Question

18.............................. .

Answer 1

$i.\: f(x) = (\sqrt{1+\sin x} + \sqrt{1-\sin x})^2 + (\sqrt{1+\sin x} - \sqrt{1-\sin x})^2=\\\\=(\sqrt{1+\sin x})^2 + 2\sqrt{1+\sin x}\cdot\sqrt{1-\sin x} + (\sqrt{1-\sin x})^2+(\sqrt{1+\sin x})^2 - 2\sqrt{1+\sin x}\cdot\sqrt{1-\sin x} + (\sqrt{1-\sin x})^2=\\\\=1+\sin x + 2\sqrt{(1+\sin x)(1-\sin x)} + 1 - \sin x + 1 + \sin x - 2\sqrt{(1+\sin x)(1-\sin x)} + 1-\sin x = 4\in\mathbb{N}$

$ii.\: f(x) = \sqrt{\sin^4x + 4\cos^2x} + \sqrt{\cos^4x+4\sin^2x} = \\=\sqrt{(\sin^2x)^2 + 4\cos^2x} + \sqrt{(\cos^2x)^2 + 4\sin^2x}=\\\sqrt{(1-\cos^2x)^2 + 4\cos^2x} + \sqrt{(1 - \sin^2x)^2 + 4\sin^2x}=\\=\sqrt{1-2\cos^2x + \cos^4x+4\cos^2x} + \sqrt{1 - 2\sin^2x+\sin^4x+4\sin^2x}=\\=\sqrt{1+2\cos^2x+\cos^4x} + \sqrt{ 1+2\sin^2x+\sin^4x}=\\=\sqrt{(1+\cos^2x)^2} + \sqrt{(1+\sin^2x)^2} = \\=1+\cos^2x+1+\sin^2x = 3$