Question

For all real $\theta$, prove cos(sin $\theta$) > sin(cos $\theta$)

Answer 1

We have to prove
$cos\left(sin\theta \right)-sin\left(cos\theta \right)>0$
or $cos\left(sin\theta \right)-cos(\frac{\pi }{2}-cos\theta )>0$
or $2sin(\frac{\pi }{4}+\frac{sin\theta -cos\theta }{2})$
$sin(\frac{\pi }{4}+\frac{sin\theta -cos\theta }{2})>0$....(1)
We now show that both the factors on the left hand side of (1) are positive. since,
$|sin\theta -cos\theta |=\left|\sqrt{2}sin(\theta -\frac{\pi }{4})\right|\le \sqrt{2}\le \frac{\pi }{2}$
We have $-\frac{\pi }{2}<sin\theta -cos\theta <\frac{\pi }{2}$
$-\frac{\pi }{4}<\frac{sin\theta -cos\theta }{2}<\frac{\pi }{4}$
Add $\frac{\pi }{4}$ so that $0<\frac{\pi }{4}<\frac{sin\theta -cos\theta }{2}<\frac{\pi }{2}$
and therefore $sin(\frac{\pi }{4}+\frac{sin\theta -cos\theta }{2})>0$
Similarly we can prove $sin(\frac{\pi }{4}-\frac{sin\theta +cos\theta }{2})>0$
Hence (1) holds which is what we wanted to prove.