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Cubic

2025/11/11 Edited to

... Read moreキュービックジルコニアネックレスは、その美しい輝きと手頃な価格で多くの人から人気を集めています。私も初めて購入した際、いくつかのポイントに気を付けたことで満足のいく買い物ができました。 まず、ネックレスを選ぶ際は、キュービックジルコニアのカットや品質に注目することが大切です。高品質なカットは光の反射を高め、よりダイヤモンドに近い輝きを出します。また、チェーンの素材も肌触りや耐久性に影響するので、アレルギー持ちの方は特に注意しましょう。 そして、日常使いに適したデザインや長さを選ぶのもポイントです。シンプルな形状のものは様々な服装に合わせやすく、特別な場面だけでなく普段使いも可能です。私自身、普段から身に着けられるシンプルなキュービックジルコニアネックレスを選び、気軽におしゃれを楽しんでいます。 価格帯については、ブランドやデザインによって幅がありますが、コスパの良さを重視するならレビューや口コミも参考にすると良いでしょう。購入後のお手入れも簡単で、長く美しい輝きを保てるのは嬉しいポイントです。 最後に、信頼できる店舗や通販サイトで購入することをおすすめします。偽物や粗悪品を避けるために、保証や返品対応の有無を確認すると安心です。私の経験をもとに、キュービックジルコニアネックレスの良さや選び方をぜひ参考にしてみてください。

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A hand-drawn diagram of an oblique triangle with an inscribed circle and excenters, accompanied by mathematical formulas for triangle area and derivations for half-angle sine and exradii IB and IC, written on lined notebook paper.
Mathematical formulas on lined paper showing the half-angle trigonometric ratios for angle B, including sin(B/2), cos(B/2), tan(B/2), csc(B/2), sec(B/2), and cot(B/2) in terms of the triangle's sides.
Mathematical formulas on lined paper showing the half-angle trigonometric ratios for angle C, including sin(C/2), cos(C/2), tan(C/2), csc(C/2), sec(C/2), and cot(C/2) in terms of the triangle's sides.
Half-angle trig ratios for the other angles
In an oblique triangle, the six trigonometric ratios can be found in terms of the sides. These are for the angles labeled B and C. #math #maths #mathematics #geometry #trigonometry
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A handwritten mathematical derivation of the sine double angle identity, sin B = 2 sin(B/2) cos(B/2), on grid paper. It features a right-angled triangle with labeled sides and angles, using geometric principles and algebraic steps to reach the final boxed identity.
Sine double angle identity derivation
Using established facts like the angle bisector theorem and the Pythagorean theorem, one can derive the sine double angle identity. #math #maths #mathematics #geometry #trigonometry
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A handwritten mathematical derivation on lined paper, showing the steps to find Pythagorean triples. It starts with a right-angled triangle labeled with sides a, b, and hypotenuse c, then derives the formulas a = m² - n², b = 2mn, and c = m² + n² from the Pythagorean theorem, with an example calculation for m=2, n=1 resulting in 3, 4, 5.
Deriving the formula to find Pythagorean triples
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A handwritten math proof on lined paper, featuring a right triangle with an inscribed circle and labeled sides/angles. The page shows trigonometric equations and algebraic steps deriving the relationship CI * AB = AI * BI.
AB × CI = AI × BI, part 2
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A handwritten math proof on lined paper shows a triangle with its altitudes intersecting at the orthocenter. Ceva's Theorem is applied using trigonometric expressions for segment lengths, demonstrating that the product equals one, thus proving the concurrency of altitudes.
Proving orthocenter concurrency
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Sides of a triangle in cosine form
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A handwritten mathematical proof on paper demonstrating that the cotangent of the Brocard angle (W) in a heptagonal triangle is equal to √7. It includes a triangle diagram with angles π/7, 2π/7, 4π/7, and step-by-step trigonometric calculations.
cot ω = √7
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cubicequation

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A handwritten page showing a geometric proof of Ceva's Theorem. It features a triangle ABC with cevians AX, BY, CZ intersecting at point P, along with auxiliary lines. Below the diagram are mathematical steps using similar triangles to derive the theorem's formula.
Proving Ceva’s theorem
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cubicequation

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csc² ω = csc² A + csc² B + csc² C
ω is the Brocard angle of the triangle. A, B, C are the angle measures of the triangle. a, b, c are the side lengths of the triangle. L₁, L₂, L₃ are the lengths between the vertices and the Brocard point (labeled D). #math #maths #mathematics #geometry #trigonometry
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A handwritten page displays a triangle diagram illustrating the Brocard angle ω, with segments L1, L2, L3 from the Brocard point to the vertices. Mathematical derivations lead to the formula 2R sin ω = ∛(L₁ × L₂ × L₃), relating the circumradius R, Brocard angle, and segment lengths.
2R sin ω = ∛(L₁ × L₂ × L₃)
The Brocard angle ω in a triangle. L₁, L₂, L₃ are the lengths of the segments from the Brocard point to the vertices. a, b, c are the side lengths of the triangle. A, B, C are the angle measures of the triangle. #math #maths #mathematics #geometry #trigonometry
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A handwritten mathematical proof on lined paper shows the derivation of csc ω = 2√2. It features a diagram of a heptagonal triangle with angles π/7, 2π/7, 4π/7, and a step-by-step derivation of the cosecant of its Brocard angle, ω, concluding with csc ω = 2√2.
csc ω = 2√2
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A handwritten page displays a triangle diagram with an inscribed circle and its incenter. It illustrates the derivation of the inradius (IA) using the triangle's area formula and side lengths (a, b, c, and semi-perimeter s), concluding with the formula IA = sqrt(bc(s-a)/s).
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Proving incenter concurrency
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Handwritten mathematical derivations on lined paper, showing the step-by-step process to derive the triangle area formula Δ = 2R² sin A sin B sin C from Δ = abc / 4R, using the sine rule substitutions for a, b, and c.
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The image displays the mathematical derivation of the Quarter Squares Rule, showing how the difference of two quarter squares, ((a+b)/2)^2 - ((a-b)/2)^2, simplifies to the product ab for real numbers a and b. The final formula is highlighted.
Quarter Squares Rule
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A diagram of triangle ABC with orthocenter O and orthic triangle DEF. Formulas for circumradius, area, and derivations for lengths OA, OB, OC are shown in terms of side lengths and area.
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Mathematical derivations for the side lengths of the orthic triangle, DE, EF, and FD, expressed in terms of the main triangle's side lengths a, b, and c.
Find lengths in terms of the sides
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AB × CI = AI × BI, part 3
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Δ = (abc)/(4R)
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Right triangle relations based on similarity
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cot A + cot B + cot C, part 1
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A hand-drawn diagram of a triangle with an inscribed circle, illustrating the inradius 'r' and deriving the formula for the area of a triangle (Δ = rs) using the semi-perimeter 's' and side lengths a, b, c.
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A handwritten mathematical proof on lined paper demonstrates that any number of the form 4n+3 cannot be expressed as the sum of two squares. It uses modular arithmetic to analyze the possible remainders of squares (even, odd, or mixed) when divided by 4.
Any number of the form 4n+3 cannot be expressed as the sum of two squares.
This is a fact that comes from elementary number theory. If you have a number that has a remainder of 3 when divided by 4, it can’t be written as a sum of two square numbers. Numbers like 7, 11, 15 etc. can’t be written in that form.
cubicequation

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A notebook page displays three sets of mathematical derivations for the area of a triangle (Δ). Each set shows the sum of two cotangent functions (cot A + cot B, cot B + cot C, cot C + cot A) expressed in terms of side lengths and area, leading to a formula for Δ.
Δ = c²/[2(cot A + cot B)]
Some more triangle area and angles identities derived from prior results. #math #maths #mathematics #geometry #trigonometry
cubicequation

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13 likes

A handwritten geometric proof of the Pythagorean theorem, a²+b²=c², is shown. It illustrates a large square formed by four right triangles with legs 'a' and 'b', and a smaller inner square with side 'c'. The area calculation (a+b)² = 4(½ab) + c² simplifies to the theorem.
Proof of the Pythagorean Theorem, part 1
This is one of my favorite geometric proofs. It’s basically a square that’s twisted into a larger square so an equating of areas occurs. #math #maths #mathematics #geometry #trigonometry
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4 likes

A handwritten derivation of the cosine double angle identity, `cos^2(B/2) - sin^2(B/2) = cos B`, on grid paper. It features a right-angled triangle with an angle bisector, applying the Pythagorean theorem and algebraic steps to reach the final identity.
Cosine double angle identity derivation
Using the angle bisector theorem and Pythagorean theorem, one can derive the cosine double angle identity. #math #maths #mathematics #geometry #trigonometry
cubicequation

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9 likes

The image displays handwritten mathematical derivations on a lined notebook page. It shows two geometric diagrams of triangles with labeled sides and angles. Calculations lead to the equation Wa² = 3c² - 2bc - b² from one triangle and 4hb² = 3c² - 2bc - b² from another, ultimately proving Wa = 2hb.
ωa = 2hb
The a, b in the formula above should be subscripts. In the regular heptagon and the heptagonal triangle, a = side length, b = length of short diagonal, c = length of long diagonal, ωa = length of angle bisector off smallest angle in heptagonal triangle and hb = length of altitude coming off side b
cubicequation

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8 likes

A handwritten mathematical proof on lined paper, showing a right triangle with an inscribed circle and its incenter. The derivation leads to the formula IC • IB = (AB - AC) • IA, demonstrating a relationship between segments connected to the incenter.
IC × IB = (AB - AC) × IA, part 1
In a right triangle, there is an interesting relationship between the segments that connect from the vertices to the incenter (labeled I). The product of the distances of the two shorter segments is equal two the distance of the longest segment multiplied by a factor that itself is the difference b
cubicequation

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11 likes

A handwritten mathematical derivation on paper, illustrating a triangle with internal lines and angles, and showing the step-by-step proof that the cotangent of the Brocard angle (ω) equals the sum of the cotangents of the triangle's angles (A, B, C).
cot ω = cot A + cot B + cot C
The cotangent of the Brocard angle in a triangle is equal to the sum of the cotangents of the triangle’s angles. #math #maths #mathematics #geometry #trigonometry
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4 × cos²(π/9) - √3/2 × csc(π/9) = 1
A proven relation from the regular nonagon. a = side length, b = short diagonal, c = medium diagonal, d = long diagonal & R = circumradius. #math #maths #mathematics #geometry #trigonometry
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3 likes

A handwritten page displays trigonometric relationships. It features a right-angled triangle defining sine, cosine, and tangent, a unit circle illustrating various ratios, and fundamental identities like sin²θ + cos²θ = 1, along with reciprocal and quotient identities.
Proving basic trigonometric relationships
In this post, the main six trigonometric ratios are shown why they interact as they do. The top half shows the traditional way. The bottom half shows how with similar triangles. #math #maths #mathematics #geometry #trigonometry
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9 likes

(r+s)/(a+b) + h/c = 1
Another relation valid in any right triangle. It’s proven using the concept of similar triangles. #math #maths #mathematics #geometry #trigonometry
cubicequation

cubicequation

5 likes

A handwritten mathematical proof on lined paper, showing a right-angled triangle ABC with its incircle and inradius 'r'. The proof uses the semiperimeter 's' and algebraic steps to derive the identity IB * IC = (AB - AC) * IA, relating segments from the incenter to the vertices.
IC × IB = (AB - AC) × IA, part 2
This proof starts with an identity about the inradius of the right triangle and an identity about a segment connecting the incenter to the medium-sized angle. This proof seems more natural given that the semiperimeter is used more with oblique triangles than right triangles. #math #maths #m
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cubicequation

10 likes

A handwritten page illustrates Garfield's Proof of the Pythagorean Theorem. It features a geometric diagram of a trapezoid formed by three right triangles, alongside algebraic steps that equate the trapezoid's area to the sum of the triangles' areas, leading to the derivation of a² + b² = c².
Proof of the Pythagorean Theorem, part 2
This is an augmented version of the first proof discovered by US President James Garfield. By slicing the original diagram in half, a trapezoid is created with three right triangles overlaying it so an equating of areas can occur. #math #maths #mathematics #geometry #trigonometry
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cubicequation

6 likes

The image displays two handwritten mathematical proofs on lined paper. The first proof demonstrates that all primes p≥3 are of the form 4n+1 or 4n+3. The second proof shows that all primes p≥5 are of the form 6n+1 or 6n+5, by eliminating even numbers and multiples of 3.
Proofs of some prime number properties
These are two proofs for why prime numbers revolve around multiples of 4 and multiples of 6. #math #maths #mathematics #algebra #arithmetic
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7 likes

A handwritten diagram illustrates a proof of the Pythagorean theorem. A large square, side 'c', contains four right triangles (legs 'a', 'b') and a central square (side 'b-a'). Equations below show the area calculation c² = 4(1/2 ab) + (b-a)² simplifying to c² = b² + a².
Proof of the Pythagorean Theorem, part 3
Here’s another classic proof of the Pythagorean Theorem where four triangles are placed in such a way that a small square in the middle of them. From there, an equating of areas occurs. #math #maths #mathematics #geometry #trigonometry
cubicequation

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8 likes

A handwritten page showing mathematical derivations of sine relations in a regular heptagon using Ptolemy's Theorem. It includes a heptagon diagram, algebraic steps, and various sine identities, with 'lemon8' and '@cubicequation' watermarks.
Sine relations in the regular heptagon
#mathematics #math #trigonometry #trig #geometry
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A handwritten mathematical derivation on lined paper, proving the trigonometric identity 4 cos²(2π/9) - √3/2 csc(2π/9) = 1. The steps involve substitutions related to geometric properties of a regular nonagon, ultimately simplifying to 1.
4 × cos²(2π/9) - √3/2 × csc(2π/9) = 1
A proven relation from the regular nonagon. a = side length, b = short diagonal, c = medium diagonal, d = long diagonal & R = circumradius. #math #maths #mathematics #geometry #trigonometry
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Proof of the First Mollweide Formula
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A handwritten page showing the initial steps of deriving Heron's formula. It includes a triangle with its circumcircle, the formula for the circumradius R, cosine rules, and the derivation of partial areas (Δa, Δb, Δc) using R and trigonometric identities.
Deriving Heron’s formula using the circumradius of the triangle
Heron’s formula is a useful formula to find the area of a triangle using only the triangle’s side lengths. This is a lesser-known derivation of the formula, but quite a cool one nonetheless. #math #maths #mathemati cs #geometry #trigonometry
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8 likes

ωa = √(a(2a+b))
The a attached to ω should be subscripted. In the heptagonal triangle, a, b, c are the side lengths in order from least to greatest. ωa = length of the angle bisector coming off the angle opposite side a. #math #maths #mathematics #trigonometry #geometry
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A hand-drawn diagram of an acute triangle ABC with altitudes and orthocenter, showing trigonometric derivations for the distance from the orthocenter to vertex A, resulting in X = 2R cos A.
A hand-drawn diagram of an acute triangle ABC with altitudes and orthocenter, showing trigonometric derivations for the distance from the orthocenter to vertex C, resulting in Y = 2R cos C.
A hand-drawn diagram of an acute triangle ABC with altitudes and orthocenter, showing trigonometric derivations for the distance from the orthocenter to vertex B, resulting in Z = 2R cos B.
Connecting the orthocenter to the vertices
In any acute triangle ABC, the distances from the orthocenter (where the altitudes meet) to the vertices are of the form 2R × cos Ε, where E is one of the angles of the triangle. R is the length of the circumradius of the triangle. #math #maths #mathematics #geometry #trigonometry
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The image shows handwritten mathematical work, including diagrams of a heptagon and an isosceles trapezoid. It illustrates the application of Ptolemy's Theorem to derive algebraic equations, leading to the conclusion that m = 2a.
2 in the regular heptagon, part 2
Since the regular heptagon is symmetric around the top vertex, two reflected heptagonal triangles can be formed near the top of the shape. Thus, from where the angle bisectors from the smallest angles in each triangle to where they meet the opposite side lengths are reflected points across the line
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A handwritten mathematical derivation on lined paper, showing a regular heptagon diagram with labeled sides 'a', 'b', 'c'. The derivation uses the Law of Cosines to prove that a specific length 'l' within the heptagon is equal to '2a', where 'a' is a side length.
2 in the regular heptagon, part 1
The distance from where the angle bisector coming off the smallest angle in the heptagonal triangle intersects with the shortest side (which is also a side length of the regular heptagon) to the opposite bottom of the vertex of the heptagon is exactly 2. Not many people know that fact. #math
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6 likes

A handwritten page in a lined notebook displays algebraic equations, specifically Brahmagupta's identity. It shows the expansion of (a²+b²)(c²+d²) into two different forms: (ac+bd)²+(ad-bc)² and (ac-bd)²+(ad+bc)², illustrating how the product of two sums of squares can be expressed as a sum of two squares.
Two squares times two squares makes two squares
The sum of two squares times the sum of another two squares equals the sum of yet another two squares. These formulae are used to find examples like 5 × 13 = (2²+1²) × (2²+3²) = 65 = 8²+1² = 7²+4². #math #maths #mathematics #algebra #arithmetic
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19 likes

mn = Δ
In any right triangle, the area of the triangle can be found using the hypotenuse when it’s spilt by the inradius. a, b & c are the side lengths of the right triangle. m, n are the lengths split by the inradius on the hypotenuse. r is the length of the inradius. Δ is the area of the right trian
cubicequation

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2 likes

A handwritten page details trigonometric relations for a regular pentagon, using side 'a', diagonal 'b', and circumradius 'R'. It presents derivations and results for various sums and products of trigonometric functions (tan, sec, csc, cot) involving angles π/5 and 2π/5, such as 10 = tan²(π/5) + tan²(2π/5).
Trigonometric relations in the regular pentagon
In the regular pentagon, a = side length, b = length of the diagonal & R = length of the circumradius. #math #maths #mathematics #geometry #trigonometry
cubicequation

cubicequation

51 likes

Proving the Law of Cosines
This post shows a standard proof of the Law of Cosines in any oblique triangle. #math #maths #mathematics #geometry #trigonometry
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cubicequation

6 likes

Handwritten mathematical notes on lined paper showing a right triangle with an inscribed circle, labeled with sides a, b, c, and inradius r. Formulas for triangle area and inradius derivation are presented, leading to r = ab/(a+b+c). The calculation then applies Pythagorean triples to simplify the inradius to r = n(m-n).
r = (a × b)/(a + b + c)
In a right triangle, the inradius can be found by finding the product of the legs and dividing by the perimeter. #math #maths #mathematics #geometry #trigonometry
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cubicequation

3 likes

A handwritten mathematical derivation on lined paper, showing the proof of the trigonometric identity 4/√7 × sin³(π/7) + 1/2 = cos(2π/7), relating it to properties of a regular heptagon with side length 'a', diagonals 'b' and 'c', and circumradius 'R'.
4/√7 × sin³(π/7) + 1/2 = cos(2π/7)
This relation comes from the regular heptagon. In that shape, a = side length, b = length of the short diagonal, c = length of the long diagonal & R = length of the circumradius. #math #maths #mathematics #geometry #trigonometry
cubicequation

cubicequation

4 likes

A handwritten math solution on lined paper, showing a geometric figure and a step-by-step derivation. The problem calculates the value of a²/b² + b²/c² + c²/a², using Ptolemy's Theorem and algebraic manipulations, concluding with the answer 6.
a²/b² + b²/c² + c²/a² = 6
#math #maths #mathematics #trig #trigonometry
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cubicequation

318 likes

b²/a² + c²/b² + a²/c² = 5
#math #mathematics #maths #trigonometry #trig
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cubicequation

23 likes

Law of Tangents
#math #maths #mathematics #trig #trigonometry
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cubicequation

7 likes

A handwritten math diagram illustrates an isosceles triangle with a cevian, labeled with sides 'c', 'a', height 'h', and base segments 'b', 'd'. Below, equations derive the relation c² = a² + bd using the Pythagorean theorem and the Quarter-Squares Rule.
c² = a² + bd
Given an isosceles triangle and a cevian that splits the base into two smaller segments, then the following relation is true. This is proven using the Pythagorean theorem twice. #math #matha #mathematics #geometry #trigonometry
cubicequation

cubicequation

3 likes

L² = a² + ab + b²
In any equilateral triangle, L is a cevian and a, b are the lengths where the cevian splits the side into two distinct lengths. s is the side length of the triangle. #math #maths #mathematics #geometry #trigonometry
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A handwritten derivation of the Law of Cotangents on graph paper. It shows the steps from individual cotangent half-angle formulas involving the semi-perimeter (s) and inradius (r) to the combined law: cot(A/2)/(s-a) = cot(B/2)/(s-b) = cot(C/2)/(s-c) = 1/r.
The law of cotangents
Deriving the law of cotangents using the inradius of a triangle. #math #maths #mathematics #geometry #trigonometry
cubicequation

cubicequation

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Handwritten math notes showing derivations for the area of an equilateral triangle, and the cosine rule for triangles with 120-degree and 60-degree angles, including diagrams and formulas.
Some formulae involving triangles
Derived here are three separate formulae involving triangles. The first one is the area for an equilateral triangle. The second one finds the side opposite of a 120° angle in a triangle. The third one finds the side opposite of a 60° angle in a triangle. #math #maths #mathematics #geom
cubicequation

cubicequation

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Handwritten mathematical derivation on lined paper, proving the identity 6 - 8 sin²(π/7) = -sec(4π/7). The steps involve geometric relations of a regular heptagon, using its side length 'a', short diagonal 'b', and circumradius 'R'.
6 - 8 × sin²(π/7) = -sec(4π/7)
This relation comes from the regular heptagon. In that shape, a = side length, b = length of the short diagonal, c = length of the long diagonal & R = length of the circumradius. #math #maths #mathematics #geometry #trigonometry
cubicequation

cubicequation

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A handwritten mathematical proof demonstrating the derivation of the Second Mollweide Formula. It starts with the Law of Sines and uses trigonometric identities to arrive at (a-b)/c = sin((A-B)/2) / cos(C/2), which is highlighted in a box.
Proof of Second Mollweide Formula
#math #maths #mathematics #trig #trigonometry
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cubicequation

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Proving the existence of the Gergonne point of a triangle
The Gergonne point is where three specific segments are concurrent. These segments come off the vertices and intersect where the incircle is tangent to the side opposite of the vertex. While not as well-known as the most well-known triangle centers, its existence is easily proven with Ceva’s theore
cubicequation

cubicequation

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A handwritten mathematical derivation of the trigonometric identity cos(A+B) = cos A cos B - sin A sin B. The derivation uses a triangle diagram, area formulas, and various trigonometric substitutions, progressing step-by-step to the final identity.
cosA × cos B - sin A × sin B = cos(A+B)
This is an obscure derivation of a well-known trigonometric identity. #math #maths #mathematics #geometry #trigonometry
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cubicequation

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