Turning trig identities into Fibonacci identities

In 2013, John Conway and Alex Ryba published a brief note [1] on how to convert identities involving sine and cosine into identities involving Fibonacci and Lucas numbers.

Fibonacci and Lucas

The Fibonacci numbers F_n are defined by F₀ = 0, F₁ = 1, and

F_n+2 = F_n + F_n+1

for n > 1. Similarly, Lucas numbers L_n are defined by the same recurrence relation

L_n+2 = L_n + L_n+1

but starting with L₀ = 2 and L₁ = 1.

The recipe

The recipe for turning trigonometric identities into Fibonacci and Lucas number identities is as follws.

1. Arguments become subscripts.
2. Replace sines by iⁿ F_n.
3. Replace cosines by iⁿ L_n.
4. Insert a factor of (-5)^k in front of every term that contains 2k or 2k + 1 terms.

The first step is admittedly unclear. It's unclear in [1] as well. If sine or cosine takes an argument of

p +q

then this becomes the subscript

pa +qb

where the Latin letters are integers and the Greek letters are angles.

There's no proof in [1] why the recipe should work, but an earlier paper [2] gives more details.

ExamplesDouble angle

For a simple example, let's start with the double angle identity

sin 2 = 2 sin cos .

This becomes

i²ⁿ F_2n = 2 ( iⁿ F_n)( iⁿ L_n)

which simplifies to

F_2n = F_nL_n.

Sum angle

For a little more involved example, let's look at the sum identity

cos( + ) = cos cos - sin sin .

This becomes

i^{a + b} L_{a + b} = i^a L_a i^b L_b -(-5) i^a F_a i^b F_b

which simplifies to

2L_{a + b} = L_a L_b + 5 F_a F_b.

Triple angle

In the previous examples the powers ofi in the recipe above cancelled out. This example will show that they are necessary.

We start with the triple angle identity

sin 3 = 3 sin - 4 sin^3 .

This becomes

i³ⁿ F_3n = 3 () iⁿ F_n - 4(-5)( iⁿ F_n)^3

which simplifies to

F_3n = 3 (-1)ⁿ F_n + 5 (F_n)^3.

Demonstration

The following Python code demonstrates that the Fibonacci / Lucas identities above are correct.

def F(n): if n == 0: return 0 if n == 1: return 1 return F(n-1) + F(n-2)def L(n): if n == 0: return 2 if n == 1: return 1 return L(n-1) + L(n-2)for n in range(10): assert(F(2*n) == F(n) * L(n))for n in range(10): for m in range(10): assert(2*L(m + n) == L(m)*L(n) + 5*F(m)*F(n))for n in range(10): assert(F(3*n) == 3*(-1)**n * F(n) + 5*F(n)**3)

Related posts

• Convert trig identities to hyperbolic identities
• Inverse Fibonacci numbers
• Lucas numbers and Lucas pseudoprimes
• Trig mnemonic diagrams

[1] John Conway and Alex Ryba. Fibonometry. The Mathematical Gazette, November 2013, pp. 494-495

[2] Barry Lewis, Trigonometry and Fibonacci Numbers, The Mathematical Gazette, July 2007, pp. 216-226

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