Metamath Proof Explorer


Theorem cos5t

Description: Five-times-angle formula for cosine, in pure cosine form. (Contributed by Ender Ting, 20-Apr-2026)

Ref Expression
Assertion cos5t ⊢ A ∈ ℂ → cos ⁡ 5 ⁢ A = 16 ⁢ cos ⁡ A 5 - 20 ⁢ cos ⁡ A 3 + 5 ⁢ cos ⁡ A

Proof

Step Hyp Ref Expression
1 picn ⊢ π ∈ ℂ
2 2cn ⊢ 2 ∈ ℂ
3 2ne0 ⊢ 2 ≠ 0
4 1 2 3 divcli ⊢ π 2 ∈ ℂ
5 4 a1i ⊢ A ∈ ℂ → π 2 ∈ ℂ
6 5cn ⊢ 5 ∈ ℂ
7 6 a1i ⊢ A ∈ ℂ → 5 ∈ ℂ
8 id ⊢ A ∈ ℂ → A ∈ ℂ
9 7 8 mulcld ⊢ A ∈ ℂ → 5 ⁢ A ∈ ℂ
10 5 9 subcld ⊢ A ∈ ℂ → π 2 − 5 ⁢ A ∈ ℂ
11 1zzd ⊢ A ∈ ℂ → 1 ∈ ℤ
12 sinper ⊢ π 2 − 5 ⁢ A ∈ ℂ ∧ 1 ∈ ℤ → sin ⁡ π 2 - 5 ⁢ A + 1 ⁢ 2 ⁢ π = sin ⁡ π 2 − 5 ⁢ A
13 10 11 12 syl2anc ⊢ A ∈ ℂ → sin ⁡ π 2 - 5 ⁢ A + 1 ⁢ 2 ⁢ π = sin ⁡ π 2 − 5 ⁢ A
14 7 5 8 subdid ⊢ A ∈ ℂ → 5 ⁢ π 2 − A = 5 ⁢ π 2 − 5 ⁢ A
15 4 mullidi ⊢ 1 ⁢ π 2 = π 2
16 15 eqcomi ⊢ π 2 = 1 ⁢ π 2
17 16 a1i ⊢ A ∈ ℂ → π 2 = 1 ⁢ π 2
18 2picn ⊢ 2 ⁢ π ∈ ℂ
19 18 mullidi ⊢ 1 ⁢ 2 ⁢ π = 2 ⁢ π
20 1 2 3 divcan2i ⊢ 2 ⁢ π 2 = π
21 20 eqcomi ⊢ π = 2 ⁢ π 2
22 21 oveq2i ⊢ 2 ⁢ π = 2 ⁢ 2 ⁢ π 2
23 2 2 4 mulassi ⊢ 2 ⋅ 2 ⁢ π 2 = 2 ⁢ 2 ⁢ π 2
24 2t2e4 ⊢ 2 ⋅ 2 = 4
25 24 oveq1i ⊢ 2 ⋅ 2 ⁢ π 2 = 4 ⁢ π 2
26 23 25 eqtr3i ⊢ 2 ⁢ 2 ⁢ π 2 = 4 ⁢ π 2
27 19 22 26 3eqtri ⊢ 1 ⁢ 2 ⁢ π = 4 ⁢ π 2
28 27 a1i ⊢ A ∈ ℂ → 1 ⁢ 2 ⁢ π = 4 ⁢ π 2
29 17 28 oveq12d ⊢ A ∈ ℂ → π 2 + 1 ⁢ 2 ⁢ π = 1 ⁢ π 2 + 4 ⁢ π 2
30 1cnd ⊢ A ∈ ℂ → 1 ∈ ℂ
31 4cn ⊢ 4 ∈ ℂ
32 31 a1i ⊢ A ∈ ℂ → 4 ∈ ℂ
33 30 32 5 adddird ⊢ A ∈ ℂ → 1 + 4 ⁢ π 2 = 1 ⁢ π 2 + 4 ⁢ π 2
34 ax-1cn ⊢ 1 ∈ ℂ
35 df-5 ⊢ 5 = 4 + 1
36 31 34 35 comraddi ⊢ 5 = 1 + 4
37 36 eqcomi ⊢ 1 + 4 = 5
38 37 a1i ⊢ A ∈ ℂ → 1 + 4 = 5
39 38 oveq1d ⊢ A ∈ ℂ → 1 + 4 ⁢ π 2 = 5 ⁢ π 2
40 29 33 39 3eqtr2d ⊢ A ∈ ℂ → π 2 + 1 ⁢ 2 ⁢ π = 5 ⁢ π 2
41 40 oveq1d ⊢ A ∈ ℂ → π 2 + 1 ⁢ 2 ⁢ π - 5 ⁢ A = 5 ⁢ π 2 − 5 ⁢ A
42 34 18 mulcli ⊢ 1 ⁢ 2 ⁢ π ∈ ℂ
43 42 a1i ⊢ A ∈ ℂ → 1 ⁢ 2 ⁢ π ∈ ℂ
44 5 43 9 addsubd ⊢ A ∈ ℂ → π 2 + 1 ⁢ 2 ⁢ π - 5 ⁢ A = π 2 - 5 ⁢ A + 1 ⁢ 2 ⁢ π
45 14 41 44 3eqtr2rd ⊢ A ∈ ℂ → π 2 - 5 ⁢ A + 1 ⁢ 2 ⁢ π = 5 ⁢ π 2 − A
46 45 fveq2d ⊢ A ∈ ℂ → sin ⁡ π 2 - 5 ⁢ A + 1 ⁢ 2 ⁢ π = sin ⁡ 5 ⁢ π 2 − A
47 sinhalfpim ⊢ 5 ⁢ A ∈ ℂ → sin ⁡ π 2 − 5 ⁢ A = cos ⁡ 5 ⁢ A
48 9 47 syl ⊢ A ∈ ℂ → sin ⁡ π 2 − 5 ⁢ A = cos ⁡ 5 ⁢ A
49 13 46 48 3eqtr3rd ⊢ A ∈ ℂ → cos ⁡ 5 ⁢ A = sin ⁡ 5 ⁢ π 2 − A
50 5 8 subcld ⊢ A ∈ ℂ → π 2 − A ∈ ℂ
51 sin5t ⊢ π 2 − A ∈ ℂ → sin ⁡ 5 ⁢ π 2 − A = 16 ⁢ sin ⁡ π 2 − A 5 - 20 ⁢ sin ⁡ π 2 − A 3 + 5 ⁢ sin ⁡ π 2 − A
52 50 51 syl ⊢ A ∈ ℂ → sin ⁡ 5 ⁢ π 2 − A = 16 ⁢ sin ⁡ π 2 − A 5 - 20 ⁢ sin ⁡ π 2 − A 3 + 5 ⁢ sin ⁡ π 2 − A
53 sinhalfpim ⊢ A ∈ ℂ → sin ⁡ π 2 − A = cos ⁡ A
54 53 oveq1d ⊢ A ∈ ℂ → sin ⁡ π 2 − A 5 = cos ⁡ A 5
55 54 oveq2d ⊢ A ∈ ℂ → 16 ⁢ sin ⁡ π 2 − A 5 = 16 ⁢ cos ⁡ A 5
56 53 oveq1d ⊢ A ∈ ℂ → sin ⁡ π 2 − A 3 = cos ⁡ A 3
57 56 oveq2d ⊢ A ∈ ℂ → 20 ⁢ sin ⁡ π 2 − A 3 = 20 ⁢ cos ⁡ A 3
58 55 57 oveq12d ⊢ A ∈ ℂ → 16 ⁢ sin ⁡ π 2 − A 5 − 20 ⁢ sin ⁡ π 2 − A 3 = 16 ⁢ cos ⁡ A 5 − 20 ⁢ cos ⁡ A 3
59 53 oveq2d ⊢ A ∈ ℂ → 5 ⁢ sin ⁡ π 2 − A = 5 ⁢ cos ⁡ A
60 58 59 oveq12d ⊢ A ∈ ℂ → 16 ⁢ sin ⁡ π 2 − A 5 - 20 ⁢ sin ⁡ π 2 − A 3 + 5 ⁢ sin ⁡ π 2 − A = 16 ⁢ cos ⁡ A 5 - 20 ⁢ cos ⁡ A 3 + 5 ⁢ cos ⁡ A
61 49 52 60 3eqtrd ⊢ A ∈ ℂ → cos ⁡ 5 ⁢ A = 16 ⁢ cos ⁡ A 5 - 20 ⁢ cos ⁡ A 3 + 5 ⁢ cos ⁡ A