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 AiA+1A2iA+1A2=1

Proof

Step Hyp Ref Expression
1 ax-icn i
2 mulcl iAiA
3 1 2 mpan AiA
4 ax-1cn 1
5 sqcl AA2
6 subcl 1A21A2
7 4 5 6 sylancr A1A2
8 7 sqrtcld A1A2
9 3 8 addcomd AiA+1A2=1A2+iA
10 mulneg2 iAiA=iA
11 1 10 mpan AiA=iA
12 sqneg AA2=A2
13 12 oveq2d A1A2=1A2
14 13 fveq2d A1A2=1A2
15 11 14 oveq12d AiA+1A2=-iA+1A2
16 3 negcld AiA
17 16 8 addcomd A-iA+1A2=1A2+iA
18 8 3 negsubd A1A2+iA=1A2iA
19 15 17 18 3eqtrd AiA+1A2=1A2iA
20 9 19 oveq12d AiA+1A2iA+1A2=1A2+iA1A2iA
21 7 sqsqrtd A1A22=1A2
22 sqmul iAiA2=i2A2
23 1 22 mpan AiA2=i2A2
24 i2 i2=1
25 24 oveq1i i2A2=-1A2
26 5 mulm1d A-1A2=A2
27 25 26 eqtrid Ai2A2=A2
28 23 27 eqtrd AiA2=A2
29 21 28 oveq12d A1A22iA2=1-A2-A2
30 subsq 1A2iA1A22iA2=1A2+iA1A2iA
31 8 3 30 syl2anc A1A22iA2=1A2+iA1A2iA
32 7 5 subnegd A1-A2-A2=1-A2+A2
33 29 31 32 3eqtr3d A1A2+iA1A2iA=1-A2+A2
34 npcan 1A21-A2+A2=1
35 4 5 34 sylancr A1-A2+A2=1
36 20 33 35 3eqtrd AiA+1A2iA+1A2=1