Metamath Proof Explorer


Theorem coseq0

Description: A complex number whose cosine is zero. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Assertion coseq0 ( 𝐴 ∈ ℂ → ( ( cos ‘ 𝐴 ) = 0 ↔ ( ( 𝐴 / π ) + ( 1 / 2 ) ) ∈ ℤ ) )

Proof

Step Hyp Ref Expression
1 picn π ∈ ℂ
2 1 a1i ( 𝐴 ∈ ℂ → π ∈ ℂ )
3 2 halfcld ( 𝐴 ∈ ℂ → ( π / 2 ) ∈ ℂ )
4 id ( 𝐴 ∈ ℂ → 𝐴 ∈ ℂ )
5 3 4 addcld ( 𝐴 ∈ ℂ → ( ( π / 2 ) + 𝐴 ) ∈ ℂ )
6 sineq0 ( ( ( π / 2 ) + 𝐴 ) ∈ ℂ → ( ( sin ‘ ( ( π / 2 ) + 𝐴 ) ) = 0 ↔ ( ( ( π / 2 ) + 𝐴 ) / π ) ∈ ℤ ) )
7 5 6 syl ( 𝐴 ∈ ℂ → ( ( sin ‘ ( ( π / 2 ) + 𝐴 ) ) = 0 ↔ ( ( ( π / 2 ) + 𝐴 ) / π ) ∈ ℤ ) )
8 sinhalfpip ( 𝐴 ∈ ℂ → ( sin ‘ ( ( π / 2 ) + 𝐴 ) ) = ( cos ‘ 𝐴 ) )
9 8 eqeq1d ( 𝐴 ∈ ℂ → ( ( sin ‘ ( ( π / 2 ) + 𝐴 ) ) = 0 ↔ ( cos ‘ 𝐴 ) = 0 ) )
10 pire π ∈ ℝ
11 pipos 0 < π
12 10 11 gt0ne0ii π ≠ 0
13 12 a1i ( 𝐴 ∈ ℂ → π ≠ 0 )
14 3 4 2 13 divdird ( 𝐴 ∈ ℂ → ( ( ( π / 2 ) + 𝐴 ) / π ) = ( ( ( π / 2 ) / π ) + ( 𝐴 / π ) ) )
15 2cnd ( 𝐴 ∈ ℂ → 2 ∈ ℂ )
16 2ne0 2 ≠ 0
17 16 a1i ( 𝐴 ∈ ℂ → 2 ≠ 0 )
18 2 15 2 17 13 divdiv32d ( 𝐴 ∈ ℂ → ( ( π / 2 ) / π ) = ( ( π / π ) / 2 ) )
19 2 13 dividd ( 𝐴 ∈ ℂ → ( π / π ) = 1 )
20 19 oveq1d ( 𝐴 ∈ ℂ → ( ( π / π ) / 2 ) = ( 1 / 2 ) )
21 18 20 eqtrd ( 𝐴 ∈ ℂ → ( ( π / 2 ) / π ) = ( 1 / 2 ) )
22 21 oveq1d ( 𝐴 ∈ ℂ → ( ( ( π / 2 ) / π ) + ( 𝐴 / π ) ) = ( ( 1 / 2 ) + ( 𝐴 / π ) ) )
23 1cnd ( 𝐴 ∈ ℂ → 1 ∈ ℂ )
24 23 halfcld ( 𝐴 ∈ ℂ → ( 1 / 2 ) ∈ ℂ )
25 4 2 13 divcld ( 𝐴 ∈ ℂ → ( 𝐴 / π ) ∈ ℂ )
26 24 25 addcomd ( 𝐴 ∈ ℂ → ( ( 1 / 2 ) + ( 𝐴 / π ) ) = ( ( 𝐴 / π ) + ( 1 / 2 ) ) )
27 14 22 26 3eqtrd ( 𝐴 ∈ ℂ → ( ( ( π / 2 ) + 𝐴 ) / π ) = ( ( 𝐴 / π ) + ( 1 / 2 ) ) )
28 27 eleq1d ( 𝐴 ∈ ℂ → ( ( ( ( π / 2 ) + 𝐴 ) / π ) ∈ ℤ ↔ ( ( 𝐴 / π ) + ( 1 / 2 ) ) ∈ ℤ ) )
29 7 9 28 3bitr3d ( 𝐴 ∈ ℂ → ( ( cos ‘ 𝐴 ) = 0 ↔ ( ( 𝐴 / π ) + ( 1 / 2 ) ) ∈ ℤ ) )