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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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    $\begingroup$ What are their values when $x=1$? $\endgroup$
    – Kavi Rama Murthy
    Jul 3, 2020 at 23:49
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    $\begingroup$ Perhaps you should try just about any number you can think of and see if it's a counter example? $\endgroup$
    – davidlowryduda ♦
    Jul 3, 2020 at 23:49
  • $\begingroup$ Why should it? $\arctan 1=\frac\pi 4\ne \arcsin 1/\arccos 1=\frac{\pi/2}0$ $\endgroup$
    – Bernard
    Jul 3, 2020 at 23:50
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    $\begingroup$ It almost never (and I'm been cautious here...) true that $$f(x)=\frac{g(x)}{h(x)}\implies f^{-1}(x)=\frac{g^{-1}(x)}{h^{-1}(x)}$$ $\endgroup$
    – DonAntonio
    Jul 3, 2020 at 23:54
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    $\begingroup$ Apart from the numerical counter examples shown in the answer, it seems to me that perhaps the OP does not exactly know the meaning of the inverse? Because once you understand the definition of inverse, this question becomes moot $\endgroup$
    – imranfat
    Jul 4, 2020 at 0:15

6 Answers 6

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

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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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    $\begingroup$ That's a great remark that $h=\frac{f}{g}$ does not imply that $h^{-1}=\frac{f^{-1}}{g^{-1}}$ in general but it may still be true for specific functions $f$, $g$ and $h$. For example, the second statement is true whenever $f_1$,...,$f_n$ all commute (e.g. linear functions with zero intercept). $\endgroup$
    – Taladris
    Jul 4, 2020 at 5:00
  • $\begingroup$ Yes, linear functions are always nicer and have much more straightforward properties than nonlinear functions. But nonlinear functions are a bit more interesting :) $\endgroup$
    – K.defaoite
    Jul 4, 2020 at 15:44
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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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  • $\begingroup$ When writing $y=\tan(x)$, I guess you are assuming $-\frac{\pi}{2}<x<\frac{\pi}{2}$ (so that $x=\arctan(y)$). But in that case, it is not obvious that $\cos(x)=\pm\sqrt{\frac{1}{y^2+1}}$ implies that $x=\arccos \left(\pm\sqrt{\frac{1}{y^2+1}} \right)$. That would be true if we know that $x\in[0,\pi]$. $\endgroup$
    – Taladris
    Jul 4, 2020 at 1:55
  • $\begingroup$ In particular, $\arccos$ is always non-negative, while $\arctan$ could be negative. I believe that the correct formula is $\arctan(y)=\pm\arccos\left(\pm\sqrt{\frac{1}{y^2+1}} \right)$ ($\pm$ are both negative when $y<0$, and positive otherwise) $\endgroup$
    – Taladris
    Jul 4, 2020 at 2:16

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