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 sin(π/3+α)sin(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)= sinαcosβcosγ+cosαsinβcosγ+cosαcosβsinγ-sinαsinβsinγ
cos(α+β+γ)= cosαcosβcosγ-cosαsinβsinγ-sinαcosβsinγ-sinαsinαsinβcosγ-sinαsinβcosγ
tan(α+β+γ)=(tanα+tanβ+tanγ-tanαtanβtanγ)/( 1-tanαtanβ-tanβtanγ-tanγtanα)
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-tanαtanβ)
tan(α-β)=(tanα-tanβ)/( 1+tanαtanβ)
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