Formulas for reducing the addition of a double angle. Reduction formulas
Trigonometric identities- these are equalities that establish a connection between sine, cosine, tangent and cotangent of one angle, which allows you to find any of these functions, provided that any other is known.
tg \alpha = \frac(\sin \alpha)(\cos \alpha), \enspace ctg \alpha = \frac(\cos \alpha)(\sin \alpha)
tg \alpha \cdot ctg \alpha = 1
This identity says that the sum of the square of the sine of one angle and the square of the cosine of one angle is equal to one, which in practice makes it possible to calculate the sine of one angle when its cosine is known and vice versa.
When converting trigonometric expressions, this identity is very often used, which allows you to replace the sum of the squares of the cosine and sine of one angle with one and also perform the replacement operation in the reverse order.
Finding tangent and cotangent using sine and cosine
tg \alpha = \frac(\sin \alpha)(\cos \alpha),\enspace
These identities are formed from the definitions of sine, cosine, tangent and cotangent. After all, if you look at it, then by definition the ordinate y is a sine, and the abscissa x is a cosine. Then the tangent will be equal to the ratio \frac(y)(x)=\frac(\sin \alpha)(\cos \alpha), and the ratio \frac(x)(y)=\frac(\cos \alpha)(\sin \alpha)- will be a cotangent.
Let us add that only for such angles \alpha at which the trigonometric functions included in them make sense, the identities will hold, ctg \alpha=\frac(\cos \alpha)(\sin \alpha).
For example: tg \alpha = \frac(\sin \alpha)(\cos \alpha) is valid for angles \alpha that are different from \frac(\pi)(2)+\pi z, A ctg \alpha=\frac(\cos \alpha)(\sin \alpha)- for an angle \alpha other than \pi z, z is an integer.
Relationship between tangent and cotangent
tg \alpha \cdot ctg \alpha=1
This identity is valid only for angles \alpha that are different from \frac(\pi)(2) z. Otherwise, either cotangent or tangent will not be determined.
Based on the above points, we obtain that tg \alpha = \frac(y)(x), A ctg \alpha=\frac(x)(y). It follows that tg \alpha \cdot ctg \alpha = \frac(y)(x) \cdot \frac(x)(y)=1. Thus, the tangent and cotangent of the same angle at which they make sense are mutually inverse numbers.
Relationships between tangent and cosine, cotangent and sine
tg^(2) \alpha + 1=\frac(1)(\cos^(2) \alpha)- the sum of the square of the tangent of the angle \alpha and 1 is equal to the inverse square of the cosine of this angle. This identity is valid for all \alpha other than \frac(\pi)(2)+ \pi z.
1+ctg^(2) \alpha=\frac(1)(\sin^(2)\alpha)- the sum of 1 and the square of the cotangent of the angle \alpha is equal to the inverse square of the sine given angle. This identity is valid for any \alpha different from \pi z.
Examples with solutions to problems using trigonometric identities
Example 1
Find \sin \alpha and tg \alpha if \cos \alpha=-\frac12 And \frac(\pi)(2)< \alpha < \pi ;
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Solution
The functions \sin \alpha and \cos \alpha are related by the formula \sin^(2)\alpha + \cos^(2) \alpha = 1. Substituting into this formula \cos \alpha = -\frac12, we get:
\sin^(2)\alpha + \left (-\frac12 \right)^2 = 1
This equation has 2 solutions:
\sin \alpha = \pm \sqrt(1-\frac14) = \pm \frac(\sqrt 3)(2)
By condition \frac(\pi)(2)< \alpha < \pi . In the second quarter the sine is positive, so \sin \alpha = \frac(\sqrt 3)(2).
In order to find tan \alpha, we use the formula tg \alpha = \frac(\sin \alpha)(\cos \alpha)
tg \alpha = \frac(\sqrt 3)(2) : \frac12 = \sqrt 3
Example 2
Find \cos \alpha and ctg \alpha if and \frac(\pi)(2)< \alpha < \pi .
Show solution
Solution
Substituting into the formula \sin^(2)\alpha + \cos^(2) \alpha = 1 given number \sin \alpha=\frac(\sqrt3)(2), we get \left (\frac(\sqrt3)(2)\right)^(2) + \cos^(2) \alpha = 1. This equation has two solutions \cos \alpha = \pm \sqrt(1-\frac34)=\pm\sqrt\frac14.
By condition \frac(\pi)(2)< \alpha < \pi . In the second quarter the cosine is negative, so \cos \alpha = -\sqrt\frac14=-\frac12.
In order to find ctg \alpha, we use the formula ctg \alpha = \frac(\cos \alpha)(\sin \alpha). We know the corresponding values.
ctg \alpha = -\frac12: \frac(\sqrt3)(2) = -\frac(1)(\sqrt 3).
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 kind 
. The explanation for this fact is quite simple: the equalities are obtained from the main trigonometric identity after dividing both of its parts by and respectively, and the equalities 
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
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 .
The following figure shows the coordinate system Oxy with the part of the unit semicircle ACB depicted in it with the center at point O. This part is the arc of the unit circle. The unit circle is described by the equation x^2+y^2 = 1.
Fundamental trigonometric identity
The ordinate y and abscissa x can be represented as the sine and cosine of the angle using the following formulas:
Substituting these values into the equations of the unit circle, we have the following equality:
(sin(a))^2 + (cos(a))^2 = 1, which will be satisfied for any value of a from 0 degrees to 180 degrees. This equality is called basic trigonometric identity.
Reduction formulas
Reduction formulas are used to express the values of trigonometric functions from arguments of the form (90˚ ±a), (180˚ ±a) through the values of sin(a), cos(a), tg(a) and ctg(a).
There are two rules for using reduction formulas.
1. If the angle can be represented as (90˚ ±a), then the name of the function changes sin to cos, cos to sin, tg to ctg, ctg to tg. If the angle can be represented in the form (180˚ ±a), then the name of the function remains unchanged.
Look at the picture below, which schematically shows when to change the sign and when not.
2. The rule “as you were, so you remain.”
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 basic trigonometric functions depending on the quarter.
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 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:
- 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).
- 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.
- 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 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)`.
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`.
