First, knowledge point induction (images and properties of trigonometric functions)
1 trigonometric function image
(omitted)
2. Properties of trigonometric functions
The domain, range and maximum value of (1) trigonometric function.
Functional domain range period
Y=sin x R [- 1， 1]2π
Y=cos x R [- 1， 1]2π
y = tan x { x/x≠kx+π/2，kz } Rπ
Domain: The set of all input values that can be regarded as a function in mathematics.
Three solutions of function definition domain
First, the resolution function is given to find its domain. Generally, the inequalities (groups) with restrictive conditions are listed first, and then solved.
2. Give the definition domain of the function and find the definition domain of the function. The solution steps are as follows: if the definition domain of a function is known, the definition domain of its composite function must be solved by inequality.
3. Given the domain, the solution of the domain is as follows: If the domain is known as, then the domain is the value domain at that time.
Range: In a function, the range of the value of the dependent variable is called the range of this function, which is mathematically the set of all the values of the dependent variable in the function definition domain.
[Edit this paragraph] Common methods for evaluating domain names
(1) reduction method; (2) Image method (combination of numbers and shapes),
(3) Monotonicity of the function,
(4) matching method, (5) substitution method, (6) inverse function method, (7) discriminant method, (8) compound function method, (9) trigonometric substitution method, (10) basic inequality method, etc.
Period: The minimum positive period t of the function f(x) must meet the following two conditions: (1) When x defines every value in the field, there is f(x+T)=f(x). (2)t is the smallest positive number that is not zero. Generally speaking, if T is the period of f(x), then n t is the so-called period of f(x), that is, f(x)=f(x+nT).
(2) The parity and monotonicity of trigonometric functions
Function Y=sin x Y=cos x Y=tan x
Odd even odd
(1) parity check
1) is an odd function (2) is an even function.
For the function f(x)
(1) If any x in the function definition domain has f (-x) =-f(x), then the function f(x) is called odd function.
(2) If any x in the function definition domain has f(-x)=f(x), the function f(x) is called an even function.
(3) If f(-x)=-f(x) and f(-x)=f(x) are true for any x in the function definition domain, then the function f(x) is both a odd function and an even function, which is called an even-even function.
(4) If f(-x)=-f(x) or f(-x)=f(x) can't hold for any x in the function definition domain, then the function f(x) is neither a odd function nor an even function, which is called an even-even function.
Description: ① Odd and even are global properties of functions, and they are global.
② The domains of odd and even functions must be symmetrical about the origin. If the domain of a function is not symmetric about the origin, then the function must not be an odd (or even) function.
③ The basis for judging or proving whether a function has parity is definition.
(2) Monotonicity
If the function y=f(x) is a increasing function or subtraction function in a certain interval, it is said that the function has (strict) monotonicity in this interval, which is called the monotonic interval of the function. At this time, it is also said that the function is monotonic in this interval.
In the monotone interval, the image of increasing function is rising, and the image of subtraction function is falling.
practise
1. If the function y=sin(2x+φ) is an even function, then a value of φ is () A.φ=-π B.φ=-π/2 C.φ=2π D.φ=π/42. The function f (x) = sin (ω x+φ) cos. 0) Take 2 as the minimum positive period and the maximum value when x=2, then a value of φ is () a.7/4 π b-5/4 π c-3/4 π d π/2.
3 Let the quadratic function f(x)=x2+bx+c(b, c∈R). It is known that no matter what α and β are, there is always f(sinα)≥0 and f(2+cosβ)≤0 in real numbers.
(1) verify that b+c =-1;
(2) verify that c ≥ 3;
(3) If the maximum value of function f(sinα) is 8, find the values of b and c..
Reference answer
1B
2.C
3 solution (1) ∫-1≤ sinα≤1and f(sinα)≥0 holds, ∴f( 1)≥0.
∫ 1≤2+cosβ≤3, f(2+cosβ)≤0 holds ∴f( 1)≤0.
So we know that f( 1)=0∴b+c+ 1=0.
(2) from f(2+cosβ)≤0, we know that f(3)≤0,∴9+3b+c≤0 and because b+c =- 1, ∴ c ≥ 3.
(3) b=-4，c=3
High school learning network www.90house.cn original link:/gaochongshuxuezt/1546.html.
formulas of trigonometric functions
Two-angle sum formula
sin(A+B) = sinAcosB+cosAsinB
sin(A-B) = sinAcosB-cosAsinB
cos(A+B) = cosAcosB-sinAsinB
cos(A-B) = cosAcosB+sinAsinB
tan(A+B)=(tanA+tanB)/( 1-tanA tanB)
tan(A-B)=(tanA-tanB)/( 1+tanA tanB)
cot(A+B)=(cotA cotB- 1)/(cot B+cotA)
cot(A-B)=(cotA cotB+ 1)/(cot b-cotA)
Double angle formula
2tanA/( 1-tan^2 A)
Sin2A=2SinA？ Kosa
Cos2A = Cos^2 A - Sin^2 A
=2Cos^2 A— 1
= 1—2sin^2 A
Triple angle formula
sin3a = 3sina-4(sina)^3；
cos3A = 4(cosA)^3 -3cosA
tan3a = tan a？ tan(π/3+a)？ tan(π/3-a)
half-angle formula
sin(A/2) = √{( 1 - cosA)/2}
cos(A/2) = √{( 1+cosA)/2}
tan(A/2)= √{( 1-cosA)/( 1+cosA)}
cot(A/2)= √{( 1+cosA)/( 1-cosA)}
Tan(A/2)=( 1-cosA)/ Sina = Sina /( 1+cosA)
Sum difference product
sin(a)+sin(b)= 2 sin[(a+b)/2]cos[(a-b)/2]
sin(a)-sin(b)= 2cos[(a+b)/2]sin[(a-b)/2]
cos(a)+cos(b)= 2cos[(a+b)/2]cos[(a-b)/2]
cos(a)-cos(b)=-2 sin[(a+b)/2]sin[(a-b)/2]
tanA+tanB=sin(A+B)/cosAcosB
Sum and difference of products
sin(a)sin(b)=- 1/2 *[cos(a+b)-cos(a-b)]
cos(a)cos(b)= 1/2 *[cos(a+b)+cos(a-b)]
sin(a)cos(b)= 1/2 *[sin(a+b)+sin(a-b)]
cos(a)sin(b)= 1/2 *[sin(a+b)-sin(a-b)]
Inductive formula
sin(-a) = -sin(a)
cos(-a) = cos(a)
sin(π/2-a) = cos(a)
cos(π/2-a) = sin(a)
sin(π/2+a) = cos(a)
cos(π/2+a) = -sin(a)
sin(π-a) = sin(a)
cos(π-a) = -cos(a)
sin(π+a) = -sin(a)
cos(π+a) = -cos(a)
tgA=tanA = sinA/cosA
General formula of trigonometric function
sin(a)=[2tan(a/2)]/{ 1+[tan(a/2)]^2}
cos(a)= { 1-[tan(a/2)]^2}/{ 1+[tan(a/2)]^2}
Tan (1) = [2 tan (a/2)]/{1-[tan (a/2)] 2}
Other formulas
Answer? Sin (a)+b? Cos(a)=[√( a2+B2)]* sin(a+c)[ where tan(c)=b/a]
Answer? Crime (A)-B? Cos(a)=[√( a2+B2)]* cos(a-c)[ where tan(c)=a/b]
1+sin(a)=[sin(a/2)+cos(a/2)]^2；
1-sin(a)=[sin(a/2)-cos(a/2)]^2； ;
Other non-critical trigonometric functions
csc(a) = 1/sin(a)
Seconds (a)= 1/ cosine (a)
Hyperbolic function
sinh(a) = [e^a-e^(-a)]/2
cosh(a) = [e^a+e^(-a)]/2
tg h(a) = sin h(a)/cos h(a)
Formula 1:
Let α be an arbitrary angle, and the values of the same trigonometric function with the same angle of the terminal edge are equal:
sin(2kπ+α)= sinα
cos(2kπ+α)= cosα
tan(2kπ+α)= tanα
cot(2kπ+α)= cotα
Equation 2:
Let α be an arbitrary angle, and the relationship between the trigonometric function value of π+α and the trigonometric function value of α;
Sine (π+α) =-Sine α
cos(π+α)= -cosα
tan(π+α)= tanα
cot(π+α)= cotα
Formula 3:
The relationship between arbitrary angle α and the value of-α trigonometric function;
Sine (-α) =-Sine α
cos(-α)= cosα
tan(-α)= -tanα
Kurt (-α) =-Kurt α
Equation 4:
The relationship between π-α and the trigonometric function value of α can be obtained by Formula 2 and Formula 3:
Sine (π-α) = Sine α
cos(π-α)= -cosα
tan(π-α)= -tanα
cot(π-α)=-coα
Formula 5:
The relationship between the trigonometric function values of 2π-α and α can be obtained by Formula-and Formula 3:
Sine (2π-α)=- Sine α
cos(2π-α)= cosα
tan(2π-α)= -tanα
Kurt (2π-α)=- Kurt α
Equation 6:
The relationship between π/2 α and 3 π/2 α and the trigonometric function value of α;
sin(π/2+α)= cosα
cos(π/2+α)= -sinα
tan(π/2+α)= -cotα
cot(π/2+α)= -tanα
sin(π/2-α)= cosα
cos(π/2-α)= sinα
tan(π/2-α)= cotα
cot(π/2-α)= tanα
sin(3π/2+α)= -cosα
cos(3π/2+α)= sinα
tan(3π/2+α)= -cotα
cot(3π/2+α)= -tanα
sin(3π/2-α)= -cosα
cos(3π/2-α)= -sinα
tan(3π/2-α)= cotα
cot(3π/2-α)= tanα
(higher than k∈Z)
Formulas of trigonometric functions daquan
Acute angle formula of trigonometric function
Opposite side/hypotenuse of sinα=∞α
Adjacent edge/hypotenuse of cosα=∞α
Opposite side of tan α = adjacent side of ∠ α/∠α.
Adjacent side of cot α = opposite side of ∠ α/∠α.
Double angle formula
Sin2A=2SinA？ Kosa
cos2a=cosa^2-sina^2= 1-2sina^2=2cosa^2- 1
tan2A=(2tanA)/( 1-tanA^2)
(Note: Sina 2 is the square of Sina 2 (a))
Triple angle formula
sin3α=4sinα？ Sine (π/3+α) Sine (π/3-α)
cos3α=4cosα？ cos(π/3+α)cos(π/3-α)
tan3a = tan a？ tan(π/3+a)？ tan(π/3-a)
Derivation of triple angle formula
sin3a
=sin(2a+a)
=sin2acosa+cos2asina
Auxiliary angle formula
Asinα+bcosα = (A2+B2) (1/2) sin (α+t), where
sint=B/(A^2+B^2)^( 1/2)
cost=A/(A^2+B^2)^( 1/2)
tant=B/A
asinα+bcosα=(a^2+b^2)^( 1/2)cos(α-t),tant=a/b
Reduced power formula
sin^2(α)=( 1-cos(2α))/2=versin(2α)/2
cos^2(α)=( 1+cos(2α))/2=covers(2α)/2
tan^2(α)=( 1-cos(2α))/( 1+cos(2α))
Derived formula
tanα+cotα=2/sin2α
tanα-cotα=-2cot2α
1+cos2α=2cos^2α
1-cos2α=2sin^2α
1+sinα=(sinα/2+cosα/2)^2
=2sina( 1-sin？ a)+( 1-2sin？ A) Sina
=3sina-4sin？ a
cos3a
=cos(2a+a)
=cos2acosa-sin2asina
=(2cos？ a- 1)cosa-2( 1-sin？ a)cosa
=4cos？ a-3cosa
sin3a=3sina-4sin？ a
=4sina(3/4-sin？ answer
=4sina[(√3/2)？ Sin? Answer]
=4sina (sin? 60- sin? answer
=4sina(sin60 +sina)(sin60 -sina)
= 4 Sina * 2 sin[(60+a)/2]cos[(60-a)/2]* 2 sin[(60-a)/2]cos[(60-a)/2]
=4sinasin(60 +a)sin(60 -a)
cos3a=4cos？ a-3cosa
=4cosa(cos？ a-3/4)
= 4c OSA【cos？ a-(√3/2)？ ]
=4cosa(cos？ a-cos？ 30 )
=4cosa(cosa+cos30 )(cosa-cos30)
= 4 cosa * 2cos[(a+30)/2]cos[(a-30)/2]* {-2 sin[(a+30)/2]sin[(a-30)/2]}
=-4 eicosapentaenoic acid (a+30) octyl (a-30)
=-4 Coxsacin [90-(60-a)] Xin [-90 +(60 +a)]
=-4 cos(60-a)[-cos(60+a)]
= 4 cos(60-a)cos(60+a)
Comparing the above two formulas, we can get
tan3a=tanatan(60 -a)tan(60 +a)
half-angle formula
tan(A/2)=( 1-cosA)/sinA = sinA/( 1+cosA)；
cot(A/2)= sinA/( 1-cosA)=( 1+cosA)/sinA。
sin^2(a/2)=( 1-cos(a))/2
cos^2(a/2)=( 1+cos(a))/2
tan(a/2)=( 1-cos(a))/sin(a)= sin(a)/( 1+cos(a))
Triangular sum
sin(α+β+γ)=sinα？ cosβ？ cosγ+cosα？ sinβ？ cosγ+cosα？ cosβ？ sinγ-sinα？ sinβ？ sinγ
cos(α+β+γ)=cosα？ cosβ？ cosγ-cosα？ sinβ？ sinγ-sinα？ cosβ？ sinγ-sinα？ sinβ？ cosγ
tan(α+β+γ)=(tanα+tanβ+tanγ-tanα？ tanβ？ tanγ)/( 1-tanα？ tanβ-tanβ？ tanγ-tanγ？ tanα)
Sum and difference of two angles
cos(α+β)=cosα？ cosβ-sinα？ sinβ
cos(α-β)=cosα？ cosβ+sinα？ sinβ
sin(α β)=sinα？ cosβ cosα？ sinβ
tan(α+β)=(tanα+tanβ)/( 1-tanα？ tanβ)
tan(α-β)=(tanα-tanβ)/( 1+tanα？ tanβ)
Sum difference product
sinθ+sinφ= 2 sin[(θ+φ)/2]cos[(θ-φ)/2]
sinθ-sinφ= 2 cos[(θ+φ)/2]sin[(θ-φ)/2]
cosθ+cosφ= 2 cos[(θ+φ)/2]cos[(θ-φ)/2]
cosθ-cosφ=-2 sin[(θ+φ)/2]sin[(θ-φ)/2]
tanA+tanB = sin(A+B)/cosa cosb = tan(A+B)( 1-tanA tanB)
tanA-tanB = sin(A-B)/cosa cosb = tan(A-B)( 1+tanA tanB)
Sum and difference of products
sinαsinβ = [cos(α-β)-cos(α+β)] /2
cosαcosβ = [cos(α+β)+cos(α-β)]/2
sinαcosβ = [sin(α+β)+sin(α-β)]/2
cosαsinβ = [sin(α+β)-sin(α-β)]/2
Inductive formula
Sine (-α) =-Sine α
cos(-α) = cosα
tan (—a)=-tanα
sin(π/2-α) = cosα
cos(π/2-α) = sinα
sin(π/2+α) = cosα
cos(π/2+α) = -sinα
Sine (π-α) = Sine α
cos(π-α) = -cosα
Sine (π+α) =-Sine α
cos(π+α) = -cosα
tanA= sinA/cosA
tan(π/2+α)=-cotα
tan(π/2-α)=cotα
tan(π-α)=-tanα
tan(π+α)=tanα
Inductive formula memory skills: odd variables are unchanged, and symbols look at quadrants.
General formula of trigonometric function
sinα=2tan(α/2)/〔 1+tan^(α/2)〕
cosα=〔 1-tan^(α/2)〕/ 1+tan^(α/2)〕
tanα=2tan(α/2)/〔 1-tan^(α/2)〕
Other formulas
( 1)(sinα)^2+(cosα)^2= 1
(2) 1+(tanα)^2=(secα)^2
(3) 1+(cotα)^2=(cscα)^2
To prove the following two formulas, just divide one formula by (sin α) 2 and the second formula by (cos α) 2.
(4) For any non-right triangle, there is always
tanA+tanB+tanC=tanAtanBtanC
Certificate:
A+B=π-C
tan(A+B)=tan(π-C)
(tanA+tanB)/( 1-tanA tanB)=(tanπ-tanC)/( 1+tanπtanC)
Surface treatment can be carried out.
tanA+tanB+tanC=tanAtanBtanC
Obtain a certificate
It can also be proved that this relationship holds when x+y+z=nπ(n∈Z).
The following conclusions can be drawn from tana+tanbtana+tanb+tanc = tanatanbtanc.
(5)cotAcotB+cotAcotC+cotbctc = 1
(6) Cost (A/2)+ Cost (B/2)+ Cost (C/2)= Cost (A/2) Cost (B/2)
(7)(cosa)^2+(cosb)^2+(cosc)^2= 1-2cosacosbcosc
(8)(sina)^2+(sinb)^2+(sinc)^2=2+2cosacosbcosc
(9)sinα+sin(α+2π/n)+sin(α+2π* 2/n)+sin(α+2π* 3/n)+……+sin[α+2π*(n- 1)/n]= 0
Cos α+cos (α+2π/n)+cos (α+2π * 2/n)+cos (α+2π * 3/n)+...+cos [α+2π * (n-1)/n] = 0 and
sin^2(α)+sin^2(α-2π/3)+sin^2(α+2π/3)=3/2
tanAtanBtan(A+B)+tanA+tan B- tan(A+B)= 0
First of all, inductive formula
Formula: (numerator) parity, symbols look at quadrants.
1.sin (α+k？ 360)=sin α
cos (α+k？ 360)=cos a
tan (α+k？ 360)=tan α
2.sin( 180 +β)=-sinα
cos( 180 +β)=-cosa
3.sin(-α)=-sina
cos(-a)=cosα
4*.tan( 180 +α)=tanα
tan(-α)=tanα
5.sin( 180 -α)=sinα
cos( 180 -α)=-cosα
6.sin(360 -α)=-sinα
cos(360 -α)=cosα
7.sin(π/2-α)=cosα
cos(π/2-α)=sinα
8*.Sin(3π/2-α)=-cosα
cos(3π/2-α)=-sinα
9*.Sin(π/2+α)=cosα
cos(π/2+a)=-sinα
10*.sin(3π/2+α)=-cosα
cos(3π/2+α)=sinα
Second, the trigonometric function of the sum and difference of two angles
1. Two-point distance formula
2.s(α+β):sin(α+β)= sinαcosβ+cosαsinβ
c(α+β):cos(α+β)= cosαcosβ-sinαsinβ
3.s(α-β):sin(α-β)= sinαcosβ-cosαsinβ
c(α-β):cos(α-β)= cosαcosβ+sinαsinβ
4.T(α+β):
T(α-β):
5*.
Triple sum angle formula
1.S2α: sin2α=2sinαcosα
2.C2a: cos2α=cos2α-sin2a
3.T2α:tan 2α=(2 tanα)/( 1-tan 2α)
4.' C2a': cos2α= 1-2sin2α
cos2α=2cos2α- 1
IV * Other miscellaneous items (none of which can be directly used)
1. Auxiliary angle formula
Asinα+bcosα=sin(a+φ), where tanφ=b/a, and its terminal edge passes through point (a, b).
Asinα+bcosα=cos(a-φ), where tanφ=a/b and its terminal edge passes through point (b, a).
2. Reduced order, formula formula
Demote:
sin2θ=( 1-cos2θ)/2
cos2θ=( 1+cos2θ)/2
formula
1 sinθ=[sin(θ/2)cos(θ/2)]2
1+cosθ=2cos2(θ/2)
1-cosθ=2sin2(θ/2)
3. Triple angle formula
sin3θ=3sinθ-4sin3θ
cos3θ=4cos3-3cosθ
4. General formula
5. Sum-difference product formula
sinα+sinβ=
sinα-sinβ=
cosα+cosβ=
cosα-cosβ=
6. Product sum and difference formula
sinαsinβ= 1/2[sin(α+β)+sin(α-β)]
cosαsinβ= 1/2[sin(α+β)-sin(α-β)]
sinαsinβ- 1/2[cos(α+β)-cos(α-β)]
cosαcosβ= 1/2[cos(α+β)+cos(α-β)]
7. Half-angle formula
Another: formulas of trigonometric functions
Keywords triangular knowledge, self-sufficiency,
Memory formula, one, two, three, four.
First, the definition of trigonometric function,
Two systems, angle radian.
Keywords three groups of formulas, firm memory,
Same angle induction, addition theorem.
Keywords same angle formula, eight groups,
Square relation, derivative quotient.
Keywords inductive formula, two categories and nine groups,
Just like a finite number, even and odd numbers are the same.
Sum and difference of two angles, find sine,
JUNG WOO, this symbol is the same as before.
Sum and difference of two angles, cosine,
JUNG WOO is positive, but the sign is opposite.
Keywords two angles are equal, double angle formula,
Reverse thrust, half angle limit.
Addition, subtraction, variable replacement,
Product sum and difference, and fraction.
Acute angle formula of trigonometric function
Opposite side/hypotenuse of sinα=∞α
Adjacent edge/hypotenuse of cosα=∞α
Opposite side of tan α = adjacent side of ∠ α/∠α.
Adjacent side of cot α = opposite side of ∠ α/∠α.
Double angle formula
Sin2A=2SinA？ Kosa
cos2a=cosa^2-sina^2= 1-2sina^2=2cosa^2- 1
tan2A=(2tanA)/( 1-tanA^2)
(Note: Sina 2 is the square of Sina 2 (a))
Triple angle formula
sin3α=4sinα？ Sine (π/3+α) Sine (π/3-α)
cos3α=4cosα？ cos(π/3+α)cos(π/3-α)
tan3a = tan a？ tan(π/3+a)？ tan(π/3-a)
Derivation of triple angle formula
sin3a
=sin(2a+a)
=sin2acosa+cos2asina
Auxiliary angle formula
Asinα+bcosα = (A2+B2) (1/2) sin (α+t), where
sint=B/(A^2+B^2)^( 1/2)
cost=A/(A^2+B^2)^( 1/2)
tant=B/A
asinα+bcosα=(a^2+b^2)^( 1/2)cos(α-t),tant=a/b
Reduced power formula
sin^2(α)=( 1-cos(2α))/2=versin(2α)/2
cos^2(α)=( 1+cos(2α))/2=covers(2α)/2
tan^2(α)=( 1-cos(2α))/( 1+cos(2α))
Derived formula
tanα+cotα=2/sin2α
tanα-cotα=-2cot2α
1+cos2α=2cos^2α
1-cos2α=2sin^2α
1+sinα=(sinα/2+cosα/2)^2
=2sina( 1-sin？ a)+( 1-2sin？ A) Sina
=3sina-4sin？ a
cos3a
=cos(2a+a)
=cos2acosa-sin2asina
=(2cos？ a- 1)cosa-2( 1-sin？ a)cosa
=4cos？ a-3cosa
sin3a=3sina-4sin？ a
=4sina(3/4-sin？ answer
=4sina[(√3/2)？ Sin? Answer]
=4sina (sin? 60- sin? answer
=4sina(sin60 +sina)(sin60 -sina)
= 4 Sina * 2 sin[(60+a)/2]cos[(60-a)/2]* 2 sin[(60-a)/2]cos[(60-a)/2]
=4sinasin(60 +a)sin(60 -a)
cos3a=4cos？ a-3cosa
=4cosa(cos？ a-3/4)
= 4c OSA【cos？ a-(√3/2)？ ]
=4cosa(cos？ a-cos？ 30 )
=4cosa(cosa+cos30 )(cosa-cos30)
= 4 cosa * 2cos[(a+30)/2]cos[(a-30)/2]* {-2 sin[(a+30)/2]sin[(a-30)/2]}
=-4 eicosapentaenoic acid (a+30) octyl (a-30)
=-4 Coxsacin [90-(60-a)] Xin [-90 +(60 +a)]
=-4 cos(60-a)[-cos(60+a)]
= 4 cos(60-a)cos(60+a)
Comparing the above two formulas, we can get
tan3a=tanatan(60 -a)tan(60 +a)
half-angle formula
tan(A/2)=( 1-cosA)/sinA = sinA/( 1+cosA)；
cot(A/2)= sinA/( 1-cosA)=( 1+cosA)/sinA。
sin^2(a/2)=( 1-cos(a))/2
cos^2(a/2)=( 1+cos(a))/2
tan(a/2)=( 1-cos(a))/sin(a)= sin(a)/( 1+cos(a))
Triangular sum
sin(α+β+γ)=sinα？ cosβ？ cosγ+cosα？ sinβ？ cosγ+cosα？ cosβ？ sinγ-sinα？ sinβ？ sinγ
cos(α+β+γ)=cosα？ cosβ？ cosγ-cosα？ sinβ？ sinγ-sinα？ cosβ？ sinγ-sinα？ sinβ？ cosγ
tan(α+β+γ)=(tanα+tanβ+tanγ-tanα？ tanβ？ tanγ)/( 1-tanα？ tanβ-tanβ？ tanγ-tanγ？ tanα)
Sum and difference of two angles
cos(α+β)=cosα？ cosβ-sinα？ sinβ
cos(α-β)=cosα？ cosβ+sinα？ sinβ
sin(α β)=sinα？ cosβ cosα？ sinβ
tan(α+β)=(tanα+tanβ)/( 1-tanα？ tanβ)
tan(α-β)=(tanα-tanβ)/( 1+tanα？ tanβ)
Sum difference product
sinθ+sinφ= 2 sin[(θ+φ)/2]cos[(θ-φ)/2]
sinθ-sinφ= 2 cos[(θ+φ)/2]sin[(θ-φ)/2]
cosθ+cosφ= 2 cos[(θ+φ)/2]cos[(θ-φ)/2]
cosθ-cosφ=-2 sin[(θ+φ)/2]sin[(θ-φ)/2]
tanA+tanB = sin(A+B)/cosa cosb = tan(A+B)( 1-tanA tanB)
tanA-tanB = sin(A-B)/cosa cosb = tan(A-B)( 1+tanA tanB)
Sum and difference of products
sinαsinβ = [cos(α-β)-cos(α+β)] /2
cosαcosβ = [cos(α+β)+cos(α-β)]/2
sinαcosβ = [sin(α+β)+sin(α-β)]/2
cosαsinβ = [sin(α+β)-sin(α-β)]/2
Inductive formula
Sine (-α) =-Sine α
cos(-α) = cosα
tan (—a)=-tanα
sin(π/2-α) = cosα
cos(π/2-α) = sinα
sin(π/2+α) = cosα
cos(π/2+α) = -sinα
Sine (π-α) = Sine α
cos(π-α) = -cosα
Sine (π+α) =-Sine α
cos(π+α) = -cosα
tanA= sinA/cosA
tan(π/2+α)=-cotα
tan(π/2-α)=cotα
tan(π-α)=-tanα
tan(π+α)=tanα
Inductive formula memory skills: odd variables are unchanged, and symbols look at quadrants.
General formula of trigonometric function
sinα=2tan(α/2)/[ 1+tan^(α/2)]
cosα=[ 1-tan^(α/2)]/ 1+tan^(α/2)]
tanα=2tan(α/2)/[ 1-tan^(α/2)]
Other formulas
( 1)(sinα)^2+(cosα)^2= 1
(2) 1+(tanα)^2=(secα)^2
(3) 1+(cotα)^2=(cscα)^2
To prove the following two formulas, just divide one formula by (sin α) 2 and the second formula by (cos α) 2.
(4) For any non-right triangle, there is always
tanA+tanB+tanC=tanAtanBtanC
Certificate:
A+B=π-C
tan(A+B)=tan(π-C)
(tanA+tanB)/( 1-tanA tanB)=(tanπ-tanC)/( 1+tanπtanC)
Surface treatment can be carried out.
tanA+tanB+tanC=tanAtanBtanC
Obtain a certificate
It can also be proved that this relationship holds when x+y+z=nπ(n∈Z).
The following conclusions can be drawn from tana+tanbtana+tanb+tanc = tanatanbtanc.
(5)cotAcotB+cotAcotC+cotbctc = 1
(6) Cost (A/2)+ Cost (B/2)+ Cost (C/2)= Cost (A/2) Cost (B/2)
(7)(cosa)^2+(cosb)^2+(cosc)^2= 1-2cosacosbcosc
(8)(sina)^2+(sinb)^2+(sinc)^2=2+2cosacosbcosc
(9)sinα+sin(α+2π/n)+sin(α+2π* 2/n)+sin(α+2π* 3/n)+……+sin[α+2π*(n- 1)/n]= 0
Cos α+cos (α+2π/n)+cos (α+2π * 2/n)+cos (α+2π * 3/n)+...+cos [α+2π * (n-1)/n] = 0 and
sin^2(α)+sin^2(α-2π/3)+sin^2(α+2π/3)=3/2
tanAtanBtan(A+B)+tanA+tan B- tan(A+B)= 0
I wonder if this can meet your requirements. Images are really difficult to process, but they will. I often see sin and cos. Many people don't know Tan's image because it is rare. This is a strange function. Regarding the symmetry of the origin, increasing function is a complete image on both sides of the origin, with a period of π and a period of -π/2~π/2. The image of cot is a decreasing function, and (0~π) is a period.