Given SiNx = 13, sin (x + y) = 1, then sin (2Y + x) = 1___ .

Given SiNx = 13, sin (x + y) = 1, then sin (2Y + x) = 1___ .

∵ sin (x + y) = 1, ∵ x + y = π 2 + 2K π, K ∈ Z, ∵ y = - x + π 2 + 2K π, ∵ sin (2Y + x) = sin (- 2x + π + 4K π + x) = sin (π - x) = - SiNx = - 13