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A Geometric Way to Visualise sin(x + y) and cos(x + y)

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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...
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The Method of Differences — A Clean Proof of the Sum of Cubes

The Method of Differences — A Clean Proof of the Sum of Cubes The method of differences is a remarkably elegant tool for evaluating finite sums. When each term of a series can be written in the form f(r+1) − f(r) , the sum “collapses” — all interior terms cancel, leaving only a boundary expression. This behaviour is called a telescoping sum . 1) Telescoping Sums Assume the general term u r can be written as: u r = f(r+1) − f(r). Then the finite sum from r = 1 to r = n becomes: Σ u r = Σ ( f(r+1) − f(r) ). To see what happens, write out a few terms: u₁ = f(2) − f(1) u₂ = f(3) − f(2) u₃ = f(4) − f(3) ⋮ uₙ = f(n+1) − f(n) When these are added, everything cancels except the first and last pieces: Σ u r = f(n+1) − f(1). This is the essence of the method: interior structure disappears, leaving just the difference between the final and initial states. 2) A Classic Application — The Sum of Cubes We will use this technique to prove the well-known ...
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The Shortest Distance Between Two Skew Lines in ℝ³

The Shortest Distance Between Two Skew Lines in ℝ³ This post derives, from first principles, a vector formula for the shortest distance between two skew lines in ℝ³. The argument uses only the definitions and basic properties of the dot product and cross product; no higher results are assumed. 1. Vector Equations of the Lines Let the two lines be given in vector form by r = a + λb r = c + μd where: a , c are position vectors of fixed points on each line, b , d are non-zero direction vectors, λ, μ ∈ ℝ are parameters. The lines are skew if they are neither parallel nor intersecting. Our goal is to find a closed-form expression for the minimum distance between them. 2. Direction of the Common Perpendicular The segment that realises the shortest distance lies along a line perpendicular to both direction vectors b and d . A vector perpendicular to both is given by their cross product: b × d. Assuming b and d are not parallel, b × d ≠ 0. A unit ...
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