\color{blue}\bf{Problem}: Show that \cos(\sin{x})>\sin(\cos{x}) for each real x.

Sol. Consider the equation \cos(\sin{x})=\sin(\cos{x}).

\Rightarrow\cos(\sin{x})=\cos(\frac{\pi}{2}−\cos{x})\Rightarrow\cos(\sin{x})=\cos(\frac{\pi}{2}−\cos{x})

\sin{x}=2n{\pi}\pm(\frac{\pi}{2}−\cos{x})

Taking +ve sign,

we get, \sin{x}+\cos{x}=2n\pi+\frac{\pi}{2},n\in\mathbb{Z} which is not possible

Taking – ve sign we get,

\sin x−\cos x=2n\pi+\frac{\pi}{2}, n\in\mathbb{Z} , which is again not possible

Hence the equation \cos(\sin x)=\sin(\cos x) has no solution.

Now consider the function

f(x)=\cos(\sin x)−\sin(\cos x)The function f is continuous on \mathbb{R} , therefore

exactly one of the statements f(x)>0 or f(x)<0 for all x\in\mathbb{R}</p> <p> must be true beacause if it is not so there would exist two real numbers a and

b (say) such that f(a)f(b)<0 which would imply that the function

f has a zero between a and b by Interemediate value

property of continuous function, which is a contradiction. Further f(0)=1−\sin(1)>0,

Hence \cos(\sin x)>\sin(\cos x) for all x\in\mathbb{R}.