A Rectangle Construction for sin(α − β) and cos(α − β)

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.

A Rectangle Construction for sin(α − β) and cos(α − β)

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 right side: ZY = ZP + PY, so

ZY = cosα cosβ + sinα sinβ.

Opposite sides of a rectangle are equal, so

OX = ZY = cosα cosβ + sinα sinβ.

On the base: OZ = sinα cosβ. The top side XY has the same length, and is split into XQ and QY. Therefore

XQ + QY = OZ = sinα cosβ.

Substituting QY = cosα sinβ gives

XQ = sinα cosβ − cosα sinβ.

Now look at triangle OQX. It is right-angled at X, with hypotenuse OQ = 1. The side OX is adjacent to the angle at O, and the side XQ is opposite to the angle at O. The angle between OQ and OX is (α − β), so:

cos(α − β) = OX / OQ = cosα cosβ + sinα sinβ,
sin(α − β) = XQ / OQ = sinα cosβ − cosα sinβ.

The rectangle construction therefore gives a clear geometric explanation of the angle–difference formulas for sine and cosine.

Comments

Post a Comment

Popular posts from this blog

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

Image
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...
Read more

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 ...
Read more

2×2 Orthogonal Matrix Mastery — A Generalised Construction

Image
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...
Read more