Metamath Proof Explorer


Theorem goldrasin

Description: Alternative trigonometric formula for the golden ratio. (Contributed by Ender Ting, 15-Mar-2026)

Ref Expression
Hypothesis goldra.val ⊢ F = 2 ⁢ cos ⁡ π 5
Assertion goldrasin ⊢ F = 2 ⁢ sin ⁡ π ⁢ 3 10

Proof

Step Hyp Ref Expression
1 goldra.val ⊢ F = 2 ⁢ cos ⁡ π 5
2 picn ⊢ π ∈ ℂ
3 5cn ⊢ 5 ∈ ℂ
4 5re ⊢ 5 ∈ ℝ
5 5pos ⊢ 0 < 5
6 4 5 gt0ne0ii ⊢ 5 ≠ 0
7 2 3 6 divreci ⊢ π 5 = π ⁢ 1 5
8 2cn ⊢ 2 ∈ ℂ
9 2ne0 ⊢ 2 ≠ 0
10 8 3 9 6 subreci ⊢ 1 2 − 1 5 = 5 − 2 2 ⋅ 5
11 3cn ⊢ 3 ∈ ℂ
12 3p2e5 ⊢ 3 + 2 = 5
13 12 eqcomi ⊢ 5 = 3 + 2
14 11 8 13 mvrraddi ⊢ 5 − 2 = 3
15 5t2e10 ⊢ 5 ⋅ 2 = 10
16 3 8 15 mulcomli ⊢ 2 ⋅ 5 = 10
17 14 16 oveq12i ⊢ 5 − 2 2 ⋅ 5 = 3 10
18 10 17 eqtri ⊢ 1 2 − 1 5 = 3 10
19 18 eqcomi ⊢ 3 10 = 1 2 − 1 5
20 10re ⊢ 10 ∈ ℝ
21 20 recni ⊢ 10 ∈ ℂ
22 10pos ⊢ 0 < 10
23 20 22 gt0ne0ii ⊢ 10 ≠ 0
24 11 21 23 divcli ⊢ 3 10 ∈ ℂ
25 24 a1i ⊢ ⊤ → 3 10 ∈ ℂ
26 3 6 reccli ⊢ 1 5 ∈ ℂ
27 26 a1i ⊢ ⊤ → 1 5 ∈ ℂ
28 halfcn ⊢ 1 2 ∈ ℂ
29 28 a1i ⊢ ⊤ → 1 2 ∈ ℂ
30 25 27 29 subexsub ⊢ ⊤ → 3 10 = 1 2 − 1 5 ↔ 1 5 = 1 2 − 3 10
31 30 mptru ⊢ 3 10 = 1 2 − 1 5 ↔ 1 5 = 1 2 − 3 10
32 19 31 mpbi ⊢ 1 5 = 1 2 − 3 10
33 32 oveq2i ⊢ π ⁢ 1 5 = π ⁢ 1 2 − 3 10
34 2 28 24 subdii ⊢ π ⁢ 1 2 − 3 10 = π ⁢ 1 2 − π ⁢ 3 10
35 33 34 eqtri ⊢ π ⁢ 1 5 = π ⁢ 1 2 − π ⁢ 3 10
36 7 35 eqtri ⊢ π 5 = π ⁢ 1 2 − π ⁢ 3 10
37 2 8 9 divreci ⊢ π 2 = π ⁢ 1 2
38 37 eqcomi ⊢ π ⁢ 1 2 = π 2
39 38 oveq1i ⊢ π ⁢ 1 2 − π ⁢ 3 10 = π 2 − π ⁢ 3 10
40 36 39 eqtri ⊢ π 5 = π 2 − π ⁢ 3 10
41 40 fveq2i ⊢ cos ⁡ π 5 = cos ⁡ π 2 − π ⁢ 3 10
42 2 24 mulcli ⊢ π ⁢ 3 10 ∈ ℂ
43 coshalfpim ⊢ π ⁢ 3 10 ∈ ℂ → cos ⁡ π 2 − π ⁢ 3 10 = sin ⁡ π ⁢ 3 10
44 42 43 ax-mp ⊢ cos ⁡ π 2 − π ⁢ 3 10 = sin ⁡ π ⁢ 3 10
45 41 44 eqtri ⊢ cos ⁡ π 5 = sin ⁡ π ⁢ 3 10
46 45 oveq2i ⊢ 2 ⁢ cos ⁡ π 5 = 2 ⁢ sin ⁡ π ⁢ 3 10
47 1 46 eqtri ⊢ F = 2 ⁢ sin ⁡ π ⁢ 3 10