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For all $\theta$ in $[0,\pi/2]$ I need to show that $\cos (\sin \theta)>\sin (\cos \theta)$.

In my book it is done like $cos (\theta)<\pi/2- sin (\theta) $.Then they took sine on both sides ? But I doubt this approach.Since $\sin $ is not an increasing function everywhere is it correct to directly take sine on both sides of inequality ? And can someone provide alternate solution to this problem too ? Thanks.

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    $\begingroup$ The sine function is not increasing everywhere but it is for sure increasing over $\left[0,\frac{\pi}{2}\right]$ or a sub-interval. $\endgroup$ – Jack D'Aurizio May 28 '16 at 17:35
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Over $I=\left[0,\frac{\pi}{2}\right]$ we have: $$ \sin(\theta)+\cos(\theta) = \sqrt{2} \sin\left(\theta+\frac{\pi}{4}\right) \leq \sqrt{2} < \frac{\pi}{2} $$ hence: $$ \forall \theta\in I,\qquad \cos\theta < \frac{\pi}{2}-\sin\theta. \tag{1}$$ The LHS of $(1)$ belongs to $[0,1]$, the RHS belongs to $\left[\frac{\pi}{2}-1,\frac{\pi}{2}\right]$; the sine function is increasing over $\left[0,\frac{\pi}{2}\right]$, hence: $$ \forall \theta\in I,\qquad \sin(\cos\theta) < \cos(\sin\theta).\tag{2}$$

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  • $\begingroup$ The RHS does not belong to [0,1] as for given $\theta $ $\sin (\theta) $ lies between [0,1] but it is written $\pi/2-sin (\theta) $ for which the range should be $[\pi/2-1,\pi/2]$.Tell me if I am wrong... $\endgroup$ – user220382 May 28 '16 at 17:45
  • $\begingroup$ @SanchayanDutta: absolutely not, you're right. Now fixed. $\endgroup$ – Jack D'Aurizio May 28 '16 at 17:47
  • $\begingroup$ Hehe :)..thanks I got it now ! $\endgroup$ – user220382 May 28 '16 at 17:50
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If $x>0$, than $\sin x<x$.

If $x \in \left[0;\frac{\pi}2\right]$ then $\cos x$ decreasing function.

Then $$\sin \cos x<\cos x< \cos \sin x.$$

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