Metamath Proof Explorer


Theorem lawcoslem1

Description: Lemma for lawcos . Here we prove the law for a point at the origin and two distinct points U and V, using an expanded version of the signed angle expression on the complex plane. (Contributed by David A. Wheeler, 11-Jun-2015)

Ref Expression
Hypotheses lawcoslem1.1 ⊢ φ → U ∈ ℂ
lawcoslem1.2 ⊢ φ → V ∈ ℂ
lawcoslem1.3 ⊢ φ → U ≠ 0
lawcoslem1.4 ⊢ φ → V ≠ 0
Assertion lawcoslem1 ⊢ φ → U − V 2 = U 2 + V 2 - 2 ⁢ U ⁢ V ⁢ ℜ ⁡ U V U V

Proof

Step Hyp Ref Expression
1 lawcoslem1.1 ⊢ φ → U ∈ ℂ
2 lawcoslem1.2 ⊢ φ → V ∈ ℂ
3 lawcoslem1.3 ⊢ φ → U ≠ 0
4 lawcoslem1.4 ⊢ φ → V ≠ 0
5 sqabssub ⊢ U ∈ ℂ ∧ V ∈ ℂ → U − V 2 = U 2 + V 2 - 2 ⁢ ℜ ⁡ U ⁢ V ‾
6 1 2 5 syl2anc ⊢ φ → U − V 2 = U 2 + V 2 - 2 ⁢ ℜ ⁡ U ⁢ V ‾
7 1 2 4 absdivd ⊢ φ → U V = U V
8 7 oveq2d ⊢ φ → ℜ ⁡ U V U V = ℜ ⁡ U V U V
9 8 oveq2d ⊢ φ → U ⁢ V ⁢ ℜ ⁡ U V U V = U ⁢ V ⁢ ℜ ⁡ U V U V
10 1 abscld ⊢ φ → U ∈ ℝ
11 2 abscld ⊢ φ → V ∈ ℝ
12 10 11 remulcld ⊢ φ → U ⁢ V ∈ ℝ
13 12 recnd ⊢ φ → U ⁢ V ∈ ℂ
14 1 2 4 divcld ⊢ φ → U V ∈ ℂ
15 14 recld ⊢ φ → ℜ ⁡ U V ∈ ℝ
16 15 recnd ⊢ φ → ℜ ⁡ U V ∈ ℂ
17 10 recnd ⊢ φ → U ∈ ℂ
18 11 recnd ⊢ φ → V ∈ ℂ
19 2 4 absne0d ⊢ φ → V ≠ 0
20 17 18 19 divcld ⊢ φ → U V ∈ ℂ
21 1 3 absne0d ⊢ φ → U ≠ 0
22 17 18 21 19 divne0d ⊢ φ → U V ≠ 0
23 13 16 20 22 div12d ⊢ φ → U ⁢ V ⁢ ℜ ⁡ U V U V = ℜ ⁡ U V ⁢ U ⁢ V U V
24 9 23 eqtrd ⊢ φ → U ⁢ V ⁢ ℜ ⁡ U V U V = ℜ ⁡ U V ⁢ U ⁢ V U V
25 13 17 18 21 19 divdiv2d ⊢ φ → U ⁢ V U V = U ⁢ V ⁢ V U
26 18 sqvald ⊢ φ → V 2 = V ⁢ V
27 26 oveq1d ⊢ φ → V 2 ⁢ U = V ⁢ V ⁢ U
28 17 18 18 mul31d ⊢ φ → U ⁢ V ⁢ V = V ⁢ V ⁢ U
29 27 28 eqtr4d ⊢ φ → V 2 ⁢ U = U ⁢ V ⁢ V
30 29 oveq1d ⊢ φ → V 2 ⁢ U U = U ⁢ V ⁢ V U
31 18 sqcld ⊢ φ → V 2 ∈ ℂ
32 31 17 21 divcan4d ⊢ φ → V 2 ⁢ U U = V 2
33 25 30 32 3eqtr2rd ⊢ φ → V 2 = U ⁢ V U V
34 33 oveq2d ⊢ φ → ℜ ⁡ U V ⁢ V 2 = ℜ ⁡ U V ⁢ U ⁢ V U V
35 16 31 mulcomd ⊢ φ → ℜ ⁡ U V ⁢ V 2 = V 2 ⁢ ℜ ⁡ U V
36 11 resqcld ⊢ φ → V 2 ∈ ℝ
37 36 14 remul2d ⊢ φ → ℜ ⁡ V 2 ⁢ U V = V 2 ⁢ ℜ ⁡ U V
38 35 37 eqtr4d ⊢ φ → ℜ ⁡ U V ⁢ V 2 = ℜ ⁡ V 2 ⁢ U V
39 1 31 2 4 div12d ⊢ φ → U ⁢ V 2 V = V 2 ⁢ U V
40 31 2 4 divrecd ⊢ φ → V 2 V = V 2 ⁢ 1 V
41 recval ⊢ V ∈ ℂ ∧ V ≠ 0 → 1 V = V ‾ V 2
42 2 4 41 syl2anc ⊢ φ → 1 V = V ‾ V 2
43 42 oveq2d ⊢ φ → V 2 ⁢ 1 V = V 2 ⁢ V ‾ V 2
44 2 cjcld ⊢ φ → V ‾ ∈ ℂ
45 sqne0 ⊢ V ∈ ℂ → V 2 ≠ 0 ↔ V ≠ 0
46 18 45 syl ⊢ φ → V 2 ≠ 0 ↔ V ≠ 0
47 19 46 mpbird ⊢ φ → V 2 ≠ 0
48 44 31 47 divcan2d ⊢ φ → V 2 ⁢ V ‾ V 2 = V ‾
49 43 48 eqtrd ⊢ φ → V 2 ⁢ 1 V = V ‾
50 40 49 eqtrd ⊢ φ → V 2 V = V ‾
51 50 oveq2d ⊢ φ → U ⁢ V 2 V = U ⁢ V ‾
52 39 51 eqtr3d ⊢ φ → V 2 ⁢ U V = U ⁢ V ‾
53 52 fveq2d ⊢ φ → ℜ ⁡ V 2 ⁢ U V = ℜ ⁡ U ⁢ V ‾
54 38 53 eqtrd ⊢ φ → ℜ ⁡ U V ⁢ V 2 = ℜ ⁡ U ⁢ V ‾
55 24 34 54 3eqtr2rd ⊢ φ → ℜ ⁡ U ⁢ V ‾ = U ⁢ V ⁢ ℜ ⁡ U V U V
56 55 oveq2d ⊢ φ → 2 ⁢ ℜ ⁡ U ⁢ V ‾ = 2 ⁢ U ⁢ V ⁢ ℜ ⁡ U V U V
57 56 oveq2d ⊢ φ → U 2 + V 2 - 2 ⁢ ℜ ⁡ U ⁢ V ‾ = U 2 + V 2 - 2 ⁢ U ⁢ V ⁢ ℜ ⁡ U V U V
58 6 57 eqtrd ⊢ φ → U − V 2 = U 2 + V 2 - 2 ⁢ U ⁢ V ⁢ ℜ ⁡ U V U V