Metamath Proof Explorer


Theorem sinhpcosh

Description: Prove that ( sinhA ) + ( coshA ) = ( expA ) using the conventional hyperbolic trigonometric functions. (Contributed by David A. Wheeler, 27-May-2015)

Ref Expression
Assertion sinhpcosh ⊢ A ∈ ℂ → sinh ⁡ A + cosh ⁡ A = e A

Proof

Step Hyp Ref Expression
1 sinhval-named ⊢ A ∈ ℂ → sinh ⁡ A = sin ⁡ i ⁢ A i
2 sinhval ⊢ A ∈ ℂ → sin ⁡ i ⁢ A i = e A − e − A 2
3 1 2 eqtrd ⊢ A ∈ ℂ → sinh ⁡ A = e A − e − A 2
4 coshval-named ⊢ A ∈ ℂ → cosh ⁡ A = cos ⁡ i ⁢ A
5 coshval ⊢ A ∈ ℂ → cos ⁡ i ⁢ A = e A + e − A 2
6 4 5 eqtrd ⊢ A ∈ ℂ → cosh ⁡ A = e A + e − A 2
7 3 6 oveq12d ⊢ A ∈ ℂ → sinh ⁡ A + cosh ⁡ A = e A − e − A 2 + e A + e − A 2
8 2cn ⊢ 2 ∈ ℂ
9 2ne0 ⊢ 2 ≠ 0
10 efcl ⊢ A ∈ ℂ → e A ∈ ℂ
11 negcl ⊢ A ∈ ℂ → − A ∈ ℂ
12 efcl ⊢ − A ∈ ℂ → e − A ∈ ℂ
13 11 12 syl ⊢ A ∈ ℂ → e − A ∈ ℂ
14 10 13 addcld ⊢ A ∈ ℂ → e A + e − A ∈ ℂ
15 10 13 subcld ⊢ A ∈ ℂ → e A − e − A ∈ ℂ
16 divdir ⊢ e A − e − A ∈ ℂ ∧ e A + e − A ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → e A − e − A + e A + e − A 2 = e A − e − A 2 + e A + e − A 2
17 15 16 syl3an1 ⊢ A ∈ ℂ ∧ e A + e − A ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → e A − e − A + e A + e − A 2 = e A − e − A 2 + e A + e − A 2
18 14 17 syl3an2 ⊢ A ∈ ℂ ∧ A ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → e A − e − A + e A + e − A 2 = e A − e − A 2 + e A + e − A 2
19 18 3anidm12 ⊢ A ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → e A − e − A + e A + e − A 2 = e A − e − A 2 + e A + e − A 2
20 8 9 19 mpanr12 ⊢ A ∈ ℂ → e A − e − A + e A + e − A 2 = e A − e − A 2 + e A + e − A 2
21 10 2timesd ⊢ A ∈ ℂ → 2 ⁢ e A = e A + e A
22 10 13 10 nppcand ⊢ A ∈ ℂ → e A − e − A + e A + e − A = e A + e A
23 15 10 13 addassd ⊢ A ∈ ℂ → e A − e − A + e A + e − A = e A − e − A + e A + e − A
24 21 22 23 3eqtr2rd ⊢ A ∈ ℂ → e A − e − A + e A + e − A = 2 ⁢ e A
25 24 oveq1d ⊢ A ∈ ℂ → e A − e − A + e A + e − A 2 = 2 ⁢ e A 2
26 7 20 25 3eqtr2d ⊢ A ∈ ℂ → sinh ⁡ A + cosh ⁡ A = 2 ⁢ e A 2
27 8 a1i ⊢ A ∈ ℂ → 2 ∈ ℂ
28 9 a1i ⊢ A ∈ ℂ → 2 ≠ 0
29 10 27 28 divcan3d ⊢ A ∈ ℂ → 2 ⁢ e A 2 = e A
30 26 29 eqtrd ⊢ A ∈ ℂ → sinh ⁡ A + cosh ⁡ A = e A