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Question Number 120029 by Ar Brandon last updated on 28/Oct/20

$$\mathrm{Montrer}\mathrm{que}\mathrm{\forall }\mathrm{x}\in \mathbb{R}$$ $$\cos \left(\mathrm{sinx}\right)>\sin \left(\mathrm{cosx}\right)$$

Commented bysoumyasaha last updated on 29/Oct/20

$$\pi /2\approx 1.57\mathrm{and}\sqrt{2}\approx 1.41$$ $$\therefore \sqrt{2}<\pi /2$$ $$\mathrm{Now},\sin (\mathrm{x}+\pi /4)⩽1$$ $$\Rightarrow \sqrt{2}\sin (\mathrm{x}+\pi /4)⩽\sqrt{2}<\pi /2$$ $$\Rightarrow \sqrt{2}(\mathrm{sinxcos}\frac{\pi }{4}+\mathrm{cosxsin}\frac{\pi }{4})<\pi /2$$ $$\Rightarrow \sqrt{2}(\mathrm{sinx}.\frac{1}{\sqrt{2}}+\mathrm{cosx}.\frac{1}{\sqrt{2}})<\pi /2$$ $$\Rightarrow \mathrm{sinx}+\mathrm{cosx}<\pi /2$$ $$\Rightarrow \mathrm{cosx}<\pi /2-\mathrm{sinx}$$ $$\Rightarrow \sin \left(\mathrm{cosx}\right)<\sin (\pi /2-\mathrm{sinx})$$ $$[\because \mathrm{sinx}\mathrm{is}\mathrm{an}\mathrm{increasing}\mathrm{function}]$$ $$\Rightarrow \sin \left(\mathrm{cosx}\right)<\cos \left(\mathrm{sinx}\right)$$ $$\Rightarrow \cos \left(\mathrm{sinx}\right)>\sin \left(\mathrm{cosx}\right)$$

Commented byAr Brandon last updated on 29/Oct/20

Thanks Sir

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