Which is Greater: $\cos(\sin\ x)$ or $\sin (\cos \ x)$

[kaliprasad]

which is greater of the 2

$\cos(\sin\ x) $ or $\sin (\cos \ x) $

 

[I like Serena]

which is greater of the 2

$\cos(\sin\ x) $ or $\sin (\cos \ x) $

Spoiler

Suppose they intersect somewhere.
Then:
\begin{array}{lcll}
\cos(\sin x)&=&\sin(\cos x) \\
\cos(\sin x)&=&\cos(\frac \pi 2 - \cos x) \\
\sin x &=& \pm(\frac \pi 2 - \cos x) + 2\pi k \\
\sin x \pm \cos x &=& \pm \frac \pi 2 + 2\pi k \\
\sqrt 2 \sin(x+\phi) &=&\pm \frac \pi 2 + 2\pi k & \text{ where }\phi =\text{atan2}(\pm 1, 1) \\
\end{array}
Since $\sqrt 2 < \frac\pi 2$ it follows that there are no intersections.

Filling in $x = 0$ tells us that:
$$\cos(\sin 0)=\cos 0 = 1 > \sin 1=\sin(\cos 0)$$
Therefore:
$$\cos(\sin x)>\sin(\cos x)$$

 

[kaliprasad]

Spoiler

Suppose they intersect somewhere.
Then:
\begin{array}{lcll}
\cos(\sin x)&=&\sin(\cos x) \\
\cos(\sin x)&=&\cos(\frac \pi 2 - \cos x) \\
\sin x &=& \pm(\frac \pi 2 - \cos x) + 2\pi k \\
\sin x \pm \cos x &=& \pm \frac \pi 2 + 2\pi k \\
\sqrt 2 \sin(x+\phi) &=&\pm \frac \pi 2 + 2\pi k & \text{ where }\phi =\text{atan2}(\pm 1, 1) \\
\end{array}
Since $\sqrt 2 < \frac\pi 2$ it follows that there are no intersections.

Filling in $x = 0$ tells us that:
$$\cos(\sin 0)=\cos 0 = 1 > \sin 1=\sin(\cos 0)$$
Therefore:
$$\cos(\sin x)>\sin(\cos x)$$

good solution. Here is another

Spoiler

we have

$\cos(\sin\, x)-\sin(\cos\, x)$

= $\cos(\sin\, x)-\cos(\dfrac{\pi}{2}- \cos\, x)$

= $\cos(\sin\, x)+\cos(\dfrac{\pi}{2}+ \cos\, x)$

= $2 \cos \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2} \cos \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$

= $2 \cos\ t \ \cos \ s$

where $t= \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$

and $s= \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$is product of 2 terms and we can show that both t and s are between 0 and $\frac{\pi}{2}$

we have $\sin\, x\pm \cos\, x= \sqrt{2}\sin ( x\pm \dfrac{\pi}{4})$ so between $- \sqrt{2}$ and $\sqrt{2}$hence s and t both are positive and less than $\frac{\pi}{2}$ so the difference $\cos(\sin\, x)-\sin(\cos\, x)\gt \ 0 $ and hence

$\cos(\sin\, x) \gt \sin(\cos\, x)$