Let A, B and C be three internal angles of a triangle.

tanA+tanB+tanC=tanAtanBtanC

cotAcotB+cotBcotC+cotCcotA = 1

(cosa)^2+(cosb)^2+(cosc)^2+2cosacosbcosc= 1

cosA+cosB+cosC = 1+4 sin(A/2)sin(B/2)sin(C/2)

tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)= 1

sin2A+sin2B+sin2C = 4 sinasinbsinc

sinA+sin B+sinC = 4 cos(A/2)cos(B/2)cos(C/2)

Double angle formula

sin2A=2sinA？ Kosa

cos2a=cos^2a-sin^2a= 1-2sin^2a=2cos^2a- 1

tan2A=(2tanA)/( 1-tan^2A)

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

=2sina( 1-sin^2a)+( 1-2sin^2a)sina

=3sina-4sin^3a

cos3a

=cos(2a+a)

=cos2acosa-sin2asina

=(2cos^2a- 1)cosa-2( 1-cos^a)cosa

=4cos^3a-3cosa

sin3a=3sina-4sin^3a

=4sina(3/4-sin^2a)

=4sina[(√3/2)^2-sin^2a]

=4sina(sin^260 -sin^2a)

=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^3a-3cosa

=4cosa(cos^2a-3/4)

=4cosa[cos^2a-(√3/2)^2]

=4cosa(cos^2a-cos^230)

=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))

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

Hyperbolic function

sinh(a) = [e^a-e^(-a)]/2

cosh(a) = [e^a+e^(-a)]/2

tanh(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)

a sin(ωt+θ)+B sin(ωt+φ)= 1

√{(A^2 +B^2 +2ABcos(θ-φ)}？ sin{ ωt + arcsin[ (A？ sinθ+B？ sinφ)/√{a^2 +b^2； +2ABcos(θ-φ)} }

√ indicates the root number, including the contents in {...}.

Inductive formula

Sine (-α) =-Sine α

cos(-α) = cosα

tan (-α)=-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.

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

Other non-critical trigonometric functions

csc(a) = 1/sin(a)

Seconds (a)= 1/ cosine (a)