 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.

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

There are two rules for using cast formulas.

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

2. The rule "what you were, so you stayed."

The 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 main trigonometric functions depending on the quarter.

Evaluate Sin (150˚)

Let's use the casting formulas:

Sin (150˚) is in the second quarter, according to the figure, we see that the sin sign in this quarter is +. This means that the given function will also have a plus sign. We applied the second rule.

If desired, all reduction formulas can be summarized in one table. But it's still easier to remember these two rules and use them.

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