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primary school to junior high school math for all formulas

1, each a few × ÷ total number of shares = total number of copies of each = total ÷ number of shares = Each number
2, 1 × multiplier = multiplier times the number of times a multiple of the number of ÷ 1 = multiple = multiple times a multiple of the number of ÷
3, distance speed × time = distance ÷ speed = time time = distance ÷ speed ” br> 4, unit price × quantity = total price = total price ÷ ÷ number of total number of = Price
5, work efficiency × working time = work ÷ total work efficiency of the total working hours of total work = work ÷ time = work efficiency
6, addend + addend = and, and – the other one addend = addend
7, minuend – subtrahend = difference minuend – subtrahend difference = difference + reduction number = minuend
8, factor × factor = plot plot ÷ a factor = another factor
9, dividend ÷ divisor = provider = providers dividend ÷ divisor = dividend divisor provider ×

Primary Mathematics Graphic formula
1, square C perimeter area of a side length of circumference S = side length × 4 C = 4a area = side length × side length S = a × a
2, cube V : Volume a: surface area = edge long edge long edge long × × 6 S form = a × a × 6 size = length × Ling Ling Ling Long Long × V = a × a × a
3, rectangular
C S area of a side length of perimeter
perimeter = (length + width) × 2
C = 2 (a + b)
area = length × width
S = ab
4, rectangular
V: Volume s: size a: Long b: W h: high
(1) surface area (length × width + length × width × high + high) × 2
S = 2 (ab + ah + bh)
(2) Volume = length × width × high
V = abh
5 triangle
s an area of a high-end
area = h × high-end of ÷ 2
s = ah ÷ 2
triangle height = area × 2 ÷ at the end of
triangle = area × 2 ÷ at the end of the high
6 area of a parallelogram
s at the end of the high
area = h × high at the end of
s = ah
7 an area of a trapezoid
s on the end of b under the high
area at the end of h = (on the bottom end + under) × high-÷ 2
s = (a + b) × h ÷ 2
8 round
S Area C perimeter Π d = diameter r = radius
(1) circumference = diameter × Π = 2 × Π × radius
C = Πd = 2Πr
(2) area = radius × radius × Π
9 cylinder
v: Volume h: high-s; bottom area of r: Bottom radius c: underside perimeter
(1 ) side of the bottom surface area = circumference × high
(2) surface area = lateral area + bottom area × 2
(3) Volume = base area × height
(4) Volume = Side area ÷ 2 × radius
10 cone
v: Volume h: high-s; bottom area of r: Bottom radius
Volume = base area × height ÷ 3
÷ total number of shares = average of the total and differential
the problem formula
(and + poor) ÷ 2 = Large Number
(and – worse) ÷ 2 = decimal
and times of the problem
and ÷ (multiplier -1) = decimal
decimal × multiplier = large numbers
(or, and – decimals = large numbers)
bad times worse problem
÷ (multiplier -1) = decimal
decimal × multiplier = large numbers
(or decimal + SD = large numbers)
tree problem
1 tree planting on non-closed circuit problems can be divided into the following three situations:
⑴ If the non-closed at both ends of the line should be planting trees, then: ” br> number of trees = number of segments + 1 = length ÷ spacing in the rows -1
length = spacing in the rows × (number of trees -1)
spacing in the rows = length ÷ (number of trees -1)
⑵ If one end of the line in the non-closed To plant trees on the other side do not plant trees, then:
number of trees = number of segments = length ÷ spacing in the rows
length = spacing in the rows × number of trees
spacing in the rows = length ÷ number of trees
⑶ If the two lines in the non-closure side are not planting trees, then:
number of trees = number of segments -1 = length ÷ spacing in the rows -1
length = spacing in the rows × (number of trees +1)
spacing in the rows = length ÷ (number of trees +1)
2 tree planting on a closed circuit relationship between the number of issues are as follows
number of trees = number of segments = length ÷ spacing in the rows
length = spacing in the rows × number of trees
spacing in the rows = length ÷ number of trees
profit and loss issues
(surplus + deficit) ÷ = twice the difference between the amount allocated to participate in the distributios air jordan(s), or simply jordans are a brand of shoes produced by nike … air jordan shoes v (5) air jordan force fusion v (5) air jordan gucci shoes, of the shares
(large surplus – a small profit) ÷ = twice the difference between the amount allocated to participate in the distribution of the shares
(burned by – a small deficit) ÷ 2 the difference betwees christian louboutin boots : christian louboutin store – buy …christian louboutin store – buy christian louboutin shoes, black, heels, sandals online : christian louboutin boots – classic christian louboutin christian, the amount of time allocated to participate is our jordan shoes and air jordan shoes are free shipping to worldwide. shoes for the best prices, guaranteed! buy michael jordan shoes at low prices. the distribution of shares =
encounter problems
encounter distance = speed × time
encounter encounter encounter time = distance ÷ speed and the speed and = phase
case of distance ÷ time to meet and question
catch
catch and distance = velocity difference × chase and time
to catch up time = distance ÷ speed of recovery and the speed difference = difference
÷ catch and distance and time to recover
flow problems ” br> Speed = downstream flow rate
hydrostatic speed + speed = hydrostatic speed counter-current – the flow rate
hydrostatic speed = (speed + upstream downstream speed) ÷ 2
flow rate = (downstream speed – upstream speed) ÷ 2
concentration problems
the weight of the solute + solvent = solution of the weight of the weight of the weight of the solute ÷ solution
weight × 100% = concentration
weight × concentration of solution = weight of the solute
solute concentration = weight ÷ weight of the solution
profit and discounts
profit = selling price – cost margin = Profit ÷
cost × 100% = (selling price ÷ cost -1) × 100%
Change Amount = Principal × Change Percentage
discount = actual sale price of the original price ÷ × 100% (discount <1)
Interest = principal × rate × time
tax After the interest = principal × rate × time × (1-20%)

length unit conversion
1 kilometer = 1000 meters 1 meter = 10 decimeters
1 decimetre = 10 cm 1 meter = 100 centimeters
1 centimeter = 10 millimeters
area unit conversion
1 square kilometer = 100 hectares
1 ha = 10,000 square meters
1 square meter = 100 square sub – m
1 square decimetre = 100 square centimeters
1 Ping Fang centimeter = 100 mm2
body (Rong) plot unit conversion
1 cubic meter = 1000 cubic decimetres
1 cubic decimetre = 1000 cubic ces jordan basketball shoes,jordan football shoes,jordan melo,jordan …the introduction of the air jordan 1 turned the athletic shoe industry upside down. before the air jordan 1, most basketball shoes were white,timeters
1 cubic decimetre = 1 liter
1 Lifanglimi = 1 ml
1 cubic meter = 1000 liters
weight unit conversion
1 tonne = 1,000 kg
1 1000 grams = 1000 grams
1 kilogram = 1 kg
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1 yuan = 10 jiao
1 angle = 10 points
1 dollar = 100 cents
time units conversion <br "1 century = 100 years 1 year = December
big month (31 days) are: 1 \ 3 \ 5 \ 7 \ 8 \ 10 & # 92; December
ozuki (30 days) were: 4 \ 6 \ 9 \ November
non-leap year February 28 days, a leap year February 29 Tian
365 days a year non-leap year, leap year 366 days
1 days = 24 hours 1 = 60 minutes
1 min = 60 seconds 1:00 = 3600 seconds

Primary Mathematics area of geometry perimeter Volume formula

1, rectangular perimeter = (length + width) × 2 C = (a + b) × 2

2, the perimeter of a square = side length × 4 C = 4a

3, rectangle area = length × width S = ab

4, square area = side length × side length S = aa = a
<br "5, triangle area = high-end × ÷ 2 S = ah ÷ 2

6, the area of parallelogram = high-end × S = ah

7, trapezoidal area = (on the at the end of + a baseline) × high-÷ 2 S = (a + b) h ÷ 2

8, diameter = radius × 2 d = 2r radius = diameter ÷ 2 r = d ÷ 2

9 , circumference of a circle = pi × diameter = pi × radius × 2 c = πd = 2πr

10, a circle area = pi × radius × radius

common junior high school mathematical formula

1 had two points and only one straight line segment between two points
2 shortest
3 with the angle or isometric
4 of the supplementary angle equal to the same angle or isometric of Yu Jiao equal ” br> 5-off point has one and only one straight line and the known point outside a straight line perpendicular to
6 points with a straight line connecting all the segments, the vertical section parallel to the shortest
7 through the line outside the axiom that has one and only a straight line and this line
8 If two straight lines parallel to both and a third line parallel to the two straight lines parallel to each other over
9 corresponding angles are equal, two straight lines parallel to
10 interior angles are equal, two a straight line parallel to
11 with the adjacent complementary angles, two straight lines parallel to two straight lines parallel to
12, corresponding angles are equal
13 two straight lines parallel alternate interior angles are equal
14 two straight lines parallel with the adjacent interior angle complementary ” br> 15 on both sides of the triangle and the theorem is greater than the third side
16 inferred the difference between both sides of the triangle is less than the third side
17 triangle angles and three angles of a triangle and the theorem is equal to 180 °
18 inference 1 the two right-angled triangle acute angle in each triangle I
19 inference of an exterior angle is equal to 2, and it is not adjacent to two angles and
20 inference of an exterior angle is greater than three triangles, and it is not adjacent to any one of the c.
21 congruent triangles correspond to the edge angle equal to the corresponding corner edge
22 axioms (SAS) has on both sides and their corresponding angles equal to two triangles congruent
23 angle corner axiom ( ASA) has corners and edges of their corresponding folders in two equal triangles congruent
24 inference (AAS) with corners and one corner of the right side of the corresponding two triangles congruent equal
25 Bian Bian Bian Axiom (SSS) with the corresponding three sides of equal two triangles congruent
26 hypotenuse, right angle side of axiom (HL) with beveled edge and a right-angle edges corresponding to equal two right-angled triangle congruent
27 Theorem 1 in the angle point of the line equally to this corner equal distances on both sides of
28 Theorem 2-1 angle distances on both sides of the same point, in this corner-sharing online
29 angle bisector is to angle equal distance on both sides of the set of all points
30 isosceles triangle isosceles triangle theorem the nature of the two bottom corners are equal (ie, an equilateral right isometric)
31 inference of an isosceles triangle vertex angle bisector split bottom and bottom edge of
32 isosceles triangle is perpendicular to the zenith angle bisector, bottom edge of the middle and bottom edge of the high overlap,
33 inferences three equilateral triangles are equal each corner, and the Each angle is equal to 60 °
34 isosceles triangle theorem of determining if a triangle has two angle are equal, then this two corners of the right side are also equal (equal angles to the other side)
35 reasoning 1 3 angle are equal equilateral triangle is a triangle
36 corollary 2 there is an angle equal to 60 ° of the isosceles triangle is the equilateral triangle
37 in the right-angled triangle, if one acute angle is equal to 30 ° so that it right The right-angle side is equal to half the hypotenuse right triangle hypotenuse
38 on the center line on the hypotenuse is equal to half of the line segment perpendicular bisector theorem
39 points and this line segment equal to the distance between the two endpoints
40 converse theorem, and a segment from the two endpoints the same point, in this segment of the perpendicular bisector line
41 vertical segments and the segment bisector can be regarded as two end points of equal distance from all points of the set of
42 Theorem 1 With regard to a particular line of the two graphics are congruent symmetrical shape
43 Theorem 2 If two graphics symmetrical about a straight line, then the symmetry axis is perpendicular to connect the corresponding points bisector
44 Theorem 3 2 symmetrical about a straight line graph if their corresponding segments or extended lines cross, then the intersection of the symmetry axis
45 Converse If two graphics with the corresponding point of connection is a straight line perpendicular bisector, then the two graphics on the This line symmetry
46 Pythagorean triangle two right angle sides a, b of the square and is equal to the square of the hypotenuse c, that is, a ^ 2 + b ^ 2 = c ^ 2
47 Pythagorean Theorem the converse theorem, if a long triangular triangle a, b, c has a relationship a ^ 2 + b ^ 2 = c ^ 2, then the triangle is a right triangle theorem
48 quadrilateral interior angle equal to 360 °
49 quadrilateral The exterior angle is equal to 360 °
50 polygons angles and theorem n-gon of the angles and is equal to (n-2) × 180 °
51 deduction equal to any multilateral exterior angle 360 °
52 parallelogram nature of the Theorem a parallelogram the diagonal equal
53 parallelogram parallelogram nature of Theorem 2 on opposite sides of the same reasoning
54 sandwiched between two parallel lines, parallel line segments are equal
55 parallelogram parallelogram nature of Theorem 3 diagonal with each other equally
56 parallelogram theorem 1 to determine the two groups were equal to the angle of quadrilateral is a parallelogram parallelogram
57 Theorem 2 to determine the two groups were equal ochristian louboutin boots ,buy the luxury christian louboutin …christian louboutin boots ,christian-louboutin-shoes.cc is a christian louboutin store,specilizing in christian louboutin shoes.sale christian louboutin the side of quadrilateral is a parallelogram
58 Theorem 3 diagonal parallelogram to determine the quadrilateral is divided equally between
59 parallelogram Parallelogram Theorem 4 to determine a set of parallel sides of the quadrilateral is a parallelogram equal to
60 rectangular nature of the theorem are the four corners of a rectangle
61 rectangular rectangular rectangular nature of the Theorem 2 are equal
62 rectangles to determine the diagonal theorem 1 there are three angle is at right achristian louboutin black leather zip boots, chritian louboutin …6 nov 2009 … christian louboutin，we can supply the best christian louboutin black leather zip boots,sale up to 75% off,free and fast shippinggles the quadrilateral is a rectangle
63 Rectangle Theorem 2 to determine the diagonal of the parallelogram is a rectangle equal to ” br> 64 diamond-shaped nature of the theorem a diamond-shaped four sides are equal
65 diamond-shaped diamond of the diagonal nature of the theorem 2 perpendicular to each other, and each one divided equally among a group of diagonal diagonal
66 diamond-shaped area = diagonal half of the product, namely S = (a × b) ÷ 2
67 diamond theorem to determine a quadrilateral is a quadrilateral are equal
68 diamond diamond Theorem 2 to determine the diagonal of the parallelogram are perpendicular to each other diamond
69 square nature of the theorem of the four corners of a square are right angles, four sides are equal
70 square nature of the theorem 2 a square equal to two diagonal, and perpendicular to each other equally, and each diagonal split between a group of diagonal
71 Theorem 1 on the central symmetry of the two graphs are congruent
72 Theorem 2 on the central symmetry of the two graphics, symmetric point connections are through the center of symmetry and is symmetric
73 reverse split Theorem If two corresponding points on the graph are connected through a certain point, and is it equally, then the two graphics on this point-symmetric nature of the theorem
74 isosceles trapezoid isosceles trapezoid in the same two corners on the end of equal
75 isosceles trapezoid two diagonal equal
76 isosceles trapezoid theorem to determine the end at the same angle on the two equal isosceles trapezoid is a trapezoid
77 diagonal of the trapezoid is equal to isosceles trapezoid
78 parallel line segments are equally Theorem If a group of parallel lines cut in a straight line segment entitled to equal, then the other straight line intercepted the line segments are equal
79 inference through a waist of a trapezoidal the midpoint and at the end of parallel straight lines, must be divided equally among the other two after deduction of Lumbar
80 side of the midpoint of the triangle parallel with the other side of the line, must be divided equally among the third side
81 triangle theorem of the median line of the triangle, the median line parallel to the third side, and is equal to its half of the median line
82 ladder ladder theorem the median line parallel to the end of two, and is equal to two and a half at the end of L = (a + b) ÷ 2 S = L × h
83 (1) If the proportion of the basic properties of a: b = c: d,supras, then ad = bc, if ad = bc, then a: b = c: d
84 (2) If the combined ratio in nature a / b = c / d, then (a ± b) / b = (c ± d) / d
85 (3) geometric in nature, if a / b = c / d = … = m / n (b + d + … + n ≠ 0), then (a + c + … + m) / (b + d + … + n) = a / b
86 sub-parallel line segments proportional to the theorem of three parallel lines cut two straight lines, derived from the corresponding line segments proportional to the inferences
87 parallel to the straight-line cut-off the other side of the triangle on either side (or both sides of the extension line), derived from the corresponding line segments proportional to the
88 theorem, if a straight cut on both sides of the triangle (or both sides of the extension), it is proportional to the corresponding segment, then this straight line parallel to the triangle’s third side
89 parallel to the side of the triangle, and the other on both sides of the intersection of the straight line intercepted by the triangle of the triangular and the original triangle Trilateral
90 proportional to the corresponding theorem of the triangle parallel to the side of the line and the other side (or both sides of the extension lines) intersect,new jordans shoes, the triangle formed similar to the original triangle
91 similar triangles to determine the corresponding theorem 1 corners are equal, two similar triangles (ASA)
92 right-angled triangle is high on the hypotenuse into two right-angled triangle and the original triangle
93 found similar to Theorem 2 is proportional to the corresponding sides and angles equal, two similar triangles (SAS) Theorem 3 triangular
94 to determine the corresponding proportion, two similar triangles (SSS)
95 Theorem If the hypotenuse of a right triangle and a right-angle side of the hypotenuse of another right triangle, and a right-angle edge is proportional to the corresponding , then the two right-angled triangle similarity theorem of a similar nature
96 triangles correspond to a high ratio, the corresponding median line than the corresponding angle bisector is equal to the ratio are similar in nature than the
97 similar triangles theorem 2 ratio of circumference Similar properties equivalent to more than
98 Theorem 3 an area of similar triangles similar to the ratio is equal to the square of
99 compared with the sine of the value of any acute angle is equal to its cosine value of Yu Jiao, any acute angle cosine value is equal to its sine Yu Jiao
100 arbitrary value of the tangent of an acute angle is equal to its Yujiao the cotangent value, the cotangent value of any acute angle is equal to its tangent
101 Yu Jiao Yuan is a fixed distance of points is equal to the set of fixed-length
102 inside a circle centered on the distance can be seen as less than the radius of the set of points
103 can be seen as outside the circle centered on the distance greater than the radius of the set of points
104 or so on the same circle
105 radius of the circle equal to the distance to the fixed point is equal to fixed-length track, is designated as the center of a circle, fixed-les supra shoes buy,supra cheap shoes online sale–free shipping!we specialize in authentic supra shoes.2009 latest design style is offered on the suprafirst.com,high quality with free shipping and no sale tax.gth for the radius of the circle
106 and the known segment of the two ends of the locus of points equidistant , is the perpendicular bisector of line segment to a known angle
107 equidistant points on both sides of the track, is the angle bisector
108 Dao two parallel lines of equal distance from the locus of points, and it is These two parallel lines parallel and equidistant in a straight line
109 theorem is not in line with the three points have been set a circle.
110 vertical diameter theorem diameter perpendicular to the string split this string by string, and split the two pairs of arc
111 deduction 1 ① split string (not diameter) of the diameter perpendicular to the strings, and split the right chord 2 arc
② string through the center of a circle perpendicular bisector, and the split between the two strings of the right arc
③ split a string by the right of the arc diameter, vertical split strings, and the split string by another one right Arc
112 inference two circle two parallel strings of the folder
113 yen the same arc as the center of symmetry is centered on the center of symmetry theorem graphics
114 yen or so in the same circle, the equivalent of the central angle The right of the arc equal, the equal right chord, the chord of the chord of the heart from the same
115 deduction in the same circle or other circular, if the two central angle, two arcs, two strings, or two strings heart strings in a group from the same amount if they correspond to the amount of the remaining groups are equal
116 theorem one arc of the circumference of the right angle of the central angle is equal to its right half
117 deduction with an arc or a other arc of the circumference of the right angle are equal; the same circle or other circular, the equivalent of the circumferential angle of the right of the arc is also equal to two semi-circular
118 inferences (or diameter) of the circumference of the right angle is right angle; 90 ° of the circumferential angle The right chord is the diameter of
119 corollary 3, if on the side of the center line of the triangle is equal to half the side, then the triangle is a right triangle
120 Theorem circle inscribed quadrilateral diagonal complement each other, and any one exterior angle is is equal to its angle within
121 ① straight line L, ⊙ O intersect d <r
② straight line L, ⊙ O tangent to d = r
③ straight line L, ⊙ O phase from the d> r
After 122 Tangent Theorem to determine the radius of the outer end and perpendicular to the radius of this tangent line is round and the nature of the theorem
123 tangent circle tangent perpendicular to the radius of the cut-off point after
124 deduction through the center of a circle and a vertical on the tangent line must be inferred through the cut-off point
125 through the cut-off point and two perpendicular to the tangent line through the center of a circle must be tangent length
126 theorem quoted from a point outside the circle round the two tangent, and their tangent appearance, etc. center point and this point split between two tangent of the angle connection
127 yen for two pairs of outer cutting edge of Quadrilateral and equivalent
128 Xianqie Jiao Xian Qiejiao theorem is equal to its folder arc pair circumferential angle
129 Xian Qiejiao inference if the two arcs of the same folder, then it is also equal to 2 Xian Qiejiao
130 intersection of the two strings intersect within the circle theorem of string has been divided into the intersection of two line segments equal to the product of a long
131 deduction if the strings and the diameter of the vertical intersection, then the string by half its diameter into the sub-segment of the ratio of the two items
132 cutting line theorem quoted from the circle outside the circle of the tangent point and the secant, tangent length is that the secant and the circle point to the point of intersection of two line segments length ratio items
133 deduction from the outside a little circle round two secant cited, which is the secant and the circle of each point of intersection of two line segments equal to the product of a long
134 if the two circles tangent, then the cut-off point is always on the line with heart
135 ① two outer circle from the d> R + r ② two round exo d = R + r
③ two round intersect Rr <d r)
④ 2 circle cut d = Rr (R> r) ⑤ two circular containing d r)
136 intersects the two circle theorem even heart split between two lines perpendicular to the circle theorem to the public string
137 Yuan is divided into n (n ≥ 3):
⑴ in turn links to the various sub-points derived from the polygon is the circle inscribed n-gon
⑵ is passing through the round-off point for the tangent to the point of intersection of tangent adjacent vertices of the polygon is the outer circle cutting
138 are n-gon theorem of any regular polygon has a circumcircle and an inscribed circle, these two circles are concentric circles
139 are n-gon of each interior angle is equal to (n-2) × 180 ° / n
140 theorem is the radius of the n-gon and edge away from the heart is a n-gon is divided into 2n congruent right-angled triangle
141 are n-gon of area Sn = pnrn / 2 p indicated that it was n-gon of perimeter
142 triangle area of √ 3a / 4 a side length
143 that if a vertex is surrounded by k in the positive n-gon of angle, because these angles, and should be 360 °, so k × (n-2) 180 ° / n = 360 ° into (n-2) (k-2) = 4
144 Arc Length formula: L = n Wu R/180
145 fan-shaped area formula: S fan = n Wu R ^ 2 / 360 = LR / 2
146 within the common tangent length = d-(Rr) grandfather tangent length = d-(R + r)

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practical Tools: common mathematical formulas

formula for classification formula expression

multiplication and coping style sub-a2-b2 = (a + b) (ab) a3 + b3 = (a + b ) (a2-ab + b2) a3-b3 = (ab (a2 + ab + b2)

triangle inequality | a + b | ≤ | a | + | b | | ab | ≤ | a | + | b | | a | ≤ b < = >-b ≤ a ≤ b

| ab | ≥ | a | – | b | – | a | ≤ a ≤ | a |

one dollar solution of quadratic equations-b + √ (b2-4ac) / 2a-b-√ (b2-4ac) / 2a

the relationship between roots and coefficients of X1 + X2 =- b / a X1 * X2 = c / a Note: Vieta theorem

discriminant
b2-4ac = 0 Note: The equation has two equal real roots
b2-4ac > 0 Comments : equation has two unequal real roots
b2-4ac < 0 Note: The equation there is no real roots, there are complex conjugate roots

trigonometric formulas

corners and Formula
sin (A + B) = sinAcosB + cosAsinB sin (AB) = sinAcosB-sinBcosA
cos (A + B) = cosAcosB-sinAsinB cos (AB) = cosAcosB + sinAsinB
tan (A + B) = (tanA + tanB) / (1-tanAtanB) tan (AB) = (tanA-tanB) / (1 + tanAtanB)
ctg (A + B) = (ctgActgB-1) / (ctgB + ctgA ) ctg (AB) = (ctgActgB +1) / (ctgB-ctgA)

fold angle formula
tan2A = 2tanA / (1-tan2A) ctg2A = (ctg2A-1) / 2ctga
cos2a = cos2a-sin2a = 2cos2a-1 = 1-2sin2a

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))

and the difference of the plot
2sinAcosB = sin (A + B) + sin (AB) 2cosAsinB = sin (A + B)-sin (AB)
2cosAcosB = cos (A + B)-sin (AB)-2sinAsinB = cos (A + B)-cos (AB)
sinA + sinB = 2sin ((A + B) / 2) cos ((AB) / 2 cosA + cosB = 2cos ((A + B) / 2) sin ((AB) / 2)
tanA + tanB = sin (A + B) / cosAcosB tanA-tanB = sin (AB) / cosAcosB
ctgA + ctgBsin (A + B) / sinAsinB-ctgA + ctgBsin ( A + B) / sinAsinB

certain number of columns before the n items and
1 +2 +3 +4 +5 +6 +7 +8 +9 + … + n = n (n +1) / 2 1 +3 +5 +7 +9 +11 +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 a / sinA = b / sinB = c / sinC = 2R Note: where R, said triangle circumcircle radius

cosine Theorem b2 = a2 + c2-2accosB Note: The angle B is a side a and side c of the angle between the

the standard equation of the circle (xa) 2 + (yb) 2 = r2 Notes: (a, b) is coordinates
circle centered on the general equation x2 + y2 + Dx + Ey + F = 0 Note: D2 + E2-4F > 0
the standard equation of the parabola y2 = 2px y2 =- 2px x2 = 2py x2 =- 2py

straight prism lateral area S = c * h oblique prism lateral area S = c ' * h
are pyramid-side area of S = 1/2c * h ' positive bevel side area of S = 1 / 2 (c + c ') h '
Yuan Taiwan side area of S = 1 / 2 (c + c ') l = pi (R + r) l the ball surface area S = 4pi * r2
cylindrical lateral area of S = c * h = 2pi * h cone lateral area of S = 1 / 2 * c * l = pi * r * l

arc length formula l = a * ra is the central angle in radians r > 0 fan-shaped area of the formula s = 1 / 2 * l * r

cone volume formula V = 1 / 3 * S * H Cone volume formula V = 1 / 3 * pi * r2h
oblique prism volume V = S ' L Note: where, S ' is a straight-sectional area, L is longer side edges
cylinder volume the formula V = s * h cylinder V = pi * r2h

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The article was from: adaptable: primary school to junior high school math for all formulas
reading room with the QQ-mail subscription adaptable
What is the reading room?