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Question Number 119246 by 1549442205PVT last updated on 23/Oct/20

$$\mathrm{Prove}\mathrm{the}\mathrm{following}\mathrm{inequalities}\mathrm{hold}$$ $$\mathrm{true}\mathrm{\forall }\mathrm{x}\in \mathrm{R}$$ $$\mathrm{a})\cos \left(\mathrm{cosx}\right)>0$$ $$\mathrm{b})\cos \left(\mathrm{sinx}\right)>\sin \left(\mathrm{cosx}\right)$$

Answered by Olaf last updated on 23/Oct/20

$$a)$$ $$-\frac{\pi }{2}<-1⩽\cos x⩽+1<+\frac{\pi }{2}$$ $$\Rightarrow \cos \left(\cos x\right)>0\left(\mathrm{trivial}\right)$$

Answered by mindispower last updated on 23/Oct/20

$$cos\left(sin\left(x\right)\right)>0,\mathrm{\forall }x\in \mathbb{R}$$ $$sin\left(cos\left(x\right)\right)<0,\mathrm{\forall }x\in [-\pi ,-\frac{\pi }{2}]\cup [\frac{\pi }{2},\pi ]$$ $$soweworckjustin[-\frac{\pi }{2},\frac{\pi }{2}]$$ $$x\to cos\left(sin(-x)\right)=cos\left(sin(-x)\right)$$ $$sin\left(cos(-x)\right)=sin\left(cos\left(x\right)\right)\Rightarrow$$ $$justx\in [0,\frac{\pi }{2}]$$ $$lets[solveinx\in [0,\frac{\pi }{2}]$$ $$cos\left(sin\left(x\right)\right)>sin\left(cos\left(x\right)\right)$$ $$\iff$$ $$sin(\frac{\pi }{2}-sin\left(x\right))>sin\left(cos\left(x\right)\right)$$ $$sincecos\left(x\right),\frac{\pi }{2}-sin\left(x\right)\in [0,\frac{\pi }{2}]andsinincrease$$ $$function$$ $$\iff \frac{\pi }{2}-sin\left(x\right)>cos\left(x\right)$$ $$\iff sin\left(x\right)+cos\left(x\right)<\frac{\pi }{2}...E$$ $$\mid sin\left(x\right)+cos\left(x\right)\mid ⩽\sqrt{1^2+1^2}.\sqrt{{cos}^2\left(x\right)+{sin}^2}=\sqrt{2}<\frac{\pi }{2}$$ $$cauchyshwartz..$$ $$byequivalentEtrue$$ $$\Rightarrow sin(\frac{\pi }{2}-sin\left(x\right))>sin\left(cos\left(x\right)\right)$$ $$\iff cos\left(sin\left(x\right)\right)>sin\left(cos\left(x\right)\right)$$ $$$$

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