Metamath Proof Explorer


Theorem sinhneg

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

Ref Expression
Assertion sinhneg ⊢ A ∈ ℂ → sinh ⁡ − A = − sinh ⁡ A

Proof

Step Hyp Ref Expression
1 ax-icn ⊢ i ∈ ℂ
2 1 jctl ⊢ A ∈ ℂ → i ∈ ℂ ∧ A ∈ ℂ
3 mulcl ⊢ i ∈ ℂ ∧ A ∈ ℂ → i ⁢ A ∈ ℂ
4 sincl ⊢ i ⁢ A ∈ ℂ → sin ⁡ i ⁢ A ∈ ℂ
5 2 3 4 3syl ⊢ A ∈ ℂ → sin ⁡ i ⁢ A ∈ ℂ
6 1 a1i ⊢ A ∈ ℂ → i ∈ ℂ
7 ine0 ⊢ i ≠ 0
8 7 a1i ⊢ A ∈ ℂ → i ≠ 0
9 5 6 8 divnegd ⊢ A ∈ ℂ → − sin ⁡ i ⁢ A i = − sin ⁡ i ⁢ A i
10 sinhval-named ⊢ A ∈ ℂ → sinh ⁡ A = sin ⁡ i ⁢ A i
11 10 negeqd ⊢ A ∈ ℂ → − sinh ⁡ A = − sin ⁡ i ⁢ A i
12 negcl ⊢ A ∈ ℂ → − A ∈ ℂ
13 sinhval-named ⊢ − A ∈ ℂ → sinh ⁡ − A = sin ⁡ i ⁢ − A i
14 12 13 syl ⊢ A ∈ ℂ → sinh ⁡ − A = sin ⁡ i ⁢ − A i
15 mulneg2 ⊢ i ∈ ℂ ∧ A ∈ ℂ → i ⁢ − A = − i ⁢ A
16 2 15 syl ⊢ A ∈ ℂ → i ⁢ − A = − i ⁢ A
17 16 eqcomd ⊢ A ∈ ℂ → − i ⁢ A = i ⁢ − A
18 17 fveq2d ⊢ A ∈ ℂ → sin ⁡ − i ⁢ A = sin ⁡ i ⁢ − A
19 18 oveq1d ⊢ A ∈ ℂ → sin ⁡ − i ⁢ A i = sin ⁡ i ⁢ − A i
20 sinneg ⊢ i ⁢ A ∈ ℂ → sin ⁡ − i ⁢ A = − sin ⁡ i ⁢ A
21 2 3 20 3syl ⊢ A ∈ ℂ → sin ⁡ − i ⁢ A = − sin ⁡ i ⁢ A
22 21 oveq1d ⊢ A ∈ ℂ → sin ⁡ − i ⁢ A i = − sin ⁡ i ⁢ A i
23 14 19 22 3eqtr2d ⊢ A ∈ ℂ → sinh ⁡ − A = − sin ⁡ i ⁢ A i
24 9 11 23 3eqtr4rd ⊢ A ∈ ℂ → sinh ⁡ − A = − sinh ⁡ A