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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. – Srivatsan Sep 10 '11 at 17:06
@Srivatsan: I think you'd have to make that argument a bit more precise to explain why it doesn't exclude $\arccos(\sin x)=\frac\pi2-x$. – joriki Sep 10 '11 at 17:20
@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.) – Srivatsan Sep 10 '11 at 17:30
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 $\pm 1$. – J. M. Sep 10 '11 at 18:01

The identities you list can all be derived by expressing the trigonometric functions in terms of each other, e.g. (glossing over sign issues)

$$\tan x=\frac{\sin x}{\cos x}=\frac{\sin x}{\sqrt{1-\sin^2x}}\;,$$

and thus

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

So these identities hold because there are identities between the trigonometric functions, which in a sense are due to Euler's formula.

You can get similar identities in the other direction for $\arcsin$ and $\arccos$ because there's a suitable relationship between these two:

$$\arccos x=\frac\pi2-\arcsin x\;,$$

and thus

$$\arccos(\sin x)=\frac\pi2-\arcsin (\sin x)=\frac\pi2-x\;.$$

That you can't get them for $\arctan\cos(x)$ and $\arcsin(\tan x)$ has to do with the fact that the identities between the corresponding inverse trigonometric functions have different arguments. For instance,

$$\arctan x=\arcsin\frac x{\sqrt{x^2+1}}\;,$$

but you can't turn this into a formula for $\arctan(\sin x)$ because the argument on the right-hand side isn't $x$.

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