Question Number 120029 by Ar Brandon last updated on 28/Oct/20
$$\mathrm{Montrer}\:\mathrm{que}\:\forall\mathrm{x}\in\mathbb{R} \\ $$ $$\mathrm{cos}\left(\mathrm{sinx}\right)>\mathrm{sin}\left(\mathrm{cosx}\right) \\ $$
Commented bysoumyasaha last updated on 29/Oct/20
![π/2 ≈ 1.57 and (√2) ≈ 1.41 ∴ (√2) < π/2 Now, sin(x+π/4) ≤ 1 ⇒ (√2) sin(x+π/4) ≤ (√2) < π/2 ⇒ (√2) (sinxcos(π/4)+cosxsin(π/4)) < π/2 ⇒ (√2) (sinx.(1/( (√2)))+cosx.(1/( (√2)))) < π/2 ⇒ sinx+cosx < π/2 ⇒ cosx < π/2−sinx ⇒ sin(cosx) < sin(π/2−sinx) [∵ sinx is an increasing function] ⇒ sin(cosx) < cos(sinx) ⇒ cos(sinx) > sin(cosx)](Q120085.png)
$$\:\:\:\:\pi/\mathrm{2}\:\approx\:\mathrm{1}.\mathrm{57}\:\:\:\mathrm{and}\:\sqrt{\mathrm{2}}\:\approx\:\mathrm{1}.\mathrm{41} \\ $$ $$\:\:\therefore\:\sqrt{\mathrm{2}}\:<\:\pi/\mathrm{2} \\ $$ $$\:\:\mathrm{Now},\:\:\mathrm{sin}\left(\mathrm{x}+\pi/\mathrm{4}\right)\:\leqslant\:\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\sqrt{\mathrm{2}}\:\mathrm{sin}\left(\mathrm{x}+\pi/\mathrm{4}\right)\:\leqslant\:\sqrt{\mathrm{2}}\:\:<\:\pi/\mathrm{2} \\ $$ $$\:\:\:\:\:\:\:\:\Rightarrow\:\sqrt{\mathrm{2}}\:\left(\mathrm{sinxcos}\frac{\pi}{\mathrm{4}}+\mathrm{cosxsin}\frac{\pi}{\mathrm{4}}\right)\:\:<\:\pi/\mathrm{2} \\ $$ $$\:\:\:\:\:\:\:\:\Rightarrow\:\sqrt{\mathrm{2}}\:\left(\mathrm{sinx}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{cosx}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:\:<\:\pi/\mathrm{2} \\ $$ $$\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{sinx}+\mathrm{cosx}\:\:<\:\pi/\mathrm{2} \\ $$ $$\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{cosx}\:\:<\:\pi/\mathrm{2}−\mathrm{sinx} \\ $$ $$\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{sin}\left(\mathrm{cosx}\right)\:\:<\:\mathrm{sin}\left(\pi/\mathrm{2}−\mathrm{sinx}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:\mathrm{sinx}\:\mathrm{is}\:\mathrm{an}\:\mathrm{increasing}\:\mathrm{function}\right] \\ $$ $$\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{sin}\left(\mathrm{cosx}\right)\:\:<\:\mathrm{cos}\left(\mathrm{sinx}\right) \\ $$ $$\:\:\:\:\:\:\:\:\Rightarrow\:\:\mathrm{cos}\left(\mathrm{sinx}\right)\:>\:\mathrm{sin}\left(\mathrm{cosx}\right) \\ $$
Commented byAr Brandon last updated on 29/Oct/20
Thanks Sir