CCFU Proof 6 — Lucas = cosh, Fibonacci = sinh (even index)

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CCFU Proof 6 — Lucas = cosh, Fibonacci = sinh (even index)

05/15/2026

Given.
Binet's formulas:

    F(n) = (φⁿ − ψⁿ)/√5
    L(n) = φⁿ + ψⁿ

where ψ = −1/φ.

Even index.  For n = 2k, k ≥ 0:

    ψ^{2k} = (−1/φ)^{2k} = (1/φ)^{2k} = φ^{−2k}

Theorem 1 (Lucas = 2·cosh).

    L(2k) = φ^{2k} + φ^{−2k} = 2·cosh(2k·ln φ)  ∎

Theorem 2 (Fibonacci = (2/√5)·sinh).

    F(2k) = (φ^{2k} − φ^{−2k})/√5 = (2/√5)·sinh(2k·ln φ)  ∎

Corollary (Hyperbolic identity).
Let r = 2k·ln φ. Then:

    L(2k)² − 5·F(2k)²
    = [2·cosh(r)]² − 5·[(2/√5)·sinh(r)]²
    = 4·cosh²(r) − 4·sinh²(r)
    = 4·(cosh²(r) − sinh²(r))
    = 4  ∎

Note.
These identities hold exactly for even indices. For odd indices the roles swap: L(2k+1) involves sinh and F(2k+1) involves cosh. The even-index identities do not apply as stated to odd indices.