|y-1|=|(1,-1)・(sin x,cos x)| ≦ \(\sqrt{(1+1)}\)・\(\sqrt{sin^{2}x +cos^{2}x}\) =\(\sqrt{2}\).
等号成立は \((1, -1) \mathrel{/\!/} (sin x, cos x)\) 即ち x ≡ \(\frac{3π}{4}\) (mod 2π).
故に -\(\sqrt{2}\) ≦ y - 1 ≦ \(\sqrt{2}\)
1 - \(\sqrt{2}\) ≦ y ≦ 1 + \(\sqrt{2}\).