Metamath Proof Explorer


Theorem asinlem2

Description: The argument to the logarithm in df-asin has the property that replacing A with -u A in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015)

Ref Expression
Assertion asinlem2 ⊢ A ∈ ℂ → i ⁢ A + 1 − A 2 ⁢ i ⁢ − A + 1 − − A 2 = 1

Proof

Step Hyp Ref Expression
1 ax-icn ⊢ i ∈ ℂ
2 mulcl ⊢ i ∈ ℂ ∧ A ∈ ℂ → i ⁢ A ∈ ℂ
3 1 2 mpan ⊢ A ∈ ℂ → i ⁢ A ∈ ℂ
4 ax-1cn ⊢ 1 ∈ ℂ
5 sqcl ⊢ A ∈ ℂ → A 2 ∈ ℂ
6 subcl ⊢ 1 ∈ ℂ ∧ A 2 ∈ ℂ → 1 − A 2 ∈ ℂ
7 4 5 6 sylancr ⊢ A ∈ ℂ → 1 − A 2 ∈ ℂ
8 7 sqrtcld ⊢ A ∈ ℂ → 1 − A 2 ∈ ℂ
9 3 8 addcomd ⊢ A ∈ ℂ → i ⁢ A + 1 − A 2 = 1 − A 2 + i ⁢ A
10 mulneg2 ⊢ i ∈ ℂ ∧ A ∈ ℂ → i ⁢ − A = − i ⁢ A
11 1 10 mpan ⊢ A ∈ ℂ → i ⁢ − A = − i ⁢ A
12 sqneg ⊢ A ∈ ℂ → − A 2 = A 2
13 12 oveq2d ⊢ A ∈ ℂ → 1 − − A 2 = 1 − A 2
14 13 fveq2d ⊢ A ∈ ℂ → 1 − − A 2 = 1 − A 2
15 11 14 oveq12d ⊢ A ∈ ℂ → i ⁢ − A + 1 − − A 2 = - i ⁢ A + 1 − A 2
16 3 negcld ⊢ A ∈ ℂ → − i ⁢ A ∈ ℂ
17 16 8 addcomd ⊢ A ∈ ℂ → - i ⁢ A + 1 − A 2 = 1 − A 2 + − i ⁢ A
18 8 3 negsubd ⊢ A ∈ ℂ → 1 − A 2 + − i ⁢ A = 1 − A 2 − i ⁢ A
19 15 17 18 3eqtrd ⊢ A ∈ ℂ → i ⁢ − A + 1 − − A 2 = 1 − A 2 − i ⁢ A
20 9 19 oveq12d ⊢ A ∈ ℂ → i ⁢ A + 1 − A 2 ⁢ i ⁢ − A + 1 − − A 2 = 1 − A 2 + i ⁢ A ⁢ 1 − A 2 − i ⁢ A
21 7 sqsqrtd ⊢ A ∈ ℂ → 1 − A 2 2 = 1 − A 2
22 sqmul ⊢ i ∈ ℂ ∧ A ∈ ℂ → i ⁢ A 2 = i 2 ⁢ A 2
23 1 22 mpan ⊢ A ∈ ℂ → i ⁢ A 2 = i 2 ⁢ A 2
24 i2 ⊢ i 2 = − 1
25 24 oveq1i ⊢ i 2 ⁢ A 2 = -1 ⁢ A 2
26 5 mulm1d ⊢ A ∈ ℂ → -1 ⁢ A 2 = − A 2
27 25 26 eqtrid ⊢ A ∈ ℂ → i 2 ⁢ A 2 = − A 2
28 23 27 eqtrd ⊢ A ∈ ℂ → i ⁢ A 2 = − A 2
29 21 28 oveq12d ⊢ A ∈ ℂ → 1 − A 2 2 − i ⁢ A 2 = 1 - A 2 - − A 2
30 subsq ⊢ 1 − A 2 ∈ ℂ ∧ i ⁢ A ∈ ℂ → 1 − A 2 2 − i ⁢ A 2 = 1 − A 2 + i ⁢ A ⁢ 1 − A 2 − i ⁢ A
31 8 3 30 syl2anc ⊢ A ∈ ℂ → 1 − A 2 2 − i ⁢ A 2 = 1 − A 2 + i ⁢ A ⁢ 1 − A 2 − i ⁢ A
32 7 5 subnegd ⊢ A ∈ ℂ → 1 - A 2 - − A 2 = 1 - A 2 + A 2
33 29 31 32 3eqtr3d ⊢ A ∈ ℂ → 1 − A 2 + i ⁢ A ⁢ 1 − A 2 − i ⁢ A = 1 - A 2 + A 2
34 npcan ⊢ 1 ∈ ℂ ∧ A 2 ∈ ℂ → 1 - A 2 + A 2 = 1
35 4 5 34 sylancr ⊢ A ∈ ℂ → 1 - A 2 + A 2 = 1
36 20 33 35 3eqtrd ⊢ A ∈ ℂ → i ⁢ A + 1 − A 2 ⁢ i ⁢ − A + 1 − − A 2 = 1