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Why $\arctan x$ not equal to $\frac{\arcsin(x)}{\arccos(x)}$? Is there a counter example that I can use to show that they are not equal? Thank!

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6 Answers

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Aside from $x=0$ and a value near $0.450116$ you can try any value you want.

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It's not terribly difficult to show if $h(x)=\frac{f(x)}{g(x)}$, then, in general,$$h^{-1}(x) \neq \frac{f^{-1}(x)}{g^{-1}(x)}$$Or, more generally, if $g(x)=f_1\circ ... \circ f_n (x)$,$$g^{-1}(x) \neq f_1^{-1}\circ ... \circ f_n^{-1}(x).$$

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Sure:

$$\arctan1=\frac\pi4\neq\frac{\cfrac\pi2}{0}=\frac{\arcsin 1}{\arccos 1}$$

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The functions $f(x)=\arctan(x)$ and $g(x)=\frac{\arcsin(x)}{\arccos(x)}$ are different for several reasons:

1. As mentioned in other answers, they take different values at many points. For example, $f(1)=\frac{\pi}{4}$ while $g(1)=\frac{\pi/2}{0}$ is undefined.
2. They have different domains: the domain of $\arctan$ is $\mathbb R$ while the domain of $\arcsin$ and $\arccos$ is $[-1,1]$, so the domain of $g$ is included in $[-1,1]$. Precisely, since $\arccos(x)=0 \iff x=1$ the domain of $g$ is $[-1,1)$.
3. The function $\arctan$ is odd, while $g$ is not. Indeed, since $\arcsin$ is odd, $f=g$ would imply that $\arccos(x)=\arcsin(x)\arctan(x)$ is even, which is known to be false.

Of course, one of these arguments is sufficient in itself.

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comment..It holds for very small values of argument.

If we take a power operation$$ z=\frac{x}{y}$$then$$ z^p=\frac{x^p}{y^p}$$

holds.

But then that is as far as it goes.

not (even) for

$$ arctan z = \frac{arctan x}{ arctan y} $$

or other operations.

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Counterexamples are useful, but knowing how to derive the inverse is also useful!

Suppose $$y= \tan(x).$$Then try to solve for x:$$y^2 = \tan^2(x) = sec^2(x)-1,$$so $$cos^2(x) = \frac{1}{y^2+1},$$$$\implies \cos(x) = \pm\sqrt{ \frac{1}{y^2+1}}.$$Thus$$\arctan(y) = \arccos\left(\pm\sqrt{\frac{1}{y^2+1}}\right)$$

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