Odd Angles Formula Trivia
10-13-2021, 03:40 AM (This post was last modified: 10-13-2021 09:05 AM by Albert Chan.)
Post: #9
 Albert Chan Senior Member Posts: 2,066 Joined: Jul 2018
RE: Odd Angles Formula Trivia
(12-20-2018 05:38 PM)Albert Chan Wrote:  Challenge: prove above pattern continues: if sin(4nx ± x) = f(s,c), then cos(4nx ± x) = ± f(c,s)

An elegant proof (induction proof work, but messy, and does not explain why ...)

Let y = pi/2 - x

Let f3(cos(x),sin(x)) = cos(3x)
→ ﻿ ﻿ f3(sin(x),cos(x)) = f3(cos(y),sin(y)) = cos(3y) = cos(3*(pi/2)-3x) = −sin(3x)

Let f5(cos(x),sin(x)) = cos(5x)
→ ﻿ ﻿ f5(sin(x),cos(x)) = f5(cos(y),sin(y)) = cos(5y) = cos(5*(pi/2)-5x) = +sin(5x)

We can extend angle to (4n ± 1)*x and get same pattern , because 4n*(pi/2) = 2n*pi ---

This is inspired by swapping of complex parts function I post here

z = x+y*i
swap(z) = y+x*i = i*conj(z)

swap(z)^3 = (i*conj(z))^3 = -i*conj(z^3) = −swap(z^3)
swap(z)^5 = (i*conj(z))^5 = i*conj(z^5) = +swap(z^5)

Since i^4 = 1, same (4n±1) pattern happened here as well.

(cos(x) + i*sin(x))^(4n±1) = ﻿ ﻿ cos((4n±1)*x) + i * sin((4n±1)*x)
(sin(x) + i*cos(x))^(4n±1) = ±sin((4n±1)*x) + i*±﻿cos((4n±1)*x)