Trigonometry. Reduction formulas.
Reduction formulas do not need to be taught; they need to be understood. Understand the algorithm for their derivation. It is very easy!
Let's take a unit circle and place all 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 clear that the function sin(a)>0
And function cos(a)>0
The first quarter can be described in terms of degrees, like (90-α) or (360+α).
In the second quarter it is clear that the function sin(a)>0, because the Y axis is positive in this quarter.
A function cos(a) because the X axis is negative in this quadrant.
The second quarter can be described in terms of degrees, like (90+α) or (180-α).
In the third quarter it is clear that the functions sin(a) The third quarter can be described in terms of degrees, like (180+α) or (270-α).
In the fourth quarter it is clear that the function sin(a) because the Y axis is negative in this quarter.
A function cos(a)>0, because the X axis is positive in this quarter.
The fourth quarter can be described in terms of degrees, like (270+α) or (360-α).
Now let's look at the reduction formulas themselves.
Let's remember simple algorithm:
1. Quarter.(Always look at what quarter you are in).
2. Sign.(For quarters, see positive or negative cosine or sine functions).
3. If you have (90° or π/2) and (270° or 3π/2) in brackets, then function changes.
And so we will begin to analyze this algorithm in quarters.
Find out what the expression cos(90-α) will be equal to
We reason according to the algorithm:
1. Quarter one.
Will cos(90-α) = sin(α)
Find out what the expression sin(90-α) will be equal to
We reason according to the algorithm:
1. Quarter one.
Will sin(90-α) = cos(α)
Find out what the expression cos(360+α) will be equal to
We reason according to the algorithm:
1. Quarter one.
2. In the first quarter, the sign of the cosine function is positive.
Will cos(360+α) = cos(α)
Find out what the expression sin(360+α) will be equal to
We reason according to the algorithm:
1. Quarter one.
2. In the first quarter, the sign of the sine function is positive.
3. There are no (90° or π/2) and (270° or 3π/2) in brackets, then the function does not change.
Will sin(360+α) = sin(α)
Find out what the expression cos(90+α) will be equal to
We reason according to the algorithm:
1. Quarter two.
3. There is (90° or π/2) in parentheses, then the function changes from cosine to sine.
Will cos(90+α) = -sin(α)
Find out what the expression sin(90+α) will be equal to
We reason according to the algorithm:
1. Quarter two.
3. There is (90° or π/2) in parentheses, then the function changes from sine to cosine.
Will sin(90+α) = cos(α)
Find out what the expression cos(180-α) will be equal to
We reason according to the algorithm:
1. Quarter two.
2. In the second quarter, the sign of the cosine function is negative.
3. There are no (90° or π/2) and (270° or 3π/2) in brackets, then the function does not change.
Will cos(180-α) = cos(α)
Find out what the expression sin(180-α) will be equal to
We reason according to the algorithm:
1. Quarter two.
2. In the second quarter, the sign of the sine function is positive.
3. There are no (90° or π/2) and (270° or 3π/2) in brackets, then the function does not change.
Will sin(180-α) = sin(α)
I’m talking about the third and fourth quarters, let’s create a table in a similar way:
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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.
And another problem B11 on the same topic - from the real Unified State Examination in mathematics.
Task. Find the meaning of the expression:
In this short video tutorial we will learn how to apply reduction formulas for solving real problems B11 from the Unified State Examination in mathematics. As you can see, we have two trigonometric expressions, each containing sines and cosines, as well as some pretty brutal numerical arguments.
Before solving these problems, let's remember what reduction formulas are. So, if we have expressions like:
Then we can get rid of the first term (of the form k · π/2) according to special rules. Let's draw a trigonometric circle and mark the main points on it: 0, π/2; π; 3π/2 and 2π. Then we look at the first term under the sign of the trigonometric function. We have:
- If the term we are interested in lies on the vertical axis of the trigonometric circle (for example: 3π/2; π/2, etc.), then the original function is replaced by a co-function: sine is replaced by cosine, and cosine, on the contrary, by sine.
- If our term lies on the horizontal axis, then the original function does not change. We simply remove the first term in the expression and that’s it.
Thus, we obtain a trigonometric function that does not contain terms of the form k · π/2. However, the work with reduction formulas does not end there. The fact is that our new function, obtained after “discarding” the first term, may have a plus or minus sign in front of it. How to identify this sign? Now we'll find out.
Let's imagine that the angle α remaining inside the trigonometric function after transformations has a very small degree measure. But what does “small measure” mean? Let's say α ∈ (0; 30°) - this is quite enough. Let's take an example of the function:
Then, following our assumptions that α ∈ (0; 30°), we conclude that the angle 3π/2 − α lies in the third coordinate quarter, i.e. 3π/2 − α ∈ (π; 3π/2). Let us remember the sign of the original function, i.e. y = sin x on this interval. Obviously, the sine in the third coordinate quarter is negative, since by definition, the sine is the ordinate of the end of the moving radius (in short, the sine is the y coordinate). Well, the y coordinate in the lower half-plane always takes negative values. This means that in the third quarter y is also negative.
Based on these reflections, we can write down the final expression:
Problem B11 - Option 1
These same techniques are quite suitable for solving problem B11 from the Unified State Examination in mathematics. The only difference is that in many real B11 problems, instead of a radian measure (i.e. numbers π, π/2, 2π, etc.) a degree measure is used (i.e. 90°, 180°, 270° and etc.). Let's look at the first task:
Let's look at the numerator first. cos 41° is a non-tabular value, so we can't do anything with it. Let's leave it like that for now.
Now let's look at the denominator:
sin 131° = sin (90° + 41°) = cos 41°
Obviously, this is a reduction formula, so the sine is replaced by a cosine. In addition, the angle 41° lies on the segment (0°; 90°), i.e. in the first coordinate quadrant - exactly as required to apply the reduction formulas. But then 90° + 41° is the second coordinate quarter. The original function y = sin x is positive there, so we put a plus sign in front of the cosine at the last step (in other words, we didn’t put anything).
It remains to deal with the last element:
cos 240° = cos (180° + 60°) = −cos 60° = −0.5
Here we see that 180° is the horizontal axis. Consequently, the function itself will not change: there was a cosine - and the cosine will also remain. But the question arises again: will plus or minus appear before the resulting expression cos 60°? Note that 180° is the third coordinate quarter. The cosine there is negative, therefore, the cosine will eventually have a minus sign in front of it. In total, we get the construction −cos 60° = −0.5 - this is a tabular value, so everything is easy to calculate.
Now we substitute the resulting numbers into the original formula and get:
As you can see, the number cos 41° in the numerator and denominator of the fraction is easily reduced, and the usual expression remains, which is equal to −10. In this case, the minus can either be taken out and placed in front of the fraction sign, or “kept” next to the second factor until the very last step of the calculations. In any case, the answer will be −10. That's it, problem B11 is solved!
Problem B14 - option 2
Let's move on to the second task. We have a fraction before us again:
Well, 27° lies in the first coordinate quarter, so we won’t change anything here. But sin 117° needs to be written (without any square for now):
sin 117° = sin (90° + 27°) = cos 27°
Obviously, before us again reduction formula: 90° is the vertical axis, therefore the sine will change to cosine. In addition, the angle α = 117° = 90° + 27° lies in the second coordinate quadrant. The original function y = sin x is positive there, therefore, after all the transformations, there is still a plus sign in front of the cosine. In other words, nothing is added there - we leave it like that: cos 27°.
We return to the original expression that needs to be calculated:
As we see, after the transformations, the main trigonometric identity arose in the denominator: sin 2 27° + cos 2 27° = 1. Total −4: 1 = −4 - so we found the answer to the second problem B11.
As you can see, with the help of reduction formulas such problems from the Unified State Examination in mathematics are solved literally in a couple of lines. No sine of the sum and cosine of the difference. All we need to remember is just the trigonometric circle.
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 assumed a 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.
Centered at a point A.
α
- angle expressed in radians.
Definition
Sine (sin α) is a trigonometric function depending on the angle α between the hypotenuse and the leg of a right triangle, equal to the ratio of the length of the opposite leg |BC| to the length of the hypotenuse |AC|.
Cosine (cos α) is a trigonometric function depending on the angle α between the hypotenuse and the leg of a right triangle, equal to the ratio of the length of the adjacent leg |AB| to the length of the hypotenuse |AC|.
Accepted notations
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Graph of the sine function, y = sin x
Graph of the cosine function, y = cos x
Properties of sine and cosine
Periodicity
Functions y = sin x and y = cos x periodic with period 2π.
Parity
The sine function is odd. The cosine function is even.
Domain of definition and values, extrema, increase, decrease
The sine and cosine functions are continuous in their domain of definition, that is, for all x (see proof of continuity). Their main properties are presented in the table (n - integer).
| y = sin x | y = cos x | |
| Scope and continuity | - ∞ < x < + ∞ | - ∞ < x < + ∞ |
| Range of values | -1 ≤ y ≤ 1 | -1 ≤ y ≤ 1 |
| Increasing | ||
| Descending | ||
| Maxima, y = 1 | ||
| Minima, y = - 1 | ||
| Zeros, y = 0 | ||
| Intercept points with the ordinate axis, x = 0 | y = 0 | y = 1 |
Basic formulas
Sum of squares of sine and cosine
Formulas for sine and cosine from sum and difference
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Formulas for the product of sines and cosines
Sum and difference formulas
Expressing sine through cosine
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Expressing cosine through sine
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Expression through tangent
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When , we have:
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At :
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Table of sines and cosines, tangents and cotangents
This table shows the values of sines and cosines for certain values of the argument. 
Expressions through complex variables
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Euler's formula
Expressions through hyperbolic functions
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Derivatives
; . Deriving formulas > > >
Derivatives of nth order:
{ -∞ <
x < +∞ }
Secant, cosecant
Inverse functions
The inverse functions of sine and cosine are arcsine and arccosine, respectively.
Arcsine, arcsin
Arccosine, arccos
References:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
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