Geometric Bites

The angle–difference identities are: sin(α − β) = sinα cosβ − cosα sinβ cos(α − β) = cosα cosβ + sinα sinβ They can be seen geometrically using a rectangle OZYX with a few right-angled triangles inside it. All side lengths can be written in terms of sinα, cosα, sinβ and cosβ. First, draw a right triangle OPQ with hypotenuse OQ = 1 and angle β at O. By definition: OP = cosβ (horizontal side), QP = sinβ (vertical side). Next, use OP as the hypotenuse of another right triangle OPZ. The right angle is at Z, and the angle at P is α. The hypotenuse is OP = cosβ, so: OZ = sinα cosβ, ZP = cosα cosβ. In a similar way, use QP as the hypotenuse of a right triangle PQY. The right angle is at Y, and the angle at Q is α. The hypotenuse is QP = sinβ, so: QY = cosα sinβ, PY = sinα sinβ. Drop a vertical line from Y to the base at Z, and a horizontal line from Y to the left side at X. This makes OZYX a rectangle with: base OZ, height ZY. On the...

Consider a cyclic quadrilateral: all four of its vertices lie on a circle. Join the centre of the circle to each vertex. This creates four isosceles triangles, each made from two radii and one side of the quadrilateral. Label the angles of the quadrilateral at the circumference by w, x, y and z. In each isosceles triangle, the angles at the base are equal, so the angle at the centre is: π − 2w, π − 2x, π − 2y, π − 2z. These four central angles meet at a point, so together they make one full turn: (π − 2w) + (π − 2x) + (π − 2y) + (π − 2z) = 2π. Rearranging gives: −2w − 2x − 2y − 2z + 4π = 2π so 2w + 2x + 2y + 2z = 2π and therefore w + x + y + z = π. From this, the opposite-angle relations in the quadrilateral follow directly: w + z = π − (x + y), z + y = π − (w + x). So in any cyclic quadrilateral, each pair of opposite angles adds up to π radians .

The angle–addition identities for sine and cosine often appear as algebraic formulas, but they can also be understood by combining two right triangles in a simple geometric construction. The calculations for the side lengths follow directly from the definitions of sine and cosine. sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y − sin x sin y Start with a right triangle of angle y and hypotenuse 1. From basic trigonometry, its horizontal and vertical sides are: cos y and sin y. Next, attach a second right triangle with angle x . Its hypotenuse is the side of length cos y from the first triangle, so its adjacent and opposite sides become: adjacent = cos x · cos y opposite = sin x · cos y Likewise, if the first triangle's vertical side sin y is used as a hypotenuse in a similar way, it contributes: adjacent = cos x · sin y opposite = sin x · sin y When the horizontal components are combined, they give the expression for cos(x + y...

When I write mathematical proofs, clarity and precision matter. A single misplaced symbol can alter the entire meaning of an argument. That is why my writing tool must be reliable, smooth, and forgiving. After years of trying different pens, pencils and fineliners, I now rely on one tool for all my handwritten proofs: the Pilot FriXion Ball 0.7 mm . This pen has become an essential part of my workflow. It writes smoothly, erases cleanly and can be refilled easily, making it ideal for long mathematical sessions where neatness and accuracy are critical. Video Review (1 Minute) Here is a short demonstration showing the pens up close and how the ink writes and erases. Why This Pen Works Perfectly for Mathematical Proofs Proof-writing is a process of refinement. You revise, adjust, correct and restructure your ideas repeatedly. With most pens, each correction introduces visual noise — scribbles, crossings-out or rewritten pages. The FriXion Ball removes that issue entirely....

A diagonal matrix is a square matrix in which every entry away from the main (leading) diagonal is zero. The leading diagonal runs from the top-left corner of the matrix to the bottom-right corner, and these diagonal entries are the only positions that may contain non-zero values. All off-diagonal entries must be zero. The diagonal entries themselves can be any real numbers, including zero. This strict structure is what makes diagonal matrices especially simple to analyse and compute with in linear algebra. Examples of Diagonal Matrices The general 2×2 diagonal matrix has the form: (a 0) (0 b) The general 3×3 diagonal matrix has the form: (a 0 0) (0 b 0) (0 0 c) In both cases, the values on the leading diagonal (a, b, c, …) are the only entries that may be non-zero. Every position above or below this diagonal is fixed at 0. The General n×n Diagonal Matrix For an n×n dia...

Why the Line ax + by = 0 Passes Through the Point (−b, a) In ℝ² , the equation ax + by = 0 describes a line that is perpendicular to the vector (a, b) . This article explains exactly why—and why that line always passes through the point (−b, a) . 1. Start with the Vector (a, b) Consider the vector (a, b) . To find a line perpendicular to it, we need a vector whose dot product with (a, b) is zero. Try the vector (−b, a) : (a, b) · (−b, a) = a(−b) + b(a) = −ab + ab = 0 Therefore, (−b, a) is perpendicular to (a, b) . 2. Any Scalar Multiple Also Works If (−b, a) is perpendicular to (a, b) , then any multiple λ(−b, a) is also perpendicular: (a, b) · [λ(−b, a)] = λ[(a, b) · (−b, a)] = λ · 0 = 0 Let this perpendicular vector be (x, y) . Then (x, y) = λ(−b, a) Every point on the line comes from a particular choice of λ . 3. Converting to an Equation Since (x, y) is perpendicular to (a, b) , we have: (a, b) · (x, y) = 0 Expanding ...

2×2 Orthogonal Matrix Mastery — A Generalised Construction Orthogonal matrices in two dimensions reveal one of the cleanest structures in linear algebra. A 2×2 matrix is orthogonal when its columns (and rows) satisfy two conditions: They are perpendicular (their dot product is zero); They have unit length (their magnitude is one). This article presents a clear generalisation: any pair of perpendicular vectors with equal magnitude can be normalised to form an orthogonal matrix. 1. Begin with two perpendicular vectors Let the first vector be: (a, b) A perpendicular vector can be chosen as: (−b, a) Their dot products confirm orthogonality: (a, b) · (−b, a) = −ab + ab = 0 (a, −b) · (b, a) = ab − ab = 0 2. Compute their shared magnitude Both vectors have the same length: |(a, b)| = √(a² + b²) We can therefore normalise each one by dividing by √(a² + b²). 3. Form the matrix using the normalised vectors Place the two normalised vectors...