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By doing some right triangle gymnastics, we can derive things like

$\cos(\arctan x) = \frac{1}{\sqrt{1+x^2}}$, for $x>0$

$\cos(\arcsin x) = \sqrt{1-x^2}$

$\tan(\arcsin x) = \frac{x}{\sqrt{1-x^2}}$

What about $\arctan\cos(x)$, $\arcsin(\tan x)$, etc? I understand that in this case $x$ is treated as an angle, not a ratio of side lengths and that it is impossible to construct the same kind of right triangle relations for these formulas. However, is there a particularly compelling non-geometric reason why the reverse application of these functions is intractable?

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    One reason why you should not hope to find an algebraic expression for $\arctan (\cos x)$ is that particular values of this function, like $\pi/4 = \arctan (\cos 0)$, happen to be transcendental.2011-09-10
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    @Srivatsan: I think you'd have to make that argument a bit more precise to explain why it doesn't exclude $\arccos(\sin x)=\frac\pi2-x$.2011-09-10
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    @joriki Well, if we allow $\pi$ on the right hand side, my argument just crumbles; after all, there's no fun in writing $\pi$ in terms of itself. But unfortunately, the argument does not tell us anything about $\arctan(\sin x)$ either. That's the part that is really unsatisfactory to me. (Also it looks like one cannot plug in anything other than $x=0$ here.)2011-09-10
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    At the very least, it is known that $\frac1{\pi}\arctan\, x$ is transcendental if $x$ is a rational number that isn't $0$ or $\pm 1$.2011-09-10

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