Question

Arătați ca cos(sin(x))>sin(cos(x)) pt oricare x real

Answer 1

$\cos(\sin x) - \sin(\cos x) > 0\\ \\ \Rightarrow \cos(\sin x) - \cos\Big(\dfrac{\pi}{2}-\cos x\Big) > 0\\ \\\Rightarrow 2\sin \Big(\dfrac{\frac{\pi}{2}-\cos x - \sin x }{2}\Big)\sin\Big(\dfrac{\frac{\pi}{2}-\cos x + \sin x}{2}\Big) > 0 \\ \\ \Rightarrow 2\sin \Big(\dfrac{\pi}{4}-\dfrac{\sin x +\cos x}{2}\Big)\sin\Big(\dfrac{\pi}{4}+\dfrac{\sin x-\cos x}{2}\Big) > 0$

$\\\boxed{1} \quad -\dfrac{\pi}{2}<\sin x \pm \cos x <\dfrac{\pi}{2}\\ \\ \Rightarrow -\dfrac{\pi}{2}<-(\sin x + \cos x) <\dfrac{\pi}{2} \\ \\ \Rightarrow 0 < \dfrac{\pi}{2}-(\sin x + \cos x) <\pi \\ \\ \Rightarrow 0 < \dfrac{\pi}{4}-\dfrac{\sin x + \cos x}{2} <\dfrac{\pi}{2} \\ \\ \Rightarrow \sin\Big(\dfrac{\pi}{4}-\dfrac{\sin x + \cos x}{2}\Big) > 0$

$\\\boxed{2} \quad -\dfrac{\pi}{2}<\sin x \pm \cos x <\dfrac{\pi}{2}\\ \\ \Rightarrow -\dfrac{\pi}{2}<\sin x - \cos x <\dfrac{\pi}{2} \\ \\ \Rightarrow 0 < \dfrac{\pi}{2}+\sin x - \cos x) <\pi \\ \\ \Rightarrow 0 < \dfrac{\pi}{4}+\dfrac{\sin x - \cos x}{2} <\dfrac{\pi}{2} \\ \\ \Rightarrow \sin\Big(\dfrac{\pi}{4}+\dfrac{\sin x - \cos x}{2}\Big) > 0$

$\\\text{Din }\,\boxed{1}\,\text{ si }\,\boxed{2}:\\ \\ \Rightarrow 2\sin \Big(\dfrac{\pi}{4}-\dfrac{\sin x +\cos x}{2}\Big)\sin\Big(\dfrac{\pi}{4}+\dfrac{\sin x-\cos x}{2}\Big) > 0\quad (A) \\ \\ \Rightarrow \cos(\sin x) - \sin(\cos x)>0\\ \\ \Rightarrow \cos(\sin x) > \sin(\cos x),\quad \forall x\in \mathbb{R}$