The angular function formula of Mathematics

The angular function formula of Mathematics

Basic relations of trigonometric functions with the same angle
Reciprocal relation: Tan α · cot α = 1 sin α · CSC α = 1 cos α · sec α = 1 quotient relation: sin α / cos α = Tan α = sec α / CSC α cos α / sin α = cot α = CSC α / sec α square relation: sin ^ 2 (α) + cos ^ 2 (α) = 1 1 + Tan ^ 2 (α) = sec ^ 2 (α) 1 + cot ^ 2 (α) = CSC ^ 2 (α)
Two commonly used formulas for different conditions
sin² α+cos² α=1 tan α *cot α=1
A special formula
(Sina + sin θ) * (Sina + sin θ) = sin (a + θ) * sin (a - θ) prove: (Sina + sin θ) * (Sina + sin θ) = 2 sin [(θ + a) / 2] cos [(a - θ) / 2] * 2 cos [(θ + a) / 2] sin [(a - θ) / 2] = sin (a + θ) * sin (a - θ)
Trigonometric function formula of acute angle
Sine: sin α = - the opposite side of α / - the hypotenuse cosine of α: cos α = - the adjacent side of α / - the hypotenuse tangent of α: Tan α = - the opposite side of α / - the adjacent side of α: cot α = - the adjacent side of α / - the opposite side of α
Double angle formula
Cosine sin2a = 2sina · cosa cosine 1. Cos2a = cos ^ 2 (a) - Sin ^ 2 (a) = 2cos ^ 2 (a) - 1 = 1-2sin ^ 2 (a) 2. Cos2a = 1-2sin ^ 2 (a) 3. Cos2a = 2cos ^ 2 (a) - 1 tangent tan2a = (2tana) / (1-tan ^ 2 (a))
Triple angle formula
3 α = 4cos α · cos (π / 3 + α) cos (3 α = 4cos α · cos (π / 3 + α) cos (π / 3 + α) cos (π / 3 + 3 + α) cos (π / 3-α) tan3a = Tan a · Tan (π / 3 + 3 + a) · Tan (π / 3 + 3 + 3 + 3 + 3 + 3 + 3 + α) sin (π / 3-α) sin (π / 3-α) cosi3 (3-α) cos3 α = 4cos α = 4cos α · cos (π / 3) the three-fold formula derivation of three-three-angle formula derivation three-three-three-three-three-fold formula derivation of three-three-three-three-three-fold formula derivation of three-three-three-three-three-three-three-three-three-fold formula derivation formula derivation formula derivation formula derivation: sin (3a) = sin (3a (3a) = sin (a (a (a (a (a) = sin (a 2asina = (2cos & # 178; a-1)cosa-2(1-cos^a)cosa =4cos^3a-3cosa sin3a=3sina-4sin^3a =4sina(3/4-sin²a) =4sina[(√3/2)²-sin²a] =4sina(sin²60°-sin²a) =4sina(sin60°+sina)(sin60°-sina) =4sina*2sin[(60+a)/2]cos[(60°-a)/2]*2sin[(60°-a)/2]cos[(60°-a)/2] =4sinasin(60°+a)sin(60°-a) cos3a=4cos^3a-3cosa =4cosa(cos² a-3/4) =4cosa[cos²a-(√3/2)^2] =4cosa(cos²a-cos² (cosa-cos 30) ((cosa-cos 30)) (cosa-cos 30 ℃) = 4cosa * 2cos [(a + 30 ℃) / 2] cos [(A-30 ℃) / 2] * {{- 2Sin [(a + 30 ℃) / 2] * {{- 2Sin [(a + 30 ℃) / 2] sin [(A-30 ℃) / 2] * {{{- 2Sin [(a + 30 ℃ / 2)] {{{- 2Sin [(a + 30 ℃ (a + 30 ℃) / 2 / 2)] {{{- 2Sin [(a + 30 ℃ (cosa + cosa + cosa + cosa + 30 (COSA + 30 ℃)))) (COSA (COSA (COSA (cosa-30 (cosa-30 (cosa-30 (cosa-30 (cosa-30 (cosa-cosa-30 (COSA) and COSA) and cosa-cosa) is: COSA) and the 60 ° - a) Tan (60 ° + a)
N-fold angle formula
sin（n a）=Rsina sin（a+π/n）…… Where r = 2 ^ (n-1) it is proved that when sin (NA) = 0, Sina = sin (π / N) or = sin (2 π / N) or = sin (3 π / N) or = Or = sin [(n-1) π / N] which means that sin (NA) = 0 and {Sina sin (π / N)} * {Sina sin (2 π / N)} * {Sina sin (3 π / N)} * *So sin (NA) and {Sina sin (π / N)} * {Sina sin (2 π / N)} * {Sina sin (3 π / N)} * *(Sina + sin θ) * (Sina + sin θ) = sin (a + θ) * sin (a - θ), so {Sina sin (π / N)} * {Sina sin (2 π / N)} * {Sina sin (3 π / N)} * *{Sina - sin [(n-1 π / N] and Sina sin (a + π / N) The coefficient of sin (2n a) is R2N = R2 * (RN) ^ 2 = RN * (R2) ^ n. It is easy to prove that R2 = 2, so RN = 2 ^ (n-1)
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)/cosAcosB=tan(A+B)(1-tanAtanB) tanA-tanB=sin(A-B)/cosAcosB=tan(A-B)(1+tanAtanB)
Formula of sum of two angles
cos(α+β)=cosαcosβ-sinαsinβcos(α-β)=cosαcosβ+sinαsinβsin(α+β)=sinαcosβ+cosαsinβsin(α-β)=sinαcosβ -cosαsinβ
Product sum difference
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 any angle, and the values of the same trigonometric function of the same angle with the same terminal edge are equal: sin (2k π + α) = sin α cos (2k π + α) = cos α Tan (2k π + α) = Tan α cot (2k π + α) = cot α formula 2: let α be any angle, The relationship between the trigonometric function value of π + α and the trigonometric function value of α: sin (π + α) = - sin α cos (π + α) = - cos α Tan (π + α) = Tan α cot (π + α) = cot α formula 3: the relationship between the trigonometric function value of any angle α and - α: sin (- α) = - sin α cos (- α) = cos α Tan (- α) = - Tan α cot (- α) = - cot α formula 4: The relationship between π - α and the trigonometric value of α can be obtained by formula 2 and formula 3: sin (π - α) = sin α cos (π - α) = - cos α Tan (π - α) = - Tan α cot (π - α) = - cot α formula 5: the relationship between 2 π - α and the trigonometric value of α can be obtained by formula 3: sin (2 π - α) = - sin α cos (2 π - α) = cos α Tan (2 π - α) = - Tan α cot (2 π - α) = - cot α formula 6: the relationship between π / 2 ± α and the trigonometric value of α: sin (π / 2 + α) = cos α cos (π / 2 + α) = - sin α Tan (π / 2 + α) = - cot α cot (π / 2 + α) = - Tan α sin (π / 2 - α) = cos α cos (π / 2 - α) = sin α Tan (π / 2 - α) = cot α cot (π / 2 - α) = Tan α Sin (3 π / 2 + α) = - cos α cos (3 π / 2 + α) = sin α Tan (3 π / 2 + α) = - cot α cot (3 π / 2 + α) = - Tan α sin (3 π / 2 - α) = - cos α cos (3 π / 2 - α) = - sin α Tan (3 π / 2 - α) = cot α cot (3 π / 2 - α) = Tan α (above K ∈ z) a · sin (ω T + θ) + B · sin (ω T + φ) = {(A & # 178; + B & # 178; +2abcos (θ - φ)} · sin {ω T + arcsin [(a · sin θ + B · sin φ) / √ {a ^ 2 + B ^ 2; + 2abcos (θ - φ)}} √ denotes the root sign, including { }Content in
Induction formula
sin(-α) = -sinα 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 sin(π-α) = sinα cos(π-α) = -cosα sin(π+α) = -sinα cos(π+α) = -cosα tanA= sinA/cosA tan（π/2＋α）＝－cotα tan（π/2－α）＝cotα tan（π－α）＝－tanα Tan (π + α) = Tan α induction formula memorize the secret: odd variable, even constant, sign look at quadrant
Universal formula
sinα=2tan(α/2)/[1+(tan(α/2))²] cosα=[1-(tan(α/2))²]/[1+(tan(α/2))²] tanα=2tan(α/2)/[1-(tan(α/2))²]
Other formulas
(1) (sin α) & # 178; + (COS α) & # 178; = 1 (2) 1 + (Tan α) & # 178; = (SEC α) & # 178; (3) 1 + (cot α) & # 178; = (CSC α) & # 178; to prove the following two formulas, we only need to divide left and right by (sin α) & # 178;, the second division (COS α) & # 178; in the same way. (4) for any non right triangle, There are always Tana + tanb + Tanc = tanatanbtanc syndrome: a + B = π - C, Tan (a + b) = Tan (π - C) (Tana + tanb) / (1-tanatanb) = (Tan π - Tanc) / (1 + Tan π Tanc) finishing, Tana + tanb + Tanc = tanatanbtanc syndrome can also be obtained, when x + y + Z = n π (n ∈ z), the same can be obtained, The following conclusions can be drawn from the Tana + tanb + tanb + Tanc = tanatan, and the following conclusions can be drawn from the Tana + Tana + tanb + Tanc (5) the following conclusions (5) cotacotbb + cotacotc + cotbtbtcoc = 1 (6) cot (A / 2) + cot (B / 2) + cot (B / 2) + cot (B / 2) + cot (B / B + Tan C = Tana + Tanc = Tana, and the following conclusions (5) cotacotbb + cotacotbb + cotacotc + cotacotc + cot btacotc + cot (c + cot (cot (c) + (cot (cot (cot (c) + (COT C) + (cot (COT) cot (cot (COT) 3535\\35\\35\3535178; =Other non key trigonometric functions CSC (a) = 1 / sin (a) sec (a) = 1 / cos (a)
Edit the rules of this paragraph
Trigonometric functions seem to be many and complex, but as long as you master the essence and internal laws of trigonometric functions, you will find that there are strong connections between the various formulas of trigonometric functions
[1] According to the figure on the right, there are sin θ = Y / R; Cos θ = x / R; Tan θ = Y / X; cot θ = x / y. with a deep understanding of this point, all the following trigonometric formulas can be derived from here. For example, take the derivation of sin (a + b) = sinacosb + cosasinb as an example: derivation: first, draw the x-axis of the unit circle at C and D, and there are any a and B points on the unit circle. Angle AOD is α, BOD is β, rotate AOB to make ob coincide with OD, A (COS α, sin α), B (COS β, sin β), a '(COS (α - β), sin (α - β)) OA' = OA = ob = od = 1, D (1, 0) ∧ [cos (α - β) - 1] ^ 2 + [sin (α - β)] ^ 2 = (COS α - cos β) ^ 2 + (sin α - sin β) ^ 2 and the difference product and the integration and difference reduction method can be combined with the above formula to deduce the definition of unit circle by (replacing (a + b) / 2 and (a-b) / 2) Six trigonometric functions can also be defined according to the unit circle whose radius is one and center is the origin. The definition of unit circle is not of great value in practical calculation. In fact, for most angles, it depends on right triangles. But the definition of unit circle does allow trigonometric functions to have definitions for all positive and negative radii, According to the Pythagorean theorem, the equation of the unit circle is as follows: some common angles measured in radians are given in the image. The counter clockwise measurement is a positive angle, while the clockwise measurement is a negative angle, The X and Y coordinates of the intersection are equal to cos θ and sin θ, respectively. The triangle in the image ensures this formula; the radius is equal to the hypotenuse and the length is 1, so there are sin θ = Y / 1 and cos θ = x / 1. The unit circle can be regarded as the unit circle by changing the length of the adjacent and opposite sides, But keeping the hypotenuse equal to 1 is a way to see an infinite number of triangles
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-tanAtanB) tan(A-B) = (tanA-tanB)/(1+tanAtanB) cot(A+B) = (cotAcotB-1)/(cotB+cotA) cot(A-B) = (cotAcotB+1)/(cotB-cotA)

A person drives 100 kilometers to a place. The speed of the first 60 kilometers is 100, and the speed of the last 40 kilometers is 120/（
Calculation formula: 100 / (60 / 100 + 40 / 120) = 750 / 7
Q: what do 750 and 7 stand for and how are they calculated?

100/（60/100+40/120）=100/(720/1200+400/1200)=100*1200/1120=750*160/7*160=750/7

Please excel formula function master to explain the meaning of this formula for me
=ROUND((B3-MAX(B5-B4,0))*LOOKUP(-(B3-MAX(B5-B4,0))/12,-1000*{9999,100,80,60,40,20,5,2,0.5,0},5%*{9,8,7,6,5,4,3,2,1,0})-LOOKUP(-(B3-MAX(B5-B4,0))/12,-1000*{9999,100,80,60,40,20,5,2,0.5,0},25*{615,415,255,135,55,15,5,1,0,0}),2)
The above formula is the calculation formula of the income tax of the year-end bonus. Cell B3 is the year-end bonus, cell B4 is the salary of the current month, and cell B5 is the starting point. At present, it is 2000 yuan
I don't know much about the definition of Excel functions, especially the big string of numbers in braces