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
2 2re 2
3 pire π
4 2 3 remulcli 2 π
5 iccssre 0 2 π 0 2 π
6 1 4 5 mp2an 0 2 π
7 6 sseli A 0 2 π A
8 7 rehalfcld A 0 2 π A 2
9 8 resincld A 0 2 π sin A 2
10 1re 1
11 recoscl A cos A
12 resubcl 1 cos A 1 cos A
13 10 11 12 sylancr A 1 cos A
14 13 rehalfcld A 1 cos A 2
15 cosbnd A 1 cos A cos A 1
16 15 simprd A cos A 1
17 subge0 1 cos A 0 1 cos A cos A 1
18 10 11 17 sylancr A 0 1 cos A cos A 1
19 halfnneg2 1 cos A 0 1 cos A 0 1 cos A 2
20 13 19 syl A 0 1 cos A 0 1 cos A 2
21 18 20 bitr3d A cos A 1 0 1 cos A 2
22 16 21 mpbid A 0 1 cos A 2
23 14 22 resqrtcld A 1 cos A 2
24 7 23 syl A 0 2 π 1 cos A 2
25 1 4 elicc2i A 0 2 π A 0 A A 2 π
26 halfnneg2 A 0 A 0 A 2
27 2pos 0 < 2
28 2 27 pm3.2i 2 0 < 2
29 ledivmul A π 2 0 < 2 A 2 π A 2 π
30 3 28 29 mp3an23 A A 2 π A 2 π
31 30 bicomd A A 2 π A 2 π
32 26 31 anbi12d A 0 A A 2 π 0 A 2 A 2 π
33 rehalfcl A A 2
34 33 rexrd A A 2 *
35 0xr 0 *
36 3 rexri π *
37 elicc4 0 * π * A 2 * A 2 0 π 0 A 2 A 2 π
38 35 36 37 mp3an12 A 2 * A 2 0 π 0 A 2 A 2 π
39 34 38 syl A A 2 0 π 0 A 2 A 2 π
40 32 39 bitr4d A 0 A A 2 π A 2 0 π
41 40 biimpd A 0 A A 2 π A 2 0 π
42 41 3impib A 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 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
49 ax-1cn 1
50 coscl A cos A
51 subcl 1 cos A 1 cos A
52 49 50 51 sylancr A 1 cos A
53 52 halfcld A 1 cos A 2
54 53 sqsqrtd A 1 cos A 2 2 = 1 cos A 2
55 halfcl A A 2
56 coscl A 2 cos A 2
57 56 sqcld A 2 cos A 2 2
58 2cn 2
59 mulcom cos A 2 2 2 cos A 2 2 2 = 2 cos A 2 2
60 57 58 59 sylancl A 2 cos A 2 2 2 = 2 cos A 2 2
61 60 oveq2d A 2 1 2 cos A 2 2 2 = 1 2 2 cos A 2 2
62 58 mulid2i 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 syl6eq A 2 1 2 cos A 2 2 2 = 1 + 1 - 2 cos A 2 2
67 subdir 1 cos A 2 2 2 1 cos A 2 2 2 = 1 2 cos A 2 2 2
68 49 58 67 mp3an13 cos A 2 2 1 cos A 2 2 2 = 1 2 cos A 2 2 2
69 57 68 syl A 2 1 cos A 2 2 2 = 1 2 cos A 2 2 2
70 mulcl 2 cos A 2 2 2 cos A 2 2
71 58 57 70 sylancr A 2 2 cos A 2 2
72 subsub3 1 2 cos A 2 2 1 1 2 cos A 2 2 1 = 1 + 1 - 2 cos A 2 2
73 49 49 72 mp3an13 2 cos A 2 2 1 2 cos A 2 2 1 = 1 + 1 - 2 cos A 2 2
74 71 73 syl A 2 1 2 cos A 2 2 1 = 1 + 1 - 2 cos A 2 2
75 66 69 74 3eqtr4d A 2 1 cos A 2 2 2 = 1 2 cos A 2 2 1
76 sincl A 2 sin A 2
77 76 sqcld A 2 sin A 2 2
78 77 57 pncand A 2 sin A 2 2 + cos A 2 2 - cos A 2 2 = sin A 2 2
79 sincossq A 2 sin A 2 2 + cos A 2 2 = 1
80 79 oveq1d A 2 sin A 2 2 + cos A 2 2 - cos A 2 2 = 1 cos A 2 2
81 78 80 eqtr3d A 2 sin A 2 2 = 1 cos A 2 2
82 81 oveq1d A 2 sin A 2 2 2 = 1 cos A 2 2 2
83 cos2t A 2 cos 2 A 2 = 2 cos A 2 2 1
84 83 oveq2d A 2 1 cos 2 A 2 = 1 2 cos A 2 2 1
85 75 82 84 3eqtr4d A 2 sin A 2 2 2 = 1 cos 2 A 2
86 55 85 syl A sin A 2 2 2 = 1 cos 2 A 2
87 2ne0 2 0
88 divcan2 A 2 2 0 2 A 2 = A
89 58 87 88 mp3an23 A 2 A 2 = A
90 89 fveq2d A cos 2 A 2 = cos A
91 90 oveq2d A 1 cos 2 A 2 = 1 cos A
92 86 91 eqtrd A sin A 2 2 2 = 1 cos A
93 92 oveq1d A sin A 2 2 2 2 = 1 cos A 2
94 55 sincld A sin A 2
95 94 sqcld A sin A 2 2
96 divcan4 sin A 2 2 2 2 0 sin A 2 2 2 2 = sin A 2 2
97 58 87 96 mp3an23 sin A 2 2 sin A 2 2 2 2 = sin A 2 2
98 95 97 syl A sin A 2 2 2 2 = sin A 2 2
99 54 93 98 3eqtr2rd A 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