Mathematical Manuscript Content: Mathematical Model A long time ago, the king of the Arabic numeral kingdom celebrated his 20th birthday, and the Roman numeral kingdom sent someone to send 20 precious trees as birthday gifts. Arabic numerals. ? 20? Minister Zhang Bangzhao, whoever can plant these 20 trees skillfully will be rewarded. But no one can design it. ? 20? The minister thought day and night, looked up a lot of information and made experiments with stones again and again. He painted thousands of patterns. Painting, trying, suddenly, his eyes lit up and he saw an extremely beautiful pattern. ? 20? The minister immediately presented the design to the king. The king was very happy to see him. 20? The minister pointed to the pattern and said to the king, Your Majesty, you see, the number of trees in the painting is not horizontal, vertical or oblique, with 4 trees in each row, and the maximum is 18 row. ?

The king said in surprise, I have never seen such a beautiful and wonderful tree planting pattern in any park. It's fantastic. I will reward you! ? . I will reward you! ? The king said in surprise, I have never seen such a beautiful and wonderful tree planting pattern in any park. It's fantastic. I will reward you! Yes, this is a man named Sam? Lloyd's mathematician invented the design, and I just applied his design to the problem of planting trees. 20? Minister, according to reason. ? Ok, ok, you can use this pattern, which is also meritorious. ? Say, did the king announce it? 20? Minister's award and name the design? 20 patterns? , is the most beautiful tree planting mode in the world. The king immediately sent someone to follow him? 20 patterns? Plant 20 trees in the palace garden. Since then, this beautiful tree planting pattern has been passed down to this day.

Data of Mathematical Manuscripts: Discriminant of Mathematical Formulas in Senior High School

B2-4ac=0 Note: This equation has two equal real roots.

B2-4ac >0 Note: The equation has two unequal real roots.

B2-4ac & lt； Note: The equation has no real root, but a complex number of the yoke.

formulas of trigonometric functions

Two-angle sum formula

sin(A+B)= Sina cosb+cosa sinb sin(A-B)= Sina cosb-sinb cosa

cos(A+B)= cosa cosb-Sina sinb cos(A-B)= cosa cosb+Sina sinb

tan(A+B)=(tanA+tanB)/( 1-tanA tanB)tan(A-B)=(tanA-tanB)/( 1+tanA tanB)

ctg(A+B)=(ctgActgB- 1)/(ctg B+ctgA)ctg(A-B)=(ctgActgB+ 1)/(ctg B-ctgA)

Double angle formula

tan2A = 2 tana/( 1-tan2A)ctg2A =(ctg2A- 1)/2c TGA

cos2a = cos2a-sin2a = 2 cos2a- 1 = 1-2 sin2a

half-angle formula

sin(A/2)=？ (( 1-cosA)/2) sin(A/2)=-？ (( 1-cosA)/2)

cos(A/2)=？ (( 1+cosA)/2) cos(A/2)=-？ (( 1+cosA)/2)

tan(A/2)=？ (( 1-cosA)/(( 1+cosA))tan(A/2)=-？ (( 1-cosA)/(( 1+cosA))

ctg(A/2)=？ (( 1+cosA)/(( 1-cosA))ctg(A/2)=-？ (( 1+cosA)/(( 1-cosA))

Sum difference product

2 Sina cosb = sin(A+B)+sin(A-B)2 cosa sinb = sin(A+B)-sin(A-B)

2 cosa cosb = cos(A+B)-sin(A-B)-2 sinasinb = cos(A+B)-cos(A-B)

sinA+sinB = 2 sin((A+B)/2)cos((A-B)/2 cosA+cosB = 2 cos((A+B)/2)sin((A-B)/2)

tanA+tanB = sin(A+B)/cosa cosb tanA-tanB = sin(A-B)/cosa cosb

ctgA+ctgBsin(A+B)/Sina sinb-ctgA+ctgBsin(A+B)/Sina sinb

The sum of the first n terms of some series

1+2+3+4+5+6+7+8+9+? +n = n(n+ 1)/2 1+3+5+7+9+ 1 1+ 13+ 15+？ +(2n- 1)=n2

2+4+6+8+ 10+ 12+ 14+? +(2n)= n(n+ 1) 12+22+32+42+52+62+72+82+？ +n2=n(n+ 1)(2n+ 1)/6

13+23+33+43+53+63+? n3 = N2(n+ 1)2/4 1 * 2+2 * 3+3 * 4+4 * 5+5 * 6+6 * 7+？ +n(n+ 1)= n(n+ 1)(n+2)/3

Sine theorem a/sinA=b/sinB=c/sinC=2R Note: where r represents the radius of the circumscribed circle of a triangle.

Cosine Theorem b2=a2+c2-2accosB Note: Angle B is the included angle between side A and side C..

The standard equation of a circle (x-a)2+(y-b)2=r2 Note: (A, B) is the center coordinate.

General equation of circle x2+y2+Dx+Ey+F=0 Note: D2+E2-4f > 0

Parabolic standard equation y2=2px y2=-2px x2=2py x2=-2py

Lateral area of a straight prism S=c*h lateral area of an oblique prism s = c' * h.

Lateral area of a regular pyramid S= 1/2c*h' lateral area of a regular prism S= 1/2(c+c')h'

The lateral area of the frustum of a cone S = 1/2(c+c')l = pi(R+R)l The surface area of the ball S=4pi*r2.

Lateral area of cylinder S=c*h=2pi*h lateral area of cone s =1/2 * c * l = pi * r * l.

The arc length formula l=a*r a is the radian number r > of the central angle; 0 sector area formula s= 1/2*l*r

Conical volume formula V= 1/3*S*H Conical volume formula V= 1/3*pi*r2h

Oblique prism volume V=S'L Note: where s' is the straight cross-sectional area and l is the side length.