od Daniel » Četvrtak, 02. April 2015, 22:30
Evo kako sam zamislio, a svakako bi me zanimalo ako može i jednostavnije.
Potrebno je da utvrdimo da li je izraz [inlmath]\sin\left(\cos x\right)-\cos\left(\sin x\right)[/inlmath] pozitivan ili negativan. Na kosinus primenimo identitet [inlmath]\cos\alpha=\sin\left(\frac{\pi}{2}-\alpha\right)[/inlmath]:
[dispmath]\sin\left(\cos x\right)-\cos\left(\sin x\right)=\sin\left(\cos x\right)-\sin\left(\frac{\pi}{2}-\sin x\right)=[/dispmath]
Zatim izvršimo transformaciju razlike sinusa u proizvod, prema formuli [inlmath]\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}[/inlmath]:
[dispmath]=2\cos\frac{\frac{\pi}{2}-\sin x+\cos x}{2}\sin\frac{\cos x+\sin x-\frac{\pi}{2}}{2}=\\
=2\cos\left(\frac{\pi}{4}-\frac{\sin x-\cos x}{2}\right)\sin\left(\frac{\sin x+\cos x}{2}-\frac{\pi}{4}\right)=\\
=2\cos\left[\frac{\pi}{4}-\frac{\sin x-\sin\left(\frac{\pi}{2}-x\right)}{2}\right]\sin\left[\frac{\sin x+\sin\left(\frac{\pi}{2}-x\right)}{2}-\frac{\pi}{4}\right]=\\
=2\cos\left[\frac{\pi}{4}-\frac{\cancel2\cos\frac{\cancel x+\frac{\pi}{2}-\cancel x}{2}\sin\frac{x-\frac{\pi}{2}+x}{2}}{\cancel2}\right]\sin\left[\frac{\cancel2\sin\frac{\cancel x+\frac{\pi}{2}-\cancel x}{2}\cos\frac{x-\frac{\pi}{2}+x}{2}}{\cancel2}-\frac{\pi}{4}\right]=\\
=2\cos\left[\frac{\pi}{4}-\cos\frac{\pi}{4}\sin\left(x-\frac{\pi}{4}\right)\right]\sin\left[\sin\frac{\pi}{4}\cos\left(x-\frac{\pi}{4}\right)-\frac{\pi}{4}\right]=\\
=2\cos\left[\frac{\pi}{4}-\frac{\sqrt2}{2}\sin\left(x-\frac{\pi}{4}\right)\right]\sin\left[\frac{\sqrt2}{2}\cos\left(x-\frac{\pi}{4}\right)-\frac{\pi}{4}\right]=[/dispmath]
Da bismo odredili predznak celog izraza, posebno ispitujemo predznak svakog njegovog činioca:
[dispmath]\sin\left(x-\frac{\pi}{4}\right)\in\left[-1,1\right]\quad\Rightarrow\quad\frac{\sqrt2}{2}\sin\left(x-\frac{\pi}{4}\right)\in\left[-\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right]\quad\Rightarrow\\
\Rightarrow\quad\frac{\pi}{4}-\frac{\sqrt2}{2}\sin\left(x-\frac{\pi}{4}\right)\in\left[\frac{\pi}{4}-\frac{\sqrt2}{2},\frac{\pi}{4}+\frac{\sqrt2}{2}\right][/dispmath]
Pošto je [inlmath]\frac{\sqrt2}{2}<\frac{\pi}{4}[/inlmath], sledi
[dispmath]\frac{\pi}{4}-\frac{\sqrt2}{2}\sin\left(x-\frac{\pi}{4}\right)\in\left(\frac{\pi}{4}-\frac{\pi}{4},\frac{\pi}{4}+\frac{\pi}{4}\right)\quad\Rightarrow\\
\Rightarrow\quad\frac{\pi}{4}-\frac{\sqrt2}{2}\sin\left(x-\frac{\pi}{4}\right)\in\left(0,\frac{\pi}{2}\right)\quad\Rightarrow\\
\Rightarrow\quad\underline{\cos\left[\frac{\pi}{4}-\frac{\sqrt2}{2}\sin\left(x-\frac{\pi}{4}\right)\right]>0}[/dispmath]
Vrlo slično se uradi i za drugi činilac:
[dispmath]\cos\left(x-\frac{\pi}{4}\right)\in\left[-1,1\right]\quad\Rightarrow\quad\frac{\sqrt2}{2}\cos\left(x-\frac{\pi}{4}\right)\in\left[-\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right]\quad\Rightarrow\\
\Rightarrow\quad\frac{\sqrt2}{2}\cos\left(x-\frac{\pi}{4}\right)-\frac{\pi}{4}\in\left[-\frac{\sqrt2}{2}-\frac{\pi}{4},\frac{\sqrt2}{2}-\frac{\pi}{4}\right]\quad\Rightarrow\\
\Rightarrow\quad\frac{\sqrt2}{2}\cos\left(x-\frac{\pi}{4}\right)-\frac{\pi}{4}\in\left(-\frac{\pi}{4}-\frac{\pi}{4},\frac{\pi}{4}-\frac{\pi}{4}\right)\quad\Rightarrow\\
\Rightarrow\quad\frac{\sqrt2}{2}\cos\left(x-\frac{\pi}{4}\right)-\frac{\pi}{4}\in\left(-\frac{\pi}{2},0\right)\quad\Rightarrow\\
\Rightarrow\quad\underline{\sin\left[\frac{\sqrt2}{2}\cos\left(x-\frac{\pi}{4}\right)-\frac{\pi}{4}\right]<0}[/dispmath]
Odatle sledi da je
[dispmath]2\cos\left[\frac{\pi}{4}-\frac{\sqrt2}{2}\sin\left(x-\frac{\pi}{4}\right)\right]\sin\left[\frac{\sqrt2}{2}\cos\left(x-\frac{\pi}{4}\right)-\frac{\pi}{4}\right]<0[/dispmath]
a odatle i da je
[dispmath]\sin\left(\cos x\right)-\cos\left(\sin x\right)<0[/dispmath]
čime je pokazano da važi
[dispmath]\enclose{box}{\sin\left(\cos x\right)<\cos\left(\sin x\right)}[/dispmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain