Metamath Proof Explorer


Theorem circum

Description: The circumference of a circle of radius R , defined as the limit as n ~> +oo of the perimeter of an inscribed n-sided isogons, is ( ( 2 x. _pi ) x. R ) . (Contributed by Paul Chapman, 10-Nov-2012) (Proof shortened by Mario Carneiro, 21-May-2014)

Ref Expression
Hypotheses circum.1 ⊢ A = 2 ⁢ π n
circum.2 ⊢ P = n ∈ ℕ ⟼ 2 ⁢ n ⁢ R ⁢ sin ⁡ A 2
circum.3 ⊢ R ∈ ℝ
Assertion circum ⊢ P ⇝ 2 ⁢ π ⁢ R

Proof

Step Hyp Ref Expression
1 circum.1 ⊢ A = 2 ⁢ π n
2 circum.2 ⊢ P = n ∈ ℕ ⟼ 2 ⁢ n ⁢ R ⁢ sin ⁡ A 2
3 circum.3 ⊢ R ∈ ℝ
4 nnuz ⊢ ℕ = ℤ ≥ 1
5 1zzd ⊢ ⊤ → 1 ∈ ℤ
6 pirp ⊢ π ∈ ℝ +
7 nnrp ⊢ n ∈ ℕ → n ∈ ℝ +
8 rpdivcl ⊢ π ∈ ℝ + ∧ n ∈ ℝ + → π n ∈ ℝ +
9 6 7 8 sylancr ⊢ n ∈ ℕ → π n ∈ ℝ +
10 9 rprene0d ⊢ n ∈ ℕ → π n ∈ ℝ ∧ π n ≠ 0
11 eldifsn ⊢ π n ∈ ℝ ∖ 0 ↔ π n ∈ ℝ ∧ π n ≠ 0
12 10 11 sylibr ⊢ n ∈ ℕ → π n ∈ ℝ ∖ 0
13 12 adantl ⊢ ⊤ ∧ n ∈ ℕ → π n ∈ ℝ ∖ 0
14 eqidd ⊢ ⊤ → n ∈ ℕ ⟼ π n = n ∈ ℕ ⟼ π n
15 eqidd ⊢ ⊤ → y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y = y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y
16 fveq2 ⊢ y = π n → sin ⁡ y = sin ⁡ π n
17 id ⊢ y = π n → y = π n
18 16 17 oveq12d ⊢ y = π n → sin ⁡ y y = sin ⁡ π n π n
19 13 14 15 18 fmptco ⊢ ⊤ → y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y ∘ n ∈ ℕ ⟼ π n = n ∈ ℕ ⟼ sin ⁡ π n π n
20 eqid ⊢ n ∈ ℕ ⟼ π n = n ∈ ℕ ⟼ π n
21 20 12 fmpti ⊢ n ∈ ℕ ⟼ π n : ℕ ⟶ ℝ ∖ 0
22 pire ⊢ π ∈ ℝ
23 22 recni ⊢ π ∈ ℂ
24 divcnv ⊢ π ∈ ℂ → n ∈ ℕ ⟼ π n ⇝ 0
25 23 24 mp1i ⊢ ⊤ → n ∈ ℕ ⟼ π n ⇝ 0
26 sinccvg ⊢ n ∈ ℕ ⟼ π n : ℕ ⟶ ℝ ∖ 0 ∧ n ∈ ℕ ⟼ π n ⇝ 0 → y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y ∘ n ∈ ℕ ⟼ π n ⇝ 1
27 21 25 26 sylancr ⊢ ⊤ → y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y ∘ n ∈ ℕ ⟼ π n ⇝ 1
28 19 27 eqbrtrrd ⊢ ⊤ → n ∈ ℕ ⟼ sin ⁡ π n π n ⇝ 1
29 2re ⊢ 2 ∈ ℝ
30 29 22 remulcli ⊢ 2 ⁢ π ∈ ℝ
31 30 3 remulcli ⊢ 2 ⁢ π ⁢ R ∈ ℝ
32 31 recni ⊢ 2 ⁢ π ⁢ R ∈ ℂ
33 32 a1i ⊢ ⊤ → 2 ⁢ π ⁢ R ∈ ℂ
34 nnex ⊢ ℕ ∈ V
35 34 mptex ⊢ n ∈ ℕ ⟼ 2 ⁢ n ⁢ R ⁢ sin ⁡ A 2 ∈ V
36 2 35 eqeltri ⊢ P ∈ V
37 36 a1i ⊢ ⊤ → P ∈ V
38 eqid ⊢ y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y = y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y
39 eldifi ⊢ y ∈ ℝ ∖ 0 → y ∈ ℝ
40 39 resincld ⊢ y ∈ ℝ ∖ 0 → sin ⁡ y ∈ ℝ
41 eldifsni ⊢ y ∈ ℝ ∖ 0 → y ≠ 0
42 40 39 41 redivcld ⊢ y ∈ ℝ ∖ 0 → sin ⁡ y y ∈ ℝ
43 38 42 fmpti ⊢ y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y : ℝ ∖ 0 ⟶ ℝ
44 fco ⊢ y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y : ℝ ∖ 0 ⟶ ℝ ∧ n ∈ ℕ ⟼ π n : ℕ ⟶ ℝ ∖ 0 → y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y ∘ n ∈ ℕ ⟼ π n : ℕ ⟶ ℝ
45 43 21 44 mp2an ⊢ y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y ∘ n ∈ ℕ ⟼ π n : ℕ ⟶ ℝ
46 19 mptru ⊢ y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y ∘ n ∈ ℕ ⟼ π n = n ∈ ℕ ⟼ sin ⁡ π n π n
47 46 feq1i ⊢ y ∈ ℝ ∖ 0 ⟼ sin ⁡ y y ∘ n ∈ ℕ ⟼ π n : ℕ ⟶ ℝ ↔ n ∈ ℕ ⟼ sin ⁡ π n π n : ℕ ⟶ ℝ
48 45 47 mpbi ⊢ n ∈ ℕ ⟼ sin ⁡ π n π n : ℕ ⟶ ℝ
49 48 ffvelcdmi ⊢ k ∈ ℕ → n ∈ ℕ ⟼ sin ⁡ π n π n ⁡ k ∈ ℝ
50 49 adantl ⊢ ⊤ ∧ k ∈ ℕ → n ∈ ℕ ⟼ sin ⁡ π n π n ⁡ k ∈ ℝ
51 50 recnd ⊢ ⊤ ∧ k ∈ ℕ → n ∈ ℕ ⟼ sin ⁡ π n π n ⁡ k ∈ ℂ
52 29 recni ⊢ 2 ∈ ℂ
53 52 a1i ⊢ ⊤ ∧ k ∈ ℕ → 2 ∈ ℂ
54 23 a1i ⊢ ⊤ ∧ k ∈ ℕ → π ∈ ℂ
55 nncn ⊢ k ∈ ℕ → k ∈ ℂ
56 55 adantl ⊢ ⊤ ∧ k ∈ ℕ → k ∈ ℂ
57 nnne0 ⊢ k ∈ ℕ → k ≠ 0
58 57 adantl ⊢ ⊤ ∧ k ∈ ℕ → k ≠ 0
59 53 54 56 58 divassd ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ π k = 2 ⁢ π k
60 59 oveq1d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ π k 2 = 2 ⁢ π k 2
61 simpr ⊢ ⊤ ∧ k ∈ ℕ → k ∈ ℕ
62 nndivre ⊢ π ∈ ℝ ∧ k ∈ ℕ → π k ∈ ℝ
63 22 61 62 sylancr ⊢ ⊤ ∧ k ∈ ℕ → π k ∈ ℝ
64 63 recnd ⊢ ⊤ ∧ k ∈ ℕ → π k ∈ ℂ
65 2ne0 ⊢ 2 ≠ 0
66 65 a1i ⊢ ⊤ ∧ k ∈ ℕ → 2 ≠ 0
67 64 53 66 divcan3d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ π k 2 = π k
68 60 67 eqtrd ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ π k 2 = π k
69 68 fveq2d ⊢ ⊤ ∧ k ∈ ℕ → sin ⁡ 2 ⁢ π k 2 = sin ⁡ π k
70 63 resincld ⊢ ⊤ ∧ k ∈ ℕ → sin ⁡ π k ∈ ℝ
71 70 recnd ⊢ ⊤ ∧ k ∈ ℕ → sin ⁡ π k ∈ ℂ
72 nnrp ⊢ k ∈ ℕ → k ∈ ℝ +
73 72 adantl ⊢ ⊤ ∧ k ∈ ℕ → k ∈ ℝ +
74 rpdivcl ⊢ π ∈ ℝ + ∧ k ∈ ℝ + → π k ∈ ℝ +
75 6 73 74 sylancr ⊢ ⊤ ∧ k ∈ ℕ → π k ∈ ℝ +
76 75 rpne0d ⊢ ⊤ ∧ k ∈ ℕ → π k ≠ 0
77 71 64 76 divcan2d ⊢ ⊤ ∧ k ∈ ℕ → π k ⁢ sin ⁡ π k π k = sin ⁡ π k
78 69 77 eqtr4d ⊢ ⊤ ∧ k ∈ ℕ → sin ⁡ 2 ⁢ π k 2 = π k ⁢ sin ⁡ π k π k
79 78 oveq2d ⊢ ⊤ ∧ k ∈ ℕ → R ⁢ sin ⁡ 2 ⁢ π k 2 = R ⁢ π k ⁢ sin ⁡ π k π k
80 3 recni ⊢ R ∈ ℂ
81 80 a1i ⊢ ⊤ ∧ k ∈ ℕ → R ∈ ℂ
82 oveq2 ⊢ n = k → π n = π k
83 82 fveq2d ⊢ n = k → sin ⁡ π n = sin ⁡ π k
84 83 82 oveq12d ⊢ n = k → sin ⁡ π n π n = sin ⁡ π k π k
85 eqid ⊢ n ∈ ℕ ⟼ sin ⁡ π n π n = n ∈ ℕ ⟼ sin ⁡ π n π n
86 ovex ⊢ sin ⁡ π k π k ∈ V
87 84 85 86 fvmpt ⊢ k ∈ ℕ → n ∈ ℕ ⟼ sin ⁡ π n π n ⁡ k = sin ⁡ π k π k
88 87 adantl ⊢ ⊤ ∧ k ∈ ℕ → n ∈ ℕ ⟼ sin ⁡ π n π n ⁡ k = sin ⁡ π k π k
89 88 51 eqeltrrd ⊢ ⊤ ∧ k ∈ ℕ → sin ⁡ π k π k ∈ ℂ
90 81 64 89 mulassd ⊢ ⊤ ∧ k ∈ ℕ → R ⁢ π k ⁢ sin ⁡ π k π k = R ⁢ π k ⁢ sin ⁡ π k π k
91 79 90 eqtr4d ⊢ ⊤ ∧ k ∈ ℕ → R ⁢ sin ⁡ 2 ⁢ π k 2 = R ⁢ π k ⁢ sin ⁡ π k π k
92 91 oveq2d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ k ⁢ R ⁢ sin ⁡ 2 ⁢ π k 2 = 2 ⁢ k ⁢ R ⁢ π k ⁢ sin ⁡ π k π k
93 mulcl ⊢ 2 ∈ ℂ ∧ k ∈ ℂ → 2 ⁢ k ∈ ℂ
94 52 56 93 sylancr ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ k ∈ ℂ
95 mulcl ⊢ R ∈ ℂ ∧ π k ∈ ℂ → R ⁢ π k ∈ ℂ
96 80 64 95 sylancr ⊢ ⊤ ∧ k ∈ ℕ → R ⁢ π k ∈ ℂ
97 94 96 89 mulassd ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ k ⁢ R ⁢ π k ⁢ sin ⁡ π k π k = 2 ⁢ k ⁢ R ⁢ π k ⁢ sin ⁡ π k π k
98 53 56 81 64 mul4d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ k ⁢ R ⁢ π k = 2 ⁢ R ⁢ k ⁢ π k
99 54 56 58 divcan2d ⊢ ⊤ ∧ k ∈ ℕ → k ⁢ π k = π
100 99 oveq2d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ R ⁢ k ⁢ π k = 2 ⁢ R ⁢ π
101 53 81 54 mul32d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ R ⁢ π = 2 ⁢ π ⁢ R
102 100 101 eqtrd ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ R ⁢ k ⁢ π k = 2 ⁢ π ⁢ R
103 98 102 eqtrd ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ k ⁢ R ⁢ π k = 2 ⁢ π ⁢ R
104 103 oveq1d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ k ⁢ R ⁢ π k ⁢ sin ⁡ π k π k = 2 ⁢ π ⁢ R ⁢ sin ⁡ π k π k
105 92 97 104 3eqtr2d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ k ⁢ R ⁢ sin ⁡ 2 ⁢ π k 2 = 2 ⁢ π ⁢ R ⁢ sin ⁡ π k π k
106 oveq2 ⊢ n = k → 2 ⁢ n = 2 ⁢ k
107 oveq2 ⊢ n = k → 2 ⁢ π n = 2 ⁢ π k
108 1 107 eqtrid ⊢ n = k → A = 2 ⁢ π k
109 108 oveq1d ⊢ n = k → A 2 = 2 ⁢ π k 2
110 109 fveq2d ⊢ n = k → sin ⁡ A 2 = sin ⁡ 2 ⁢ π k 2
111 110 oveq2d ⊢ n = k → R ⁢ sin ⁡ A 2 = R ⁢ sin ⁡ 2 ⁢ π k 2
112 106 111 oveq12d ⊢ n = k → 2 ⁢ n ⁢ R ⁢ sin ⁡ A 2 = 2 ⁢ k ⁢ R ⁢ sin ⁡ 2 ⁢ π k 2
113 ovex ⊢ 2 ⁢ k ⁢ R ⁢ sin ⁡ 2 ⁢ π k 2 ∈ V
114 112 2 113 fvmpt ⊢ k ∈ ℕ → P ⁡ k = 2 ⁢ k ⁢ R ⁢ sin ⁡ 2 ⁢ π k 2
115 114 adantl ⊢ ⊤ ∧ k ∈ ℕ → P ⁡ k = 2 ⁢ k ⁢ R ⁢ sin ⁡ 2 ⁢ π k 2
116 88 oveq2d ⊢ ⊤ ∧ k ∈ ℕ → 2 ⁢ π ⁢ R ⁢ n ∈ ℕ ⟼ sin ⁡ π n π n ⁡ k = 2 ⁢ π ⁢ R ⁢ sin ⁡ π k π k
117 105 115 116 3eqtr4d ⊢ ⊤ ∧ k ∈ ℕ → P ⁡ k = 2 ⁢ π ⁢ R ⁢ n ∈ ℕ ⟼ sin ⁡ π n π n ⁡ k
118 4 5 28 33 37 51 117 climmulc2 ⊢ ⊤ → P ⇝ 2 ⁢ π ⁢ R ⋅ 1
119 118 mptru ⊢ P ⇝ 2 ⁢ π ⁢ R ⋅ 1
120 32 mulridi ⊢ 2 ⁢ π ⁢ R ⋅ 1 = 2 ⁢ π ⁢ R
121 119 120 breqtri ⊢ P ⇝ 2 ⁢ π ⁢ R