Metamath Proof Explorer


Theorem sin2h

Description: Half-angle rule for sine. (Contributed by Brendan Leahy, 3-Aug-2018)

Ref Expression
Assertion sin2h ⊢ A ∈ 0 2 ⁢ π → sin ⁡ A 2 = 1 − cos ⁡ A 2

Proof

Step Hyp Ref Expression
1 0re ⊢ 0 ∈ ℝ
2 2re ⊢ 2 ∈ ℝ
3 pire ⊢ π ∈ ℝ
4 2 3 remulcli ⊢ 2 ⁢ π ∈ ℝ
5 iccssre ⊢ 0 ∈ ℝ ∧ 2 ⁢ π ∈ ℝ → 0 2 ⁢ π ⊆ ℝ
6 1 4 5 mp2an ⊢ 0 2 ⁢ π ⊆ ℝ
7 6 sseli ⊢ A ∈ 0 2 ⁢ π → A ∈ ℝ
8 7 rehalfcld ⊢ A ∈ 0 2 ⁢ π → A 2 ∈ ℝ
9 8 resincld ⊢ A ∈ 0 2 ⁢ π → sin ⁡ A 2 ∈ ℝ
10 1re ⊢ 1 ∈ ℝ
11 recoscl ⊢ A ∈ ℝ → cos ⁡ A ∈ ℝ
12 resubcl ⊢ 1 ∈ ℝ ∧ cos ⁡ A ∈ ℝ → 1 − cos ⁡ A ∈ ℝ
13 10 11 12 sylancr ⊢ A ∈ ℝ → 1 − cos ⁡ A ∈ ℝ
14 13 rehalfcld ⊢ A ∈ ℝ → 1 − cos ⁡ A 2 ∈ ℝ
15 cosbnd ⊢ A ∈ ℝ → − 1 ≤ cos ⁡ A ∧ cos ⁡ A ≤ 1
16 15 simprd ⊢ A ∈ ℝ → cos ⁡ A ≤ 1
17 subge0 ⊢ 1 ∈ ℝ ∧ cos ⁡ A ∈ ℝ → 0 ≤ 1 − cos ⁡ A ↔ cos ⁡ A ≤ 1
18 10 11 17 sylancr ⊢ A ∈ ℝ → 0 ≤ 1 − cos ⁡ A ↔ cos ⁡ A ≤ 1
19 halfnneg2 ⊢ 1 − cos ⁡ A ∈ ℝ → 0 ≤ 1 − cos ⁡ A ↔ 0 ≤ 1 − cos ⁡ A 2
20 13 19 syl ⊢ A ∈ ℝ → 0 ≤ 1 − cos ⁡ A ↔ 0 ≤ 1 − cos ⁡ A 2
21 18 20 bitr3d ⊢ A ∈ ℝ → cos ⁡ A ≤ 1 ↔ 0 ≤ 1 − cos ⁡ A 2
22 16 21 mpbid ⊢ A ∈ ℝ → 0 ≤ 1 − cos ⁡ A 2
23 14 22 resqrtcld ⊢ A ∈ ℝ → 1 − cos ⁡ A 2 ∈ ℝ
24 7 23 syl ⊢ A ∈ 0 2 ⁢ π → 1 − cos ⁡ A 2 ∈ ℝ
25 1 4 elicc2i ⊢ A ∈ 0 2 ⁢ π ↔ A ∈ ℝ ∧ 0 ≤ A ∧ A ≤ 2 ⁢ π
26 halfnneg2 ⊢ A ∈ ℝ → 0 ≤ A ↔ 0 ≤ A 2
27 2pos ⊢ 0 < 2
28 2 27 pm3.2i ⊢ 2 ∈ ℝ ∧ 0 < 2
29 ledivmul ⊢ A ∈ ℝ ∧ π ∈ ℝ ∧ 2 ∈ ℝ ∧ 0 < 2 → A 2 ≤ π ↔ A ≤ 2 ⁢ π
30 3 28 29 mp3an23 ⊢ A ∈ ℝ → A 2 ≤ π ↔ A ≤ 2 ⁢ π
31 30 bicomd ⊢ A ∈ ℝ → A ≤ 2 ⁢ π ↔ A 2 ≤ π
32 26 31 anbi12d ⊢ A ∈ ℝ → 0 ≤ A ∧ A ≤ 2 ⁢ π ↔ 0 ≤ A 2 ∧ A 2 ≤ π
33 rehalfcl ⊢ A ∈ ℝ → A 2 ∈ ℝ
34 33 rexrd ⊢ A ∈ ℝ → A 2 ∈ ℝ *
35 0xr ⊢ 0 ∈ ℝ *
36 3 rexri ⊢ π ∈ ℝ *
37 elicc4 ⊢ 0 ∈ ℝ * ∧ π ∈ ℝ * ∧ A 2 ∈ ℝ * → A 2 ∈ 0 π ↔ 0 ≤ A 2 ∧ A 2 ≤ π
38 35 36 37 mp3an12 ⊢ A 2 ∈ ℝ * → A 2 ∈ 0 π ↔ 0 ≤ A 2 ∧ A 2 ≤ π
39 34 38 syl ⊢ A ∈ ℝ → A 2 ∈ 0 π ↔ 0 ≤ A 2 ∧ A 2 ≤ π
40 32 39 bitr4d ⊢ A ∈ ℝ → 0 ≤ A ∧ A ≤ 2 ⁢ π ↔ A 2 ∈ 0 π
41 40 biimpd ⊢ A ∈ ℝ → 0 ≤ A ∧ A ≤ 2 ⁢ π → A 2 ∈ 0 π
42 41 3impib ⊢ A ∈ ℝ ∧ 0 ≤ A ∧ A ≤ 2 ⁢ π → A 2 ∈ 0 π
43 25 42 sylbi ⊢ A ∈ 0 2 ⁢ π → A 2 ∈ 0 π
44 sinq12ge0 ⊢ A 2 ∈ 0 π → 0 ≤ sin ⁡ A 2
45 43 44 syl ⊢ A ∈ 0 2 ⁢ π → 0 ≤ sin ⁡ A 2
46 14 22 sqrtge0d ⊢ A ∈ ℝ → 0 ≤ 1 − cos ⁡ A 2
47 7 46 syl ⊢ A ∈ 0 2 ⁢ π → 0 ≤ 1 − cos ⁡ A 2
48 7 recnd ⊢ A ∈ 0 2 ⁢ π → A ∈ ℂ
49 ax-1cn ⊢ 1 ∈ ℂ
50 coscl ⊢ A ∈ ℂ → cos ⁡ A ∈ ℂ
51 subcl ⊢ 1 ∈ ℂ ∧ cos ⁡ A ∈ ℂ → 1 − cos ⁡ A ∈ ℂ
52 49 50 51 sylancr ⊢ A ∈ ℂ → 1 − cos ⁡ A ∈ ℂ
53 52 halfcld ⊢ A ∈ ℂ → 1 − cos ⁡ A 2 ∈ ℂ
54 53 sqsqrtd ⊢ A ∈ ℂ → 1 − cos ⁡ A 2 2 = 1 − cos ⁡ A 2
55 halfcl ⊢ A ∈ ℂ → A 2 ∈ ℂ
56 coscl ⊢ A 2 ∈ ℂ → cos ⁡ A 2 ∈ ℂ
57 56 sqcld ⊢ A 2 ∈ ℂ → cos ⁡ A 2 2 ∈ ℂ
58 2cn ⊢ 2 ∈ ℂ
59 mulcom ⊢ cos ⁡ A 2 2 ∈ ℂ ∧ 2 ∈ ℂ → cos ⁡ A 2 2 ⋅ 2 = 2 ⁢ cos ⁡ A 2 2
60 57 58 59 sylancl ⊢ A 2 ∈ ℂ → cos ⁡ A 2 2 ⋅ 2 = 2 ⁢ cos ⁡ A 2 2
61 60 oveq2d ⊢ A 2 ∈ ℂ → 1 ⋅ 2 − cos ⁡ A 2 2 ⋅ 2 = 1 ⋅ 2 − 2 ⁢ cos ⁡ A 2 2
62 58 mullidi ⊢ 1 ⋅ 2 = 2
63 df-2 ⊢ 2 = 1 + 1
64 62 63 eqtri ⊢ 1 ⋅ 2 = 1 + 1
65 64 oveq1i ⊢ 1 ⋅ 2 − 2 ⁢ cos ⁡ A 2 2 = 1 + 1 - 2 ⁢ cos ⁡ A 2 2
66 61 65 eqtrdi ⊢ A 2 ∈ ℂ → 1 ⋅ 2 − cos ⁡ A 2 2 ⋅ 2 = 1 + 1 - 2 ⁢ cos ⁡ A 2 2
67 subdir ⊢ 1 ∈ ℂ ∧ cos ⁡ A 2 2 ∈ ℂ ∧ 2 ∈ ℂ → 1 − cos ⁡ A 2 2 ⋅ 2 = 1 ⋅ 2 − cos ⁡ A 2 2 ⋅ 2
68 49 58 67 mp3an13 ⊢ cos ⁡ A 2 2 ∈ ℂ → 1 − cos ⁡ A 2 2 ⋅ 2 = 1 ⋅ 2 − cos ⁡ A 2 2 ⋅ 2
69 57 68 syl ⊢ A 2 ∈ ℂ → 1 − cos ⁡ A 2 2 ⋅ 2 = 1 ⋅ 2 − cos ⁡ A 2 2 ⋅ 2
70 mulcl ⊢ 2 ∈ ℂ ∧ cos ⁡ A 2 2 ∈ ℂ → 2 ⁢ cos ⁡ A 2 2 ∈ ℂ
71 58 57 70 sylancr ⊢ A 2 ∈ ℂ → 2 ⁢ cos ⁡ A 2 2 ∈ ℂ
72 subsub3 ⊢ 1 ∈ ℂ ∧ 2 ⁢ cos ⁡ A 2 2 ∈ ℂ ∧ 1 ∈ ℂ → 1 − 2 ⁢ cos ⁡ A 2 2 − 1 = 1 + 1 - 2 ⁢ cos ⁡ A 2 2
73 49 49 72 mp3an13 ⊢ 2 ⁢ cos ⁡ A 2 2 ∈ ℂ → 1 − 2 ⁢ cos ⁡ A 2 2 − 1 = 1 + 1 - 2 ⁢ cos ⁡ A 2 2
74 71 73 syl ⊢ A 2 ∈ ℂ → 1 − 2 ⁢ cos ⁡ A 2 2 − 1 = 1 + 1 - 2 ⁢ cos ⁡ A 2 2
75 66 69 74 3eqtr4d ⊢ A 2 ∈ ℂ → 1 − cos ⁡ A 2 2 ⋅ 2 = 1 − 2 ⁢ cos ⁡ A 2 2 − 1
76 sincl ⊢ A 2 ∈ ℂ → sin ⁡ A 2 ∈ ℂ
77 76 sqcld ⊢ A 2 ∈ ℂ → sin ⁡ A 2 2 ∈ ℂ
78 77 57 pncand ⊢ A 2 ∈ ℂ → sin ⁡ A 2 2 + cos ⁡ A 2 2 - cos ⁡ A 2 2 = sin ⁡ A 2 2
79 sincossq ⊢ A 2 ∈ ℂ → sin ⁡ A 2 2 + cos ⁡ A 2 2 = 1
80 79 oveq1d ⊢ A 2 ∈ ℂ → sin ⁡ A 2 2 + cos ⁡ A 2 2 - cos ⁡ A 2 2 = 1 − cos ⁡ A 2 2
81 78 80 eqtr3d ⊢ A 2 ∈ ℂ → sin ⁡ A 2 2 = 1 − cos ⁡ A 2 2
82 81 oveq1d ⊢ A 2 ∈ ℂ → sin ⁡ A 2 2 ⋅ 2 = 1 − cos ⁡ A 2 2 ⋅ 2
83 cos2t ⊢ A 2 ∈ ℂ → cos ⁡ 2 ⁢ A 2 = 2 ⁢ cos ⁡ A 2 2 − 1
84 83 oveq2d ⊢ A 2 ∈ ℂ → 1 − cos ⁡ 2 ⁢ A 2 = 1 − 2 ⁢ cos ⁡ A 2 2 − 1
85 75 82 84 3eqtr4d ⊢ A 2 ∈ ℂ → sin ⁡ A 2 2 ⋅ 2 = 1 − cos ⁡ 2 ⁢ A 2
86 55 85 syl ⊢ A ∈ ℂ → sin ⁡ A 2 2 ⋅ 2 = 1 − cos ⁡ 2 ⁢ A 2
87 2ne0 ⊢ 2 ≠ 0
88 divcan2 ⊢ A ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → 2 ⁢ A 2 = A
89 58 87 88 mp3an23 ⊢ A ∈ ℂ → 2 ⁢ A 2 = A
90 89 fveq2d ⊢ A ∈ ℂ → cos ⁡ 2 ⁢ A 2 = cos ⁡ A
91 90 oveq2d ⊢ A ∈ ℂ → 1 − cos ⁡ 2 ⁢ A 2 = 1 − cos ⁡ A
92 86 91 eqtrd ⊢ A ∈ ℂ → sin ⁡ A 2 2 ⋅ 2 = 1 − cos ⁡ A
93 92 oveq1d ⊢ A ∈ ℂ → sin ⁡ A 2 2 ⋅ 2 2 = 1 − cos ⁡ A 2
94 55 sincld ⊢ A ∈ ℂ → sin ⁡ A 2 ∈ ℂ
95 94 sqcld ⊢ A ∈ ℂ → sin ⁡ A 2 2 ∈ ℂ
96 divcan4 ⊢ sin ⁡ A 2 2 ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → sin ⁡ A 2 2 ⋅ 2 2 = sin ⁡ A 2 2
97 58 87 96 mp3an23 ⊢ sin ⁡ A 2 2 ∈ ℂ → sin ⁡ A 2 2 ⋅ 2 2 = sin ⁡ A 2 2
98 95 97 syl ⊢ A ∈ ℂ → sin ⁡ A 2 2 ⋅ 2 2 = sin ⁡ A 2 2
99 54 93 98 3eqtr2rd ⊢ A ∈ ℂ → sin ⁡ A 2 2 = 1 − cos ⁡ A 2 2
100 48 99 syl ⊢ A ∈ 0 2 ⁢ π → sin ⁡ A 2 2 = 1 − cos ⁡ A 2 2
101 9 24 45 47 100 sq11d ⊢ A ∈ 0 2 ⁢ π → sin ⁡ A 2 = 1 − cos ⁡ A 2