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.

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If there is a more appropriate tag please feel free to suggest :) –  Antonio Vargas Apr 15 '12 at 18:34
Have you tried to calculate the values with a computer calculator? If yes, then verifying the computation shouldn't be such a difficult task. –  Beni Bogosel Apr 15 '12 at 19:16
Look it up in a table. If just this one value is at stake and it is not $0.000000013$ then anyone could do an estimate with pencil and paper if (s)he is not convinced. –  Christian Blatter Apr 15 '12 at 19:20

Recall that $$1-\frac{1}{2}x^2 \;<\; \cos x \;<\; 1-\frac{1}{2}x^2 + \frac{1}{24}x^4$$ for $x>0$. From the upper bound, we see that $\cos 1 < 13/24$. 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 $$\sin x < x$$ for all $x>0$. Since $\sin x$ is increasing, this gives us $$\sin\sin\sin 1 \;<\; \sin \sin 1 \;<\; \sin 1.$$ But we also know that $$\sin x \;<\; x - \frac{1}{6}x^3 + \frac{1}{120}x^5$$ for all $x>0$. It follows that $$\sin\sin\sin 1 \;<\; \sin 1 \;<\; \frac{101}{120} \;<\; 0.85.$$

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Of course, this kind of estimation proof isn't really necessary. If you want to know whether $\cos\cos 1$ or $\sin\sin\sin 1$ is bigger, you can just plug them into a calculator and get an answer. –  Jim Belk Apr 15 '12 at 19:26

I guess you have written your question as you want it, since $$\cos\cos\cos 1= 0.65< \sin \sin \sin 1=0.67$$

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 $f(1)<\pi/2$ and using the monotonicity of $\sin$ on $[0,\pi/2]$ you are done.

It remains to notice that $$\sin 1 >\sin \sin 1>\sin \sin \sin 1$$

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