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 ½ iⁿ F_n.
3. Replace cosines by ½ 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

pα + qβ

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.

Examples

Double angle

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

sin 2θ = 2 sin θ cos θ.

This becomes

½ i²ⁿ F_2n = 2 (½ iⁿ F_n)(½ iⁿ L_n)

which simplifies to

F_2n = F_n L_n.

Sum angle

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

cos(θ + φ) = cos θ cos φ − sin θ sin φ.

This becomes

½ i^{a + b} L_{a + b} = ½ i^a L_a ½ i^b L_b −(−5) ½ i^a F_a ½ 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 of i in the recipe above cancelled out. This example will show that they are necessary.

We start with the triple angle identity

sin 3θ = 3 sin θ − 4 sin³ θ.

This becomes

½ i³ⁿ F_3n = 3 (½) iⁿ F_n − 4(−5)(½ 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