Trigonometry! #34

For theta [0,pi/2] ,sin increases from 0 to 1 , while for cos it decreases from 1 to 0. Hence cos(sin theta) and sin(cos theta) both are decreasing functions with initial values at theta = 0 is 1 and sin1 and value at theta = pi/2 are cos1 and 0 respectively. Therefore we can say that the left hand side decreases from 1 -> cos1 while right hand side decreases from sin1 -> 0. hence '>' is the answer!

We have

$\cos\theta+\sin\theta=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos\theta+\frac{1}{\sqrt{2}}\sin\theta\right)$

$\cos\theta\sin\theta= \sqrt{2}\left(\sin\frac{\pi}{4}\cos\theta+\cos\frac{\pi}{4}\sin\theta\right)$

$\cos\theta+\sin\theta= \sqrt{2}\sin\left(\frac{\pi}{4}+\theta\right)$

$\therefore \cos\theta+\sin\theta\le\sqrt{2}<\frac{\pi}{2}$

$\left(as~\sqrt{2}\approx1.4141,~\frac{\pi}{2}\approx1.57\right)$

$\therefore \cos\theta+\sin\theta<\frac{\pi}{2}$

$\cos\theta<\frac{\pi}{2}-\sin\theta$

$\sin(\cos\theta)<\sin\left(\frac{\pi}{2}-\sin\theta\right)$

$\sin(\cos\theta)<\cos(\sin\theta)$

$\boxed{\cos(\sin\theta)>\sin(\cos\theta)}$

Note that we do not have an equality, because if we did, then we would have the result $\cos\theta+\sin\theta>\sqrt{2}$ , which is not possible because $\cos\theta+\sin\theta=\sin\left(\frac{\pi}{4}+\theta\right)$ .