Is $\cos \cos 1 - \sin \sin \sin 1$ positive? How can I show that
$$\cos \cos 1 - \sin \sin \sin 1$$
is positive?
This is motivated by this question.  If
$$\begin{align}
f(x) &= \cos \cos \cos \cos(\pi/2 + ix) - \sin \sin \sin \sin(\pi/2+ix)\\
     &= \cos \cos \cos \sinh x - \sin \sin \sin \cosh x,
\end{align}$$
then it looks like $f(x)$ has a zero in the interval $(0,1)$.  This would imply that
$$
\cos \cos \cos \cos(z) - \sin \sin \sin \sin(z)
$$
has infinitely-many zeros in the strip $0 < \Im(z) < 1$.
One way to show that such a zero exists would be to show that $f(1) < 0 < f(0)$, the right-side of which is the current question.  I don't know how to show the left side either, but now I'm interested in this question for its own sake.
 A: Recall that
$$
1-\frac{1}{2}x^2 \;&lt\; \cos x \;&lt\; 1-\frac{1}{2}x^2 + \frac{1}{24}x^4
$$
for $x>0$.  From the upper bound, we see that $\cos 1 &lt 13/24$.  Since $\cos$ is decreasing, it follows that
$$
\cos(\cos 1) \;>\; \cos\left(\frac{13}{24}\right) \;>\; 1-\frac{1}{2}\left(\frac{13}{24}\right)^2 \;>\; 0.85.
$$
Next, recall that
$$
\sin x &lt x
$$
for all $x>0$.  Since $\sin x$ is increasing, this gives us
$$
\sin\sin\sin 1 \;&lt\; \sin \sin 1 \;&lt\; \sin 1.
$$
But we also know that
$$
\sin x \;&lt\; x - \frac{1}{6}x^3 + \frac{1}{120}x^5
$$
for all $x>0$.  It follows that
$$
\sin\sin\sin 1 \;&lt\; \sin 1 \;&lt\; \frac{101}{120} \;&lt\; 0.85.
$$
A: I guess you have written your question as you want it, since 
$$ \cos\cos\cos 1= 0.65&lt \sin \sin \sin 1=0.67$$
If you really want only two $\cos$ iterations in the left then the inequality is true and you can prove it like this:
$$\cos(\cos(1))=\sin(\pi/2-\cos 1)>\sin 1$$
since $f(x)=\cos x+x$ is strictly increasing and $f(\pi/2)=\pi/2$. Therefore $f(1)&lt\pi/2$ and using the monotonicity of $\sin $ on $[0,\pi/2]$ you are done. 
It remains to notice that $$ \sin 1 >\sin \sin 1>\sin \sin \sin 1$$
