Metamath Proof Explorer


Theorem sin2pi

Description: The sine of 2 _pi is 0. (Contributed by Paul Chapman, 23-Jan-2008)

Ref Expression
Assertion sin2pi ( sin ‘ ( 2 · π ) ) = 0

Proof

Step Hyp Ref Expression
1 picn π ∈ ℂ
2 sin2t ( π ∈ ℂ → ( sin ‘ ( 2 · π ) ) = ( 2 · ( ( sin ‘ π ) · ( cos ‘ π ) ) ) )
3 1 2 ax-mp ( sin ‘ ( 2 · π ) ) = ( 2 · ( ( sin ‘ π ) · ( cos ‘ π ) ) )
4 sinpi ( sin ‘ π ) = 0
5 cospi ( cos ‘ π ) = - 1
6 4 5 oveq12i ( ( sin ‘ π ) · ( cos ‘ π ) ) = ( 0 · - 1 )
7 neg1cn - 1 ∈ ℂ
8 7 mul02i ( 0 · - 1 ) = 0
9 6 8 eqtri ( ( sin ‘ π ) · ( cos ‘ π ) ) = 0
10 9 oveq2i ( 2 · ( ( sin ‘ π ) · ( cos ‘ π ) ) ) = ( 2 · 0 )
11 2t0e0 ( 2 · 0 ) = 0
12 10 11 eqtri ( 2 · ( ( sin ‘ π ) · ( cos ‘ π ) ) ) = 0
13 3 12 eqtri ( sin ‘ ( 2 · π ) ) = 0