Basic formulas of trigonometry. Basic trigonometric identity
Reduction 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 at the exam, exams, tests. But we will 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 the angles are written in radians:
To use it, you need to select the row with the function we need, and the column with the desired argument. For example, to use a table to find out what ` sin(\pi + \alpha)` will be, 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 the angles are written in degrees:
Mnemonic rule of casting formulas or how to remember them
As we already mentioned, it is not necessary to memorize all the above ratios. If you looked closely at them, you probably noticed some patterns. They allow us to formulate a mnemonic rule (mnemonic - memorize), with which you can easily get any of the reduction formulas.
We note right away that in order to apply this rule, one must be well able to determine (or remember) the signs of trigonometric functions in different quarters of the unit circle. 
The graft itself contains 3 stages:
- 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`, where `\alpha` is always an acute angle (from 0 to 90 degrees).
- For arguments `\frac (\pi)2 \pm \alpha`, `\frac (3\pi)2 \pm \alpha` trigonometric function of the converted expression 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 a few expressions:
1. `cos(\pi + \alpha)`.
The function is not reversed. The angle ` \pi + \alpha` is in the third quadrant, the cosine in this quadrant has a "-" sign, so the converted function will also have a "-" sign.
Answer: ` cos(\pi + \alpha)= - cos \alpha`
2. `sin(\frac (3\pi)2 - \alpha)`.
According to mnemonic rule function will be reversed. The angle `\frac (3\pi)2 - \alpha` is in the third quadrant, 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))`. 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 above mnemonic rule 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 points, 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 the vertical y-axis.
We ask ourselves the question: “Does the 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 memorize reduction formulas without memorizing them.
Practical Examples of Using Casting Formulas
The use of reduction formulas begins in the 9th and 10th grades. A lot of tasks with their use are submitted to the exam. Here are some of the tasks where you will need to apply these formulas:
- tasks for solving a right-angled triangle;
- conversion of numerical and alphabetic trigonometric expressions, calculation of their values;
- stereometric problems.
Example 1. Use the reduction formulas to calculate 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 the cosine through 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 `sin \frac (\pi)8 We 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 a unit circle and point A on it with coordinates (1,0). Let after turning on From 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 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)`. In this article, we will take a comprehensive look at . Basic trigonometric identities are equalities that establish a relationship between the sine, cosine, tangent and cotangent of one angle, and allow you to find any of these trigonometric functions through a known other. We immediately list the main trigonometric identities, which we will analyze in this article. We write them down in a table, and below we give the derivation of these formulas and give the necessary explanations. Page navigation. Sometimes they talk not about the main trigonometric identities listed in the table above, but about one single basic trigonometric identity kind That is, it is the equality that is of particular interest, which was given the name of the main trigonometric identity. Before proving the basic 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 in transformation of 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 reverse order: the unit is replaced by the sum of the squares of the sine and cosine of any angle. Identities connecting the tangent and cotangent with the sine and cosine of one angle of the form and Due to this obviousness of the identities and To conclude this section, it should be noted that the identities and An even more obvious trigonometric identity than the previous two is the identity connecting the tangent and cotangent of one angle of the form Proof of the formula So, the tangent and cotangent of one angle, at which they make sense, is. The ratios between the main trigonometric functions - sine, cosine, tangent and cotangent - are given trigonometric formulas. And since there are quite a lot of connections between trigonometric functions, this also explains the abundance of trigonometric formulas. Some formulas connect the trigonometric functions of the same angle, others - the functions of a multiple angle, others - allow you to lower the degree, the fourth - to express all functions through the tangent of a half angle, etc. In this article, we list in order all the basic trigonometric formulas, which are enough to solve the vast majority of trigonometry problems. For ease of memorization and use, we will group them according to their purpose, and enter them into tables. Page navigation. Basic trigonometric identities set the relationship between the sine, cosine, tangent and cotangent of one angle. They follow from the definition of sine, cosine, tangent and cotangent, as well as the concept of the unit circle. They allow you to express one trigonometric function through any other. For a detailed description of these trigonometry formulas, their derivation and application examples, see the article. Cast formulas follow from the properties of sine, cosine, tangent and cotangent, that is, they reflect the property of periodicity of trigonometric functions, the property of symmetry, and also the property of shift by a given angle. These trigonometric formulas allow you to move from working with arbitrary angles to working with angles ranging from zero to 90 degrees. The rationale for these formulas, a mnemonic rule for memorizing them, and examples of their application can be studied in the article. Trigonometric addition formulas show how the trigonometric functions of the sum or difference of two angles are expressed in terms of the trigonometric functions of these angles. These formulas serve as the basis for the derivation of the following trigonometric formulas. Formulas for double, triple, etc. angle (they are also called multiple angle formulas) show how the trigonometric functions of double, triple, etc. angles () are expressed in terms of trigonometric functions of a single angle. Their derivation is based on addition formulas. More detailed information is collected in the article formulas for double, triple, etc. angle . Half Angle Formulas show how the trigonometric functions of a half angle are expressed in terms of the cosine of an integer angle. These trigonometric formulas follow from the double angle formulas. Their conclusion and examples of application can be found in the article. Trigonometric formulas for decreasing degrees are designed to facilitate the transition from natural powers of trigonometric functions to sines and cosines in the first degree, but multiple angles. In other words, they allow one to reduce the powers of trigonometric functions to the first. main destination sum and difference formulas for trigonometric functions consists in the transition to the product of functions, which is very useful when simplifying trigonometric expressions. These formulas are also widely used in solving trigonometric equations, as they allow factoring the sum and difference of sines and cosines. The transition from the product of trigonometric functions to the sum or difference is carried out through the formulas for the product of sines, cosines and sine by cosine. Copyright by clever students All rights reserved. Trigonometric identities are equalities that establish a relationship between the 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 reverse order. 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, then by definition, the ordinate of y is the sine, and the abscissa of x is the 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. We add that only for such angles \alpha for which the trigonometric functions included in them make sense, the identities will take place, ctg \alpha=\frac(\cos \alpha)(\sin \alpha). For example: tg \alpha = \frac(\sin \alpha)(\cos \alpha) is valid for \alpha angles that are different from \frac(\pi)(2)+\pi z, but ctg \alpha=\frac(\cos \alpha)(\sin \alpha)- for an angle \alpha other than \pi z , z is an integer. 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 points above, we get that tg \alpha = \frac(y)(x), but ctg\alpha=\frac(x)(y). Hence it follows that tg \alpha \cdot ctg \alpha = \frac(y)(x) \cdot \frac(x)(y)=1. Thus, the tangent and cotangent of one angle at which they make sense are mutually reciprocal numbers. 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 , equals the inverse square of the sine of the given angle. This identity is valid for any \alpha other than \pi z . Find \sin \alpha and tg \alpha if \cos \alpha=-\frac12 And \frac(\pi)(2)< \alpha < \pi
; Show Solution Solution The functions \sin \alpha and \cos \alpha are linked 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). To find tg \alpha , we use the formula tg \alpha = \frac(\sin \alpha)(\cos \alpha) tg \alpha = \frac(\sqrt 3)(2) : \frac12 = \sqrt 3 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 conditional 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). This is the last and most important lesson needed to solve problems B11. We already know how to convert angles from a radian measure to a degree measure (see lesson “ Radian and degree measure of an angle"), and we also know how to determine the sign of a trigonometric function, focusing on coordinate quarters (see lesson " Signs of trigonometric functions"). The matter remains small: to calculate the value of the function itself - the very number that is written in the answer. Here the basic trigonometric identity comes to the rescue. Basic trigonometric identity. For any angle α, the 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: Notice 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 is an even function that "burns" all the minuses (if any). That is why in all B11 tasks that are found in the USE 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 surely ask: “What about the tangent and cotangent?” It is impossible to directly calculate these functions from the above formulas. However, there are important corollaries from the basic trigonometric identity that already contain tangents and cotangents. Namely: An important corollary: for any angle α, the basic trigonometric identity can be rewritten as follows: These equations are easily deduced from the basic identity - it is enough to divide both sides by cos 2 α (to get the tangent) or by sin 2 α (for the cotangent). Let's look at all this with specific examples. The following are actual B11 problems taken from the 2012 Mathematics USE trials. 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°). Therefore, the angle α lies in the II coordinate quarter - all the 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. We 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, the angle α belongs to the interval (π 3π /2). Let's convert the angles from radian measure to degree measure - we get: α ∈ (180°; 270°). Obviously, this is the III coordinate quarter, where all cosines are negative. Therefore cosα = −0.5. A task. Find tg α if you know the following: Tangent and cosine are related by an equation following from the basic trigonometric identity: We get: tg 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 the radian measure to the degree measure - we get α ∈ (270°; 360°). Obviously, this is the IV coordinate quarter, where all tangents are negative. Therefore, tgα = −3. A task. Find cos α if you know the following: Again, the sine is known and the cosine is unknown. We 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 are positive there. Therefore, cos α = 0.6. A task. Find sin α if you know the following: Let's write down the formula that follows from the basic trigonometric identity and directly connects the 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 the angle α ∈ (0; π /2). In degrees, this is written as follows: α ∈ (0°; 90°) - I coordinate quarter. So, the angle is in the I coordinate quarter - all trigonometric functions are positive there, therefore sin α \u003d 0.2.
corner `\alpha` it will go to the point `A_1(x, y)`, and after turning through the angle `\frac (\pi)2 + \alpha` to the 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`. How can one write 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`.Relationship between sine and cosine of one angle
. The explanation for this fact is quite simple: the equalities are obtained from the basic 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 will discuss this in more detail in the following paragraphs.Tangent and cotangent through sine and cosine
immediately follow from the definitions of sine, cosine, tangent and cotangent. Indeed, by definition, the sine is the ordinate of y, the cosine is the abscissa of x, the 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, 
.often the definitions of tangent and cotangent are given not through the ratio of the abscissa and the ordinate, but through the ratio of the 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.hold for all such angles for which the trigonometric functions in them make sense. So the formula is valid for any other than (otherwise the denominator will be 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
. It is clear that it takes place for any angles other than , otherwise either the tangent or the cotangent is not defined.very simple. By definition and from where 
. The proof could have been carried out in a slightly different way. Since and , then 
.Basic trigonometric identities
Cast formulas
Addition Formulas
Formulas for double, triple, etc. angle
Half Angle Formulas
Reduction Formulas
Formulas for the sum and difference of trigonometric functions
Formulas for the product of sines, cosines and sine by cosine
Protected by copyright law. No part of www.site, including internal materials and external design, may be reproduced in any form or used without the prior written permission of the copyright holder.Finding tangent and cotangent through sine and cosine
Relationship between tangent and cotangent
Relationships between tangent and cosine, cotangent and sine
Examples with solutions to problems using trigonometric identities
Example 1
Example 2
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