Metamath Proof Explorer


Theorem coshneg

Description: The cosh function is even. (Contributed by Jon Pennant, 9-Sep-2026)

Ref Expression
Assertion coshneg ⊢ A ∈ ℂ → cosh ⁡ − A = cosh ⁡ A

Proof

Step Hyp Ref Expression
1 negcl ⊢ A ∈ ℂ → − A ∈ ℂ
2 coshval-named ⊢ − A ∈ ℂ → cosh ⁡ − A = cos ⁡ i ⁢ − A
3 1 2 syl ⊢ A ∈ ℂ → cosh ⁡ − A = cos ⁡ i ⁢ − A
4 ax-icn ⊢ i ∈ ℂ
5 mulcl ⊢ i ∈ ℂ ∧ − A ∈ ℂ → i ⁢ − A ∈ ℂ
6 4 1 5 sylancr ⊢ A ∈ ℂ → i ⁢ − A ∈ ℂ
7 cosneg ⊢ i ⁢ − A ∈ ℂ → cos ⁡ − i ⁢ − A = cos ⁡ i ⁢ − A
8 6 7 syl ⊢ A ∈ ℂ → cos ⁡ − i ⁢ − A = cos ⁡ i ⁢ − A
9 3 8 eqtr4d ⊢ A ∈ ℂ → cosh ⁡ − A = cos ⁡ − i ⁢ − A
10 4 a1i ⊢ A ∈ ℂ → i ∈ ℂ
11 id ⊢ A ∈ ℂ → A ∈ ℂ
12 10 11 mulneg2d ⊢ A ∈ ℂ → i ⁢ − A = − i ⁢ A
13 12 negeqd ⊢ A ∈ ℂ → − i ⁢ − A = − − i ⁢ A
14 10 11 mulcld ⊢ A ∈ ℂ → i ⁢ A ∈ ℂ
15 14 negnegd ⊢ A ∈ ℂ → − − i ⁢ A = i ⁢ A
16 13 15 eqtrd ⊢ A ∈ ℂ → − i ⁢ − A = i ⁢ A
17 16 fveq2d ⊢ A ∈ ℂ → cos ⁡ − i ⁢ − A = cos ⁡ i ⁢ A
18 9 17 eqtrd ⊢ A ∈ ℂ → cosh ⁡ − A = cos ⁡ i ⁢ A
19 coshval-named ⊢ A ∈ ℂ → cosh ⁡ A = cos ⁡ i ⁢ A
20 18 19 eqtr4d ⊢ A ∈ ℂ → cosh ⁡ − A = cosh ⁡ A