Metamath Proof Explorer


Theorem cos2t

Description: Double-angle formula for cosine. (Contributed by Paul Chapman, 24-Jan-2008)

Ref Expression
Assertion cos2t ⊢ A ∈ ℂ → cos ⁡ 2 ⁢ A = 2 ⁢ cos ⁡ A 2 − 1

Proof

Step Hyp Ref Expression
1 coscl ⊢ A ∈ ℂ → cos ⁡ A ∈ ℂ
2 1 sqcld ⊢ A ∈ ℂ → cos ⁡ A 2 ∈ ℂ
3 ax-1cn ⊢ 1 ∈ ℂ
4 subsub3 ⊢ cos ⁡ A 2 ∈ ℂ ∧ 1 ∈ ℂ ∧ cos ⁡ A 2 ∈ ℂ → cos ⁡ A 2 − 1 − cos ⁡ A 2 = cos ⁡ A 2 + cos ⁡ A 2 - 1
5 3 4 mp3an2 ⊢ cos ⁡ A 2 ∈ ℂ ∧ cos ⁡ A 2 ∈ ℂ → cos ⁡ A 2 − 1 − cos ⁡ A 2 = cos ⁡ A 2 + cos ⁡ A 2 - 1
6 2 2 5 syl2anc ⊢ A ∈ ℂ → cos ⁡ A 2 − 1 − cos ⁡ A 2 = cos ⁡ A 2 + cos ⁡ A 2 - 1
7 cosadd ⊢ A ∈ ℂ ∧ A ∈ ℂ → cos ⁡ A + A = cos ⁡ A ⁢ cos ⁡ A − sin ⁡ A ⁢ sin ⁡ A
8 7 anidms ⊢ A ∈ ℂ → cos ⁡ A + A = cos ⁡ A ⁢ cos ⁡ A − sin ⁡ A ⁢ sin ⁡ A
9 2times ⊢ A ∈ ℂ → 2 ⁢ A = A + A
10 9 fveq2d ⊢ A ∈ ℂ → cos ⁡ 2 ⁢ A = cos ⁡ A + A
11 1 sqvald ⊢ A ∈ ℂ → cos ⁡ A 2 = cos ⁡ A ⁢ cos ⁡ A
12 sincl ⊢ A ∈ ℂ → sin ⁡ A ∈ ℂ
13 12 sqvald ⊢ A ∈ ℂ → sin ⁡ A 2 = sin ⁡ A ⁢ sin ⁡ A
14 11 13 oveq12d ⊢ A ∈ ℂ → cos ⁡ A 2 − sin ⁡ A 2 = cos ⁡ A ⁢ cos ⁡ A − sin ⁡ A ⁢ sin ⁡ A
15 8 10 14 3eqtr4d ⊢ A ∈ ℂ → cos ⁡ 2 ⁢ A = cos ⁡ A 2 − sin ⁡ A 2
16 12 sqcld ⊢ A ∈ ℂ → sin ⁡ A 2 ∈ ℂ
17 16 2 addcomd ⊢ A ∈ ℂ → sin ⁡ A 2 + cos ⁡ A 2 = cos ⁡ A 2 + sin ⁡ A 2
18 sincossq ⊢ A ∈ ℂ → sin ⁡ A 2 + cos ⁡ A 2 = 1
19 17 18 eqtr3d ⊢ A ∈ ℂ → cos ⁡ A 2 + sin ⁡ A 2 = 1
20 subadd ⊢ 1 ∈ ℂ ∧ cos ⁡ A 2 ∈ ℂ ∧ sin ⁡ A 2 ∈ ℂ → 1 − cos ⁡ A 2 = sin ⁡ A 2 ↔ cos ⁡ A 2 + sin ⁡ A 2 = 1
21 3 2 16 20 mp3an2i ⊢ A ∈ ℂ → 1 − cos ⁡ A 2 = sin ⁡ A 2 ↔ cos ⁡ A 2 + sin ⁡ A 2 = 1
22 19 21 mpbird ⊢ A ∈ ℂ → 1 − cos ⁡ A 2 = sin ⁡ A 2
23 22 oveq2d ⊢ A ∈ ℂ → cos ⁡ A 2 − 1 − cos ⁡ A 2 = cos ⁡ A 2 − sin ⁡ A 2
24 15 23 eqtr4d ⊢ A ∈ ℂ → cos ⁡ 2 ⁢ A = cos ⁡ A 2 − 1 − cos ⁡ A 2
25 2 2timesd ⊢ A ∈ ℂ → 2 ⁢ cos ⁡ A 2 = cos ⁡ A 2 + cos ⁡ A 2
26 25 oveq1d ⊢ A ∈ ℂ → 2 ⁢ cos ⁡ A 2 − 1 = cos ⁡ A 2 + cos ⁡ A 2 - 1
27 6 24 26 3eqtr4d ⊢ A ∈ ℂ → cos ⁡ 2 ⁢ A = 2 ⁢ cos ⁡ A 2 − 1