# 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}

$

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 $x\in\mathbb{R}$.