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.
Finding tangent and cotangent through 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, 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 instance: tg \alpha = \frac(\sin \alpha)(\cos \alpha) is valid for \alpha angles 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 points above, we get that tg \alpha = \frac(y)(x), a 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.
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 , equals the inverse square of the sine of the given angle. This identity is valid for any \alpha other than \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 ;
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
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 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).
Example 2 Prove Identity
We will prove this identity by transforming the expression on the right side.
Method 1.
So 
Method 2.
First of all, note that ctg α =/= 0; otherwise, the expression tg would not make sense α = 1/ctg α . But if ctg α =/= 0, then the numerator and denominator of the radical expression can be multiplied by ctg α without changing the value of the fraction. Hence,
Using the identities tg α ctg α = 1 and 1+ ctg 2 α = cosec 2 α , we get
So 
Q.E.D.
Comment. Attention should be paid to the fact that the left-hand side of the proved identity (sin α ) is defined for all values α , and the right one - only when α =/= π / 2 n.
Therefore, only when all admissible values α In general, these expressions are not equivalent to each other.
Example 3 Prove Identity
sin (3 / 2 π + α ) + cos( π - α ) = cos(2 π + α )-3sin( π / 2 - α )
We transform the left and right parts of this identity using the reduction formulas:
sin (3 / 2 π + α ) + cos( π - α ) = - cos α - cos α = - 2 cos α ;
cos (2 π + α )-3sin( π / 2 - α ) = cos α - 3cos α = - 2 cos α .
So, the expressions in both parts of this identity are reduced to the same form. Thus, the identity is proved.
Example 4 Prove Identity
sin 4 α + cos 4 α - 1 = - 2 sin 2 α cos 2 α .
Let us show that the difference between the left and right parts. of this identity is zero.
(sin 4 α + cos 4 α - 1) - (- 2 sin 2 α cos 2 α ) = (sin 4 α +2sin2 α cos 2 α + cos 4 α ) - 1 =
= (sin 2 α + cos2 α ) 2 - 1 = 1 - 1 = 0.
Thus, the identity is proved.
Example 5 Prove Identity
This identity can be viewed as a proportion. But to prove the validity of the proportion a / b = c / d, it is enough to show that the product of its extreme terms ad is equal to the product of its middle terms bc. So we will do in this case. Let us show that (1 - sin α ) (1+ sin α ) = cos α cos α .
Indeed, (1 - sin α ) (1 + sin α ) = 1-sin 2 α = cos2 α .
"Trigonometric identities". 10th grade“Mathematical truth, whatever
whether in Paris or in Toulouse, the same
B. Pascal
Lesson type: Lesson in the formation of skills and abilities.
Lesson of general methodological orientation.
activity goal : formation of students' ability to a new mode of action associated with the construction of the structure of the studied concepts and algorithms.
Lesson Objectives:
didactic : to teach how to apply previously acquired knowledge, skills and abilities to simplify expressions and prove trigonometric identities.
developing: develop logical thinking, memory, cognitive interest, continue the formation of mathematical speech, develop the ability to analyze and compare.
educational: show that mathematical concepts are not isolated from each other, but represent a certain system of knowledge, all links of which are interconnected, continue the formation of aesthetic skills when making notes, skills of control and self-control.
To successfully solve problems in trigonometry, you need to be confident in numerous formulas. Trigonometric formulas must be remembered. But this does not mean that they need to be memorized by heart, the main thing is to memorize not the formulas themselves, but the algorithms for their derivation. Any trigonometric formula can be obtained fairly quickly if you firmly know the definitions and basic properties of the functions sinα, cosα, tgα, ctgα, the ratio sin 2 α+ cos 2 α =1 etc.
Learning trigonometric formulas at school is not for you to calculate sines and cosines for the rest of your life, but for your brain to acquire the ability to work. ( . slide 2 )
“ The roads are not the knowledge that is deposited in the brain like fat; the roads are those that turn into mental muscles,” wrote G. Speser, an English philosopher and sociologist.
We will pump up and train the mental muscles. Therefore, we repeat the basic trigonometric formulas.TEST (Slide 4) (Slide 5)
We repeated the formulas, now we can help two friends, let's call them Islam and Mohammed.
After transforming some very complex trigonometric expressionA they got the following expressions:(Slide 6)
(Slide 7) Each defended his answer. How to find out which one is right? We turned to Artyom, who is friends with Peter"Plato is my friend but the truth is dearer": Artyom said and suggested several ways to resolve their dispute. And what ways can you suggest to establish the truth?Suggest ways to establish the truth (Slide 8):
1) Transform, simplify A P and A With , i.e. led to one expression
2) A P - A With = 0
3) …..
That is, both were right. And their answers are equal for all possible valuesα and β .
What are such expressions called?Identities. What identities do you know?
Identity , the basic concept of logic, philosophy and mathematics; used in the languages of scientific theories to formulate defining relations, laws and theorems.
Identity is a philosophical category that expresses equality, the sameness of an object, a phenomenon with itself, or the equality of several objects.
In mathematics identity is an equality that is valid for any admissible values of the variables included in it.(Slide 9)
Lesson topic : "Trigonometric identities".
Objectives: find ways.
Two people work at the blackboard.
№ 2. Prove the identity.
P.h. \u003d L.h.
The identity has been proven.
№ 3. Prove the identity:
1 way:
2 way:
Ways to prove identities.
the right side of the identity. If in the end we get the left side, then the identity is considered proven.
Perform equivalent transformationsleft and right sides of the identity. If we get the same result as a result, then the identity is considered proven.
Subtract the left side from the right side of the identity.
Subtract the right side from the left side of the identity. We perform equivalent transformations on the difference. And if in the end we get zero, then the identity is considered proven.
It should also be remembered that the identity is valid only for admissible values of variables.
Why is it necessary to be able to prove trigonometric identities? In the exam, task C1 is trigonometric equations!
Decided No. 465-467
So, let's sum up the lesson. (Slide 10)
What was the topic of the lesson?
What methods of proving identities do you know?
1. Convert left to right or right to left.
2. Converting the left and right parts to the same expression.
3. Drawing up the difference between the left and right parts and proving that this difference is equal to zero.
What formulas are used for this?
1. Formulas for abbreviated multiplication.
2. 6 trigonometric identities.
Lesson reflection. (Slide 11)
Continue the phrases:
– Today in class I learned...
Today in class I learned...
- Today at the lesson I repeated ...
Today in class I met...
I enjoyed my lesson today...
Homework. №№465-467 (Slide 12)
Creative task: Prepare a presentation about the famous identities of mathematics. (For example, the Euler identity.)(Slide
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.
\[ \sin^(2)\alpha + \cos^(2) \alpha = 1 \]
\[ tg \alpha = \dfrac(\sin \alpha)(\cos \alpha), \enspace ctg \alpha = \dfrac(\cos \alpha)(\sin \alpha) \]
\[ tg \alpha \cdot ctg \alpha = 1 \]
Relationship between sine and cosine
\[ \sin^(2) \alpha+\cos^(2) \alpha=1 \]
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.
Finding tangent and cotangent through sine and cosine
\[ tg \alpha = \dfrac(\sin \alpha)(\cos \alpha),\enspace ctg \alpha=\dfrac(\cos \alpha)(\sin \alpha) \]
These identities are formed from the definitions of sine, cosine, tangent and cotangent. After all, if you look, then by definition the ordinate \(\dfrac(y)(x)=\dfrac(\sin \alpha)(\cos \alpha) \), and the ratio \(\dfrac(x)(y)=\dfrac(\cos \alpha)(\sin \alpha) \)- will be a cotangent.
We add that only for such angles \(\alpha \) , for which the trigonometric functions included in them make sense, will the identities , .
For instance: \(tg \alpha = \dfrac(\sin \alpha)(\cos \alpha) \) is valid for angles \(\alpha \) that are different from \(\dfrac(\pi)(2)+\pi z \) , and \(ctg \alpha=\dfrac(\cos \alpha)(\sin \alpha) \)- 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 \(\dfrac(\pi)(2) z \) . Otherwise, either cotangent or tangent will not be determined.
Based on the above points, we get that \(tg \alpha = \dfrac(y)(x) \) , and \(ctg \alpha=\dfrac(x)(y) \) . Hence it follows that \(tg \alpha \cdot ctg \alpha = \dfrac(y)(x) \cdot \dfrac(x)(y)=1 \). Thus, the tangent and cotangent of one angle at which they make sense are mutually reciprocal numbers.
Relationships between tangent and cosine, cotangent and sine
\(tg^(2) \alpha + 1=\dfrac(1)(\cos^(2) \alpha) \)- the sum of the squared tangent of the angle \(\alpha \) and \(\alpha \) , other than \(\dfrac(\pi)(2)+ \pi z \) .
\(1+ctg^(2) \alpha=\dfrac(1)(\sin^(2)\alpha) \)- sum \(\alpha \) , equals the inverse square of the sine of the given angle. This identity is valid for any \(\alpha \) other than \(\pi z \) .
Javascript is disabled in your browser.ActiveX controls must be enabled in order to make calculations!
Identity examples:
\(2(x+5)=2x+10\);
\(a^2-b^2=(a+b)(a-b)\);
\(1-\sin^2x=\cos^2x\).
But the expression \(\frac(x^2)(x)=x\) is an identity only under the condition \(x≠0\) (otherwise the left side does not exist).
How to prove identity?
The recipe is crazy simple:
To prove an identity, you need to prove that its right and left parts are equal, i.e. reduce it to the form "expression" = "same expression".
For instance,
\(5=5\);
\(\sin^2x=\sin^2x\);
\(\cosx-4=\cosx-4\).
In order to do this you can:
- Convert only the right side or only the left side.
- Convert both parts at the same time.
- Use any valid mathematical transformations (for example, give similar; open brackets; transfer terms from one part to another by changing the sign; multiply or divide the left and right parts by the same number or expression that is not equal to zero, etc.) .
- Use any mathematical formulas.
It is the fourth point that is used most often when proving identities, so you need to know, remember and be able to use everything.
Example
. Prove the trigonometric identity \(\sin2x=2\sinx\cdot \cos(x)\)
Solution
:
Example
. Prove that the expression \(\frac (\cos^2(t))(1-\sin(t))\)\(-\sin(t)=1\) is an identity.
Solution
:
Example
. Prove the trigonometric identity \(1-tg^2 t=\) \(\frac(\cos2t)(\cos^2t)\)
Solution
:
|
\(1-tg^2 t=\) \(\frac(\cos2t)(\cos^2t)\) |
Here we will transform only the right side, trying to reduce it to the left. We leave the left one unchanged. We remember. |
|
|
\(1-tg^2 t=\) |
Now let's do a term-by-term division in a fraction (i.e. apply in the opposite direction): \(\frac(a+c)(b)\) \(=\) \(\frac(a)(b)\) \( +\)\(\frac(c)(b)\) |
|
|
\(1-tg^2 t=\) \(\frac(\cos^2t)(\cos^2t)\)\(-\)\(\frac(\sin^2t)(\cos^2t)\) |
We cancel the first fraction on the right side, and apply to the second: \(\frac(a^n)(b^n)\) \(=\)\((\frac(a)(b))^n\) . |
|
|
\(1-tg^2 t=1-\) \((\frac(\sint)(\cost))^2\) |
Well, the sine divided by the cosine is equal to the same angle: \(\frac(\sinx)(\cosx)\) \(=tg x\) |
|
|
\(1-tg^2 t=1-tg^2 t\) |
Example
. Prove the trigonometric identity \(=ctg(π+t)-1\)
Solution
:
|
\(\frac(\cos2t)(\sint\cdot\cost+\sin^2t)\)\(=ctg(π+t)-1\) |
Here we will transform both parts: |
|
|
\(\frac(\cos^2t-\sin^2t)(\sint\cdot\cost+\sin^2t)\)\(=ctg\:t-1\) |
Now we work only with the left side. |
|
|
\(\frac((\cost-\sin(t))(\cost+\sin(t)))(\sint(\cost+\sin(t)))\)\(=ctg\:t-1\) |
Reduce the fraction by \(\cos(t)+\sin(t)\). |
|
|
\(\frac(\cost-\sin(t))(\sint)\)\(=ctg\:t-1\) |
We divide the fraction term by term, turning it into two separate fractions. |
|
|
\(\frac(\cost)(\sin(t))-\frac(\sin(t))(\sin(t))\)\(=ctg\:t-1\) |
The first fraction is , and the second is equal to one. |
|
|
\(ctg\:t-1=ctg\:t-1\) |
The left side is equal to the right side, the identity is proved. |
As you can see, everything is quite simple, but you need to know all the formulas and properties.
How to prove the basic trigonometric identity
Two easy ways to derive the formula \(\sin^2x+\cos^2x=1\). You only need to know the Pythagorean theorem and the definition of sine and cosine.
Answers to frequently asked questions:
Question:
How to determine what needs to be transformed in an identity - the left side, the right side, or both together?
Answer:
There is no difference - in any case, you will get the same result. For example, in the third example, we could easily get from the left side \(1-tg^2 t\) the right \(\frac(cos2t)(cos^2t)\)(try to do it yourself). Or transform both, so that they "meet in the middle", somewhere in the area \(\frac(\cos^2t-\sin^2t)(\cos^2t)\)\(=\)\(\frac(\cos^2t-\sin^2t)(\cos^2t)\). Therefore, you can prove in any way convenient for you. Whichever path you see, follow that one. The only main thing is to transform “legally”, that is, understand on the basis of what property, rule or formula you are doing the next transformation.