Show that $\forall x \in \mathbb{R}\ \cos (\cos x) > \sin (\sin x)$
I've tried the obvious thing to do:
$\varphi: x \mapsto \cos(\cos x) - \sin (\sin x)$
$\varphi ' (x) = \sin (\cos x) \sin x - \cos(\sin x) \cos x$
I'd like to show now that $\varphi$ is a positive function for all $x \in \mathbb{R}$, but the derivative does not look friendly so either there's something with $\varphi '$. I'm not seeing or it's not the right way. Any hint or idea?
$\endgroup$ 23 Answers
$\begingroup$We know that $|\sin(x)|<|x|$, which I assume is taken to be known, since you differentiated $\sin(x)$. Hence, it follows from this and Pythagorean identities that
$$\cos^2(\cos(x))=1-\sin^2(\cos(x))>1-\cos^2(x)=\sin^2(x)$$
$$\sin^2(\sin(x))<\sin^2(x)$$
Thus, it follows that
$$\cos^2(\cos(x))>\sin^2(x)>\sin^2(\sin(x))$$
And since $\cos(\cos(x))\in[\cos(1),1]\implies\cos(\cos(x))>0$ and $a^2>b^2\implies|a|>|b|\ge b$, it follows that
$$\cos^2(\cos(x))>\sin^2(\sin(x))\\\implies|\cos(\cos(x))|>|\sin(\sin(x))|\ge\sin(\sin(x))$$
$\endgroup$ 10 $\begingroup$We need to prove that $$\cos\cos{x}-\cos\left(\frac{\pi}{2}-\sin{x}\right)>0$$ or $$2\sin\frac{\frac{\pi}{2}-\sin{x}-\cos{x}}{2}\sin\frac{\frac{\pi}{2}-\sin{x}+\cos{x}}{2}>0,$$ which is true because By C-S $$\sin{x}\pm\cos{x}\leq\sqrt{(1^2+1^2)(\sin^2x+\cos^2x)}=\sqrt2<\frac{\pi}{2},$$ which gives $$0<\frac{\frac{\pi}{2}-\sin{x}\pm\cos{x}}{2}<\frac{\frac{\pi}{2}+\sqrt2}{2}<\pi$$
$\endgroup$ 5 $\begingroup$$\sin(\sin(x))=\cos(\pi/2-\sin(x))$, write $f(x)=\pi/2-\sin(x)-\cos(x)$, $f'(x)=-\cos(x)+\sin(x)$, we study $f$ in $[0,\pi/2]$, $f'(x)=0$ implies $x=\pi/4$, $f(\pi/4)>0$ $f(0)>0, f(\pi/2)>0$, implies that $f$ decreases from $0$ to $\pi/4$ and increases from $\pi/4$ to $\pi/2$, and $f>0$ on $[0,\pi/2]$.
this implies that $\pi/2-\sin(x)>\cos(x)$, since $\cos$ decreases on $[0,\pi/2]$ we deduce that $\cos(\cos(x))>\cos(\pi/2-\sin(x))=\sin(\sin(x))$.
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