Complete set of basic formulas of senior high school mathematics

I have been studying hard at the cold window for more than ten years, and now the exam is on the verge; Think calmly and calmly, and write like a genius exhibition; Calm and confident, customs clearance will be like running water; Be careful, be patient, study hard and prepare for the exam, and you will be admitted to the ideal university. Next, I sorted out the basic formulas of high school mathematics for you. I hope you like it!

A Complete Collection of Basic Mathematics Formulas in Senior High School 1

How to derive the compound function f[g(x)], let g(x)=u, then f[g(x)]=f(u),

Therefore (formula): f'[g(x)]=f'(u)_'(x)

Hehe, when our teacher wrote on the blackboard, I couldn't understand it at first. Let's give an example. Be patient!

F[g(x)]=sin(2x), then let g(x)=2x, let g(x)=2x=u, then f(u)=sin(u).

So f'[g(x)]=[sin(u)]'_2x)'=2cos(u), and then replace u with 2x to get f'[g(x)]=2cos(2x).

And so on y'=[cos(3x)]'=-3sin(x)

y'={sin(3-x)]'=-cos(x)

I can't do it well at first, but I always have to compare formulas with examples.

But as long as you practice and recite formulas, the most important thing is to remember one or two examples and practice more.

Proof of derivative law of compound function: First, prove a lemma.

The necessary and sufficient condition for f(x) to be derivable at point x0 is that there exists a function H(x) continuous at point x0 in the neighborhood U(x0) of x0, so that f(x)-f(x0)=H(x)(x-x0), and thus f'(x0)=H(x0).

It is proved that f(x) is differentiable at x0, and h (x) = [f (x)-f (x0)]/(x-x0), x ∈ u' (x0) (the eccentric neighborhood of x0); H(x)=f'(x0)，x=x0

Because lim (x->; x0)H(x)= lim(x-& gt； x0)[f(x)-f(x0)]/(x-x0)= f '(x0)= H(x0)

So H(x) is continuous at point x0, and f(x)-f(x0)=H(x)(x-x0), x∈U(x0).

On the other hand, there are H(x), x∈U(x0), which are continuous at point x0, and f(x)-f(x0)=H(x)(x-x0), x∈U(x0).

Because the limit lim (x->; x0)H(x)= lim(x-& gt； x0)[f(x)-f(x0)]/(x-x0)= lim(x-& gt； x0)f'(x)=H(x0)

So f(x) is derivable at point x0, and f'(x0)=H(x0).

Transitive lemma

Let u=φ(x) be derivable at point u0 and y=f(u) be derivable at point u0=φ(x0), then the composite function F(x)=f(φ(x)) is derivable at x0 and f' (x0) = f' (u0) φ' (x0).

It is proved that f(u) is derivable at u0, and there exists a function H(u) that is continuous at point u0, so that f'(u0)=H(u0), f(u)-f(u0)=H(u)(u-u0).

And u=φ(x) is derivable at x0. Similarly, there is a continuous function G(x) at x0, so that φ'(x0)=G(x0), φ(x)-φ(x0)=G(x)(x-x0).

So f (φ (x))-f (φ (x0)) = h (φ (x)) (φ (x)-φ (x0)) = h (φ (x)) g (x) (x-x0).

Because φ, g is continuous at x0, h is continuous at u0=φ(x0), and H(φ(x))G(x) is continuous at x0, it is known from the sufficiency of lemma that F(x) is derivable at x0, and

f '(x0)= f '(u0)φ'(x0)= f '(φ(x0))φ'(x0)

It is proved that 2: y=f(u) is derivable at u point and u=g(x) is derivable at x point, then the composite function y=f(g(x)) is derivable at x0 point and dy/dx=(dy/du)_du/dx).

It is proved that because y=f(u) is differentiable in u, then lim (δ u->; 0) δ y/δ u = f' (u) or δ y/δ u = f' (u)+α (lim (δ u->; 0)α=0)

When Δ u ≠ 0, multiply Δ u by both sides of the equation, Δ y = f' (u) Δ u+α Δ u.

But when δu = 0, δy = f(u+δu)-f(u)= 0, so the above equation still holds.

Because Δ x ≠ 0, divide Δ x by both sides of the equation to find Δ x->; It is also the limit of 0.

dy/dx = lim(δx-& gt； 0)δy/δx = lim(δx-& gt； 0)[f '(u)δu+αδu]/δx = f '(u)lim(δx-& gt； 0)δu/δx+lim(δx-& gt； 0)αδu/δx

And g(x) is continuous at x (because it is differentiable), so when δ x->; 0，δu = g(x+δx)-g(x)->； 0

Then lim (δ x->; 0)α=0

Eventually there is dy/dx=(dy/du)_du/dx).

Basic formulas of senior high school mathematics II

1 There is only one straight line at two points.

The line segment between two points is the shortest.

The complementary angles of the same angle or equal angle are equal.

The complementary angles of the same angle or the same angle are equal.

One and only one straight line is perpendicular to the known straight line.

Of all the line segments connecting a point outside the straight line with points on the straight line, the vertical line segment is the shortest.

7 Parallel axiom passes through a point outside a straight line, and there is only one straight line parallel to this straight line.

If both lines are parallel to the third line, the two lines are also parallel to each other.

The same angle is equal and two straight lines are parallel.

The internal dislocation angles of 10 are equal, and the two straight lines are parallel.

1 1 are complementary and two straight lines are parallel.

12 Two straight lines are parallel and have the same angle.

13 two straight lines are parallel, and the internal dislocation angles are equal.

14 Two straight lines are parallel and complementary.

Theorem 15 The sum of two sides of a triangle is greater than the third side.

16 infers that the difference between two sides of a triangle is smaller than the third side.

The sum of the internal angles of 17 triangle is equal to 180.

18 infers that the two acute angles of 1 right triangle are complementary.

19 Inference 2 An outer angle of a triangle is equal to the sum of two non-adjacent inner angles.

Inference 3 The outer angle of a triangle is greater than any inner angle that is not adjacent to it.

2 1 congruent triangles has equal sides and angles.

Axiom of Angular (SAS) has two triangles with equal angles.

The Axiom of 23 Angles (ASA) has the congruence of two triangles, which have two angles and their sides correspond to each other.

The inference (AAS) has two angles, and the opposite side of one angle corresponds to the congruence of two triangles.

The axiom of 25 sides (SSS) has two triangles with equal sides.

Axiom of hypotenuse and right angle (HL) Two right angle triangles with hypotenuse and right angle are congruent.

Theorem 1 The distance between a point on the bisector of an angle and both sides of the angle is equal.

Theorem 2 is a point with equal distance on both sides of an angle, which is on the bisector of this angle.

The bisector of an angle 29 is the set of all points with equal distance to both sides of the angle.

The nature theorem of isosceles triangle 30 The two base angles of isosceles triangle are equal (that is, equilateral and equiangular).

3 1 Inference 1 The bisector of the vertices of an isosceles triangle bisects the base and is perpendicular to the base.

The bisector of the top angle, the median line on the bottom edge and the height on the bottom edge of the isosceles triangle coincide with each other.

Inference 3 All angles of an equilateral triangle are equal, and each angle is equal to 60.

34 Judgment Theorem of an isosceles triangle If a triangle has two equal angles, then the opposite sides of the two angles are also equal (equal angles and equal sides).

Inference 1 A triangle with three equal angles is an equilateral triangle.

Inference 2 An isosceles triangle with an angle equal to 60 is an equilateral triangle.

In a right triangle, if an acute angle is equal to 30, the right side it faces is equal to half of the hypotenuse.

The center line of the hypotenuse of a right triangle is equal to half of the hypotenuse.

Theorem 39 The distance between the point on the vertical line of a line segment and the two endpoints of the line segment is equal.

A Complete Collection of Basic Mathematics Formulas in Senior High School Ⅲ

Commonly used inductive formulas have the following groups:

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(k∈Z)

cos(2kπ+α)=cosα(k∈Z)

tan(2kπ+α)=tanα(k∈Z)

cot(2kπ+α)=cotα(k∈Z)

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

Kurt (π-α) =-Kurt α

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)

Note: When doing the problem, it is best to regard A as an acute angle.

Inductive formula memory formula

Summary of the law. ※。

These inductive formulas can be summarized as follows:

For the trigonometric function value of 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, 2 π-α ∈ (270,360), sin (2π-α)

So sin(2π-α)=-sinα.

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 sine (cotangent); Cut in twos and threes; Four cosines (secant) ".

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 "-".

The above memory formulas are all positive, sine, inscribed and cosine.

There is another way to define positive and negative according to the function type:

Function Type First Quadrant Second Quadrant Third Quadrant Fourth Quadrant

Sine ...........+............+............-............- ........

Cosine ...........+............-............-............+ ........

Tangent ...........+............-............+............- ........

I cut ...........+............-............+............- ........

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.

Basic formulas of high school mathematics daquan 4

1, straight line

Linear equation of the distance between two points and a fixed fractional point

|AB|=| |

|P 1P2|=

y-y 1=k(x-x 1)

y=kx+b

The positional relationship between two straight lines, including angle and distance.

Or k 1=k2, and b 1≠b2.

L 1 coincides with l2.

Or k 1=k2 and b 1=b2.

L 1 intersects with l2.

Or k 1≠k2

l2⊥l2

Or k 1k2=- 1 l 1 to l2.

The angle between l 1 and l2.

Distance from point to straight line

2. Conic curve

Circular ellipse

The standard equation (x-a)2+(y-b)2=r2.

The center of the circle is (a, b) and the radius is r.

The general equation x2+y2+Dx+Ey+F=0.

Where the center of the circle is (),

Radius r

(1) Use the distance d from the center of the circle to the straight line and the radius r of the circle to judge or use the discriminant to judge the positional relationship between the straight line and the circle.

(2) Use the sum and difference of center distance d and radius to judge the positional relationship between two circles.

Focus F 1(-c, 0), F2(c, 0)

(b2=a2-c2)

weird

collinearity equation

The focal radius |MF 1|=a+ex0, |MF2|=a-ex0.

Hyperbolic parabola

hyperbola

Focus F 1(-c, 0), F2(c, 0)

(a，b & gt0，b2=c2-a2)

weird

collinearity equation