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Trigonometric Inequalities

\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}.

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