<s>[INST] For all real $\theta$ prove that $ \cos(\sin\theta) \gt \sin(\cos\theta)$ How do I prove this? Im not able to even start it. Help please!
 [/INST] Try graphing both of the functions.


Also, this answer may help:


$$
\begin{align}
&\cos (\sin x) - \sin (\cos x) > 0\\ 
\implies &\cos (\sin x) - \cos ( π/2 - \cos x ) > 0\\ 
\implies &2 \sin \left[ \frac{\pi}{4} + \frac{1}{2}(\sin x - \cos x) \right]\cdot \sin \left[ \frac{\pi}{4} - \frac{1}{2}(\sin x - \cos x) \right] > 0 \tag{1} 
\end{align}
$$
  If we could prove that both the factors on the left hand side of $(1)$ are positive then the result obtained above $(1)$ is proved. 
  Since: $$\left| \sin x - \cos x \right| = \left| √2 \sin (x- \frac{\pi}{4}) \right| ≤ √2 < \frac{\pi}{2} $$
  We have,
  $$- \frac{\pi}{2} < ( \sin x - \cos x ) < \frac{\pi}{2}\\ 
\implies - \frac{\pi}{4} < ( \sin x - \cos x )/2 < \frac{\pi}{4} $$
  So that, $$0 < \frac{\pi}{4}+ \frac{1}{2}( \sin x - \cos x ) < \frac{\pi}{2}$$ 
  $\therefore \space \sin [ \frac{\pi}{4} + \frac{1}{2} (\sin x - \cos x) ] > 0 \quad\text{(ie, Positive)}$
  Similarly we can prove that 
  $\sin [ \frac{\pi}{4} - \frac{1}{2}(\sin x - \cos x) ] > 0$ 
  Hence $(1)$ is true. QED 

Reference: Yahoo Answers
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