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 2π-α and the trigonometric function value of α can be obtained by formula 1 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)

Inductive formula memory formula

Summary of the law. ※。

The above inductive formula can be summarized as follows:

For the trigonometric function value of k π/2 α (k ∈ z),

① When k is an even number, the function value of α with the same name is obtained, that is, the function name is unchanged;

② When k is an odd number, the cofunction value corresponding to α is obtained, that is, sin→cos；; cos→sin； Tan → Kurt, Kurt → Tan.

(Odd and even numbers remain the same)

Then when α is regarded as an acute angle, the sign of the original function value is added.

(Symbols look at quadrants)

For example:

Sin (2π-α) = sin (4 π/2-α), and k = 4 is an even number, so we take sinα.

When α is an acute angle, 2π-α ∈ (270,360), sin (2π-α) < 0, and the symbol is "-".

So sin (2 π-α) =-sin α.

The above memory formula is:

Odd couples, symbols look at quadrants.

The symbols on the right side of the formula are angles k 360+α (k ∈ z),-α, 180 α, and when α is regarded as an acute angle, it is 360-α.

The sign of the original trigonometric function value in the quadrant can be remembered.

The name of horizontal induction remains unchanged; Symbols look at quadrants.

How to judge the symbols of various trigonometric functions in four quadrants, you can also remember the formula "a full pair; Two sinusoids; The third is cutting; Four cosines ".

The meaning of this 12 formula is:

The four trigonometric functions at any angle in the first quadrant are "+";

In the second quadrant, only the sine is "+",and the rest are "-";

The tangent function of the third quadrant is+and the chord function is-.

In the fourth quadrant, only cosine is "+",others are "-".

Other trigonometric function knowledge:

Basic relations of trigonometric functions with the same angle

1. Basic relations of trigonometric functions with the same angle.

Reciprocal relationship:

tanα cotα= 1

sinα cscα= 1

cosα secα= 1

Relationship between businesses:

sinα/cosα=tanα=secα/cscα

cosα/sinα=cotα=cscα/secα

Square relation:

sin^2(α)+cos^2(α)= 1

1+tan^2(α)=sec^2(α)

1+cot^2(α)=csc^2(α)

Hexagon memory method of equilateral trigonometric function relationship

Hexagonal mnemonics: (see pictures or links to resources)

The structure is "winding, cutting and cutting; Zuo Zheng, the right remainder and the regular hexagon of the middle 1 "are models.

(1) Reciprocal relation: The two functions on the diagonal are reciprocal;

(2) Quotient relation: the function value at any vertex of a hexagon is equal to the product of the function values at two adjacent vertices.

(Mainly the product of trigonometric function values at both ends of two dotted lines). From this, the quotient relation can be obtained.

(3) Square relation: In a triangle with hatched lines, the sum of squares of trigonometric function values on the top two vertices is equal to the square of trigonometric function values on the bottom vertex.

Two-angle sum and difference formula

2. The sum and difference of formulas of trigonometric functions.

sin(α+β)=sinαcosβ+cosαsinβ

sin(α-β)=sinαcosβ-cosαsinβ

cos(α+β)=cosαcosβ-sinαsinβ

cos(α-β)=cosαcosβ+sinαsinβ

tanα+tanβ

tan(α+β)=———

1-tanα tanβ

tanα-tanβ

tan(α-β)=———

1+tanα tanβ

Double angle formula

13. Double angle sine, cosine and tangent formulas (increasing power and decreasing angle formula)

sin2α=2sinαcosα

cos2α=cos^2(α)-sin^2(α)=2cos^2(α)- 1= 1-2sin^2(α)

2tanα

tan2α=———

1-tan^2(α)

half-angle formula

4. Sine, Cosine and Tangent Formulas of Half Angle (Power Decreasing and Angle Expanding Formulas)

1-cosα

sin^2(α/2)=—————

2

1+cosα

cos^2(α/2)=—————

2

1-cosα

tan^2(α/2)=—————

1+cosα

General formula of trigonometric function

⒌ General formula

2 tons (α/2)

sinα=————

1+tan^2(α/2)

1-tan^2(α/2)

cosα=————

1+tan^2(α/2)

2 tons (α/2)

tanα=————

1-tan^2(α/2)

Derivation of universal formula

Additional derivation:

sin2α=2sinαcosα=2sinαcosα/(cos^2(α)+sin^2(α))......*,

(Because cos 2 (α)+sin 2 (α) = 1)

Divide the * fraction up and down by COS 2 (α) to get SIN 2 α = TAN 2 α/( 1+TAN 2 (α)).

Then replace α with α/2.

Similarly, the universal formula of cosine can be derived. By comparing sine and cosine, a general formula of tangent can be obtained.

Triple angle formula

Sine, cosine and tangent formulas of triple angle

sin3α=3sinα-4sin^3(α)

cos3α=4cos^3(α)-3cosα

3tanα-tan^3(α)

tan3α=—————

1-3tan^2(α)

Derivation of triple angle formula

Additional derivation:

tan3α=sin3α/cos3α

=(sin 2αcosα+cos 2αsinα)/(cos 2αcosα-sin 2αsinα)

=(2sinαcos^2(α)+cos^2(α)sinα-sin^3(α))/(cos^3(α)-cosαsin^2(α)-2sin^2(α)cosα)

Divided by COS 3 (α), we get:

tan3α=(3tanα-tan^3(α))/( 1-3tan^2(α))

sin 3α= sin(2α+α)= sin 2αcosα+cos 2αsinα

=2sinαcos^2(α)+( 1-2sin^2(α))sinα

=2sinα-2sin^3(α)+sinα-2sin^2(α)

=3sinα-4sin^3(α)

cos 3α= cos(2α+α)= cos 2αcosα-sin 2αsinα

=(2cos^2(α)- 1)cosα-2cosαsin^2(α)

=2cos^3(α)-cosα+(2cosα-2cos^3(α))

=4cos^3(α)-3cosα

that is

sin3α=3sinα-4sin^3(α)

cos3α=4cos^3(α)-3cosα

Associative memory of triangle formula

Memory methods: homophonic and associative.

Sine Triangle: 3 yuan minus 4 yuan Triangle (debt (minus negative number), so "making money" (sounds like "sine").

Cosine triple angle: 4 yuan minus 3 yuan (there is a "remainder" after subtraction).

☆☆ Pay attention to the function name, that is, the triple angle of sine is expressed by sine, and the triple angle of cosine is expressed by cosine.

Sum-difference product formula

Χ sum and product of trigonometric functions

α+β α-β

sinα+sinβ=2sin— - cos— -

2 2

α+β α-β

sinα-sinβ=2cos— - sin— -

2 2

α+β α-β

cosα+cosβ=2cos— - cos— -

2 2

α+β α-β

cosα-cosβ=-2sin— - sin— -

2 2

Sum and difference formula of product

⒏ Formula of product and difference of trigonometric functions.

sinαcosβ= 0.5[sin(α+β)+sin(α-β)]

cosαsinβ= 0.5[sin(α+β)-sin(α-β)]

cosαcosβ= 0.5[cos(α+β)+cos(α-β)]

sinαsinβ=-0.5[cos(α+β)-cos(α-β)]

Derivation of sum-difference product formula

Additional derivation:

First of all, we know that SIN (a+b) = Sina * COSB+COSA * SINB, SIN (a-b) = Sina * COSB-COSA * SINB.

We add these two expressions to get sin(a+b)+sin(a-b)=2sina*cosb.

So sin a * cosb = (sin (a+b)+sin (a-b))/2.

Similarly, if you subtract the two expressions, you get COSA * SINB = (SIN (A+B)-SIN (A-B))/2.

Similarly, we also know that COS (a+b) = COSA * COSB-SINA * SINB, COS (a-b) = COSA * COSB+SINA * SINB.

Therefore, by adding the two expressions, we can get cos(a+b)+cos(a-b)=2cosa*cosb.

So we get, COSA * COSB = (COS (A+B)+COS (A-B))/2.

Similarly, by subtracting two expressions, Sina * sinb =-(cos (a+b)-cos (a-b))/2 can be obtained.

In this way, we get the formulas of the sum and difference of four products:

Sina * cosb =(sin(a+b)+sin(a-b))/2

cosa * sinb =(sin(a+b)-sin(a-b))/2

cosa * cosb =(cos(a+b)+cos(a-b))/2

Sina * sinb =-(cos(a+b)-cos(a-b))/2

Well, with four formulas of sum and difference, we can get four formulas of sum and difference product with only one deformation.

Let a+b be X and A-B be Y in the above four formulas, then A = (X+Y)/2 and B = (X-Y)/2.

If a and b are represented by x and y respectively, we can get four sum-difference product formulas:

sinx+siny = 2 sin((x+y)/2)* cos((x-y)/2)

sinx-siny = 2cos((x+y)/2)* sin((x-y)/2)

cosx+cosy = 2cos((x+y)/2)* cos((x-y)/2)

cosx-cosy =-2 sin((x+y)/2)* sin((x-y)/2)

vector operation

Add operation

AB+BC = AC, this calculation rule is called triangle rule of vector addition.

It is known that the two vectors OA and OB starting from the same point O are parallelogram OACB, and the diagonal OC starting from O is the sum of the vectors OA and OB. This calculation method is called parallelogram rule of vector addition.

For zero vector and arbitrary vector a, there are: 0+a = a+0 = a.

|a+b|≤|a|+|b| .

The addition of vectors satisfies all the laws of addition.

subtraction

The vector with the same length and opposite direction as A is called the inverse quantity of A, -(-a) = A, and the inverse quantity of zero vector is still zero vector.

( 1)a+(-a)=(-a)+a = 0(2)a-b = a+(-b).

multiply operation

The product of real number λ and vector A is a vector, and this operation is called vector multiplication, which is denoted as λa, | λ A | = | λ| | A |. When λ > 0, the direction of λ A is the same as that of A. When λ

Let λ and μ be real numbers, then: (1) (λ μ) a = λ (μ a) (2) (λ+μ) a = λ a+μ a (3) λ (ab) = λ a λ b (4) (-λ) a =-(.

Chemistry: summary of knowledge points of organic chemistry in senior high school

1. The reactions requiring water bath heating are:

(1), silver mirror reaction (2), hydrolysis of ethyl acetate (3) nitration of benzene (4) hydrolysis of sugar.

(5) Preparation of Phenolic Resin (6) Determination of Solid Solubility

Any reaction with a temperature not higher than 100℃ can be heated in a water bath, and the temperature changes smoothly without ups and downs, which is beneficial to the reaction.

2. The experiments that need a thermometer are:

(1), laboratory ethylene (170℃) (2), distillation (3), determination of solid solubility.

(4) Hydrolysis of ethyl acetate (70-80℃) (5) Determination of neutralization heat.

(6) generating nitrobenzene (50-60 DEG C)

[Description]: (1) Anyone who needs to control the temperature accurately must use a thermometer. (2) Pay attention to the position of the mercury ball of the thermometer.

3. Organic substances that can react with Na are: alcohol, phenol, carboxylic acid, etc. -All compounds containing hydroxyl groups.

4. The substances that can react with the silver mirror are:

Aldehyde, formic acid, formate, formate, glucose, maltose-all substances containing aldehyde groups.

5. The substances that can make the acidic solution of potassium permanganate fade are:

(1) hydrocarbons containing carbon-carbon double bonds and carbon-carbon triple bonds, derivatives of hydrocarbons and homologues of benzene.

(2) Compounds containing hydroxyl groups, such as alcohols and phenols

(3) Compounds containing aldehyde groups

(4) Reducing inorganic substances (such as SO2, FeSO4, KI, HCl, H2O2, etc.). )

6. Substances that can make bromine water fade are:

(1) hydrocarbons and hydrocarbon derivatives with carbon-carbon double bonds and carbon-carbon triple bonds (addition)

(2) Phenols, such as phenol (substitute)

(3) Aldehyde-containing substances (oxidation)

(4) Alkaline substances (such as NaOH and Na2CO3) (redox-disproportionation reaction)

(5) Strong inorganic reducing agents (such as SO2, KI, FeSO4, etc.). ) (oxidation)

(6) Organic solvents (such as benzene and benzene homologues, carbon tetrachloride, gasoline, hexane, etc.). ) belongs to extraction, which makes the water layer fade and the organic layer is orange-red. )

7. Liquid organics with density greater than water include: bromoethane, bromobenzene, nitrobenzene, carbon tetrachloride, etc.

8. Liquid organics with lower density than water are hydrocarbons, most esters and monochloroalkanes.

9. The substances that can undergo hydrolysis reaction are

Halogenated hydrocarbon, ester (fat), disaccharide, polysaccharide, protein (peptide), salt.

10. The water-insoluble organic matter is:

Hydrocarbons, halogenated hydrocarbons, esters, starch, cellulose

1 1. Organic substances that are gaseous at room temperature are:

Hydrocarbons (except neopentane), methyl chloride and formaldehyde with less than or equal to 4 carbon atoms in the molecule.

12. The reaction under the conditions of concentrated sulfuric acid and heating is as follows:

Nitration, sulfonation, alcohol dehydration, cellulose esterification and hydrolysis of benzene and its homologues.

13. The substances that may be oxidized are:

Unsaturated compound containing carbon-carbon double bond or carbon-carbon triple bond (KMnO4), homologues of benzene, alcohol, aldehyde and phenol.

Most organic matter can be burned, and combustion is oxidized by oxygen.

14. acidic organic compounds are compounds containing phenolic hydroxyl groups and carboxyl groups.

15. The substances that can denature protein are: strong acid, strong alkali, heavy metal salt, formaldehyde, phenol, strong oxidant, strong alcohol, hydrogen peroxide, iodine, trichloroacetic acid, etc.

16. Organic substances that can react with acid and alkali: organic substances with acid-base bifunctional groups (amino acids, protein, etc. ).

17. Organic substances that can react with NaOH solution:

(1) phenol:

(2) Carboxylic acid:

(3) halogenated hydrocarbon (aqueous solution: hydrolysis; Alcohol solution: eliminate)

(4) Ester: (Hydrolysis, slow reaction without heating, fast reaction with heating)

(5) protein (hydrolysis)

18, organic reaction with obvious color change:

1. Phenol reacts with ferric chloride solution to be purple;

2.2 fading. Potassium permanganate solution;

3. Bromine water fades;

Starch turns blue when it comes into contact with iodine.

5. protein will turn yellow when it encounters concentrated nitric acid (color reaction).

I. Physical characteristics

Methane: colorless, odorless and insoluble.

Ethylene: colorless, slightly odorous and insoluble.

Acetylene: colorless, tasteless, slightly soluble.

(Calcium carbide generation: particularly unpleasant smell containing H2S and PH3)

Benzene: Colorless liquid, with special smell, insoluble and toxic.

Ethanol: colorless, with special fragrance, easily miscible and volatile.

Acetic acid: colorless, irritating, soluble and volatile.

Second, the laboratory method

Methane: CH3COONa+NaOH →(CaO, heating) → CH4↑+Na2CO3.

Note: anhydrous sodium acetate: alkaline lime = 1: 3.

Solid-solid heating (same as O2 and NH3)

Anhydrous (NaAc crystals cannot be used)

CaO: Absorbs moisture and dilutes NaOH, not a catalyst.

Ethylene: C2H5OH → (concentrated H2SO4, 170℃)→ CH2=CH2↑+H2O.

Note: V alcohol: V concentrated sulfuric acid = 1: 3 (dehydrated, the mixed solution is brown).

The temperature of drainage and collection (the same as that of Cl2 and HCl) is controlled at 170℃( 140℃: ether).

Removal of sulfur dioxide and carbon dioxide from alkali lime

Broken porcelain: prevent boiling

Acetylene: CaC2+2H2O → C2H2↑+Ca(OH)2

Note: Drainage and collection are not impurity-free.

You can't use kip generator.

Saturated sodium chloride: reducing the reaction rate

Put cotton at the catheter mouth: to prevent slightly soluble Ca(OH)2 foam from blocking the catheter.

Ethanol: CH2=CH2+H2O → (catalyst, heating, pressurizing) →CH3CH2OH

I don't know if this is an industry or a laboratory. . . )

Note: Test water with anhydrous copper sulfate (white → blue)

Thickening: adding CaO, and then heating and distilling.

Third, the burning phenomenon.

Alkanes: The flame is light blue and not bright.

Olefin: Bright flame with black smoke.

Acetylene: bright flame with strong black smoke (in pure oxygen above 3000℃: oxyacetylene flame)

Benzene: bright flame and a lot of black smoke (homoyne)

Alcohol: The flame is light blue, releasing a lot of heat.

4. Acid KMnO4 &; bromine water

Alkanes: none of them fade.

Alkynes: all faded (former oxidized, latter added)

Decolorization of Bromine Water by Benzene Extraction with Potassium Permanganate without Decolorization

Verb (abbreviation of verb) is an important reaction equation.

Alkanes: substitution

CH4+Cl2 → (illumination) → CH3Cl (g) +HCl

CH3Cl+Cl2 → (lighting) → ch2cl2 (L)+HCl

CH2Cl+Cl2 → (lighting) → chcl3 (l)+HCl

CHCl3+Cl2 → (lighting) → CCl4(l)+HCl

Phenomenon: The color becomes lighter, and there is oily liquid on the device wall.

Note: Of the four products, only methyl chloride is a gas.

Chloroform = chloroform

Carbon tetrachloride as fire extinguishing agent

Olefin: 1, addition

CH2=CH2 + Br2 → CH2BrCH2Br

CH2=CH2+HCl → (catalyst )→ ch3ch2cl

CH2=CH2+H2 → (catalyst, heating )→ CH3H3 ethane

CH2=CH2+H2O → (catalyst, heating and pressurizing )→ CH3 H2OH ethanol.

2. Polymerization (addition polymerization)

NCH2=CH2 → (under some conditions )→ [-CH2-CH2-] n

(monomer → polymer)

Note: Breaking the Double Bond → Two "Half Bonds"

Polymers (macromolecular compounds) are all mixtures.

Acetylene: basically the same olefin. . .

Benzene: 1. 1, substituted (bromine)

◎+Br2 →(Fe or FeBr3)→ ◎-Br+HBr

Note: V benzene: V bromine = 4: 1.

Long conduit: condensate reflux gas conduit.

Reverse suction

Removal of NaOH impurities

Phenomenon: there is white fog at the mouth of the catheter, light yellow precipitate (AgBr), CCl4: brown water-insoluble liquid (bromobenzene).

1.2, substitution-nitration (nitric acid)

◎+HNO3 → (concentrated H2SO4, 60℃)→ ◎-NO2+H2O

Note: Add concentrated nitric acid first, then concentrated sulfuric acid, cool to room temperature, and then add benzene.

Insert a 50℃-60℃ water bath thermometer into a beaker.

Except mixed acid: NaOH

Nitrobenzene: colorless oily liquid, insoluble bitter almond, toxic.

1.3, substitution-sulfonation (concentrated sulfuric acid)

◎+H2SO4 (concentrated) → (70-80 degrees) → ◎-SO3H+H2O

Step 2 add

◎+3H2 → (Nickel, heated) → ◎ (cyclohexane)

Alcohol: 1, displacement (active metal)

2CH3CH2OH + 2Na → 2CH3CH2ONa + H2↑

The density of sodium is higher than that of alcohol and the reaction is stable.

{cf.} When the density of sodium is less than that of water, a violent reaction will occur.

2. Elimination (intramolecular dehydration)

C2H5OH → (concentrated H2SO4, 170℃)→ CH2=CH2↑+H2O

3. Substitution (intermolecular dehydration)

2CH3CH2OH → (concentrated H2SO4, 140 degree) → CH3CH2OCH2CH3 (ether) +H2O.

Ether: colorless, nontoxic and volatile liquid anesthetic.

4. Catalytic oxidation

2CH3CH2OH+O2 →(Cu, heating) → 2CH3CHO (acetaldehyde) +2H2O

Phenomenon: The surface of copper wire turns black and red after being soaked in alcohol. This liquid has a special pungent smell.

Acid: substitution (esterification)

CH3COOH+C2H5OH → (concentrated H2SO4, heated) → CH3COOC2H5+H2O

(Ethyl acetate: odorous colorless oily liquid)

Note: Hydroxyl alcohol is dehydrogenated by acid (isotope tracer method).

Broken porcelain: prevent boiling

Concentrated sulfuric acid: catalytic dehydration and water absorption

Saturated Na2CO3: easy to separate and purify.

Halogenated hydrocarbon: 1, substituted (hydrolyzed) NaOH aqueous solution

CH3CH2X+NaOH →(H2O, heating) → CH3CH2OH+NaX

Note: NaOH function: neutralize HBr and speed up the reaction.

Experiment X: Add AgNO3 acidified by nitric acid and observe the precipitation.

2, removing the NaOH alcohol solution

CH3CH2Cl+NaOH → (alcohol, heating )→ CH2 = CH2 ↑+NaCl+H2O.

Note: it can only be eliminated if there is H on the adjacent C atom.

Add H in many places and remove H in few places (Mahalanobis's Law).

Alcohol solution: inhibition of hydrolysis (inhibition of NaOH ionization)

General formula of intransitive verbs

CnH2n+2 alkane

CnH2n olefins/cycloalkanes

CnH2n-2 alkynes/dienes

CnH2n-6 benzene and its homologues

CnH2n+2O monohydric alcohol/alkyl ether

CnH2nO saturated monoaldehyde/ketone

CnH2n-6O aromatic alcohol/phenol

CnH2nO2 carboxylic acid/ester

Seven, other knowledge points

1, Tiangan name: a, b, c, d, e, g, n, n, n.

2. Combustion formula: CxHy+(x+y/4)O2 → (ignition) → x CO2+y/2 H2O.

CxHyOz+(x+y/4-z/2)O2 → (ignition) → x CO2+y/2 H2O.

3. The pressure/volume is unchanged before and after the reaction: y = 4.

Smaller: y

Get bigger: y

4. Oxygen consumption: the amount of the same substance (equal to V): The more C, the more oxygen consumption.

Equal mass: The higher C%, the less oxygen consumption.

5. Unsaturation (ω ~) = (number of C atoms × 2+2–number of H atoms) /2

Double bond/ring = 1, triple bond = 2, which can be superimposed.

6. Industrial olefin production: cracking (non-cracking)

7. Medical alcohol: 75%

Industrial alcohol: 95% (methanol is toxic)

Anhydrous alcohol: 99%

8. Glycerol: Glycerol

9. The acidity of acetic acid is between HCl and H2CO3.

Vinegar: 3%~5%

Glacial acetic acid: pure acetic acid.

10, and alkyl is not a functional group.