Odd Angles Formula Trivia

[Albert Chan]

(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

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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)

This already proved it !
Think of LHS the "f". For effect of swapping parts, read RHS vertically.