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Define a function $f(x)$ such that:

$$f(\sin x)+f(\cos x)=\frac{\tan x}2$$

What is $f(x)?$

My attempt: I hypothesized the denominator of the function to be like of the form $x+(1-x^2)^{1/2}$

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  • $\begingroup$ @Benjamin Moss: Thanks for editing $\endgroup$ – Chen Guo Dec 6 '17 at 17:09
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    $\begingroup$ Maybe It would be more correct to create a new OP. Now what of my answer. I think it's not a good way to change questions in this manner. $\endgroup$ – gimusi Dec 6 '17 at 17:34
  • $\begingroup$ @gimusi i already told you my attempt, you should have given other suggested solve. It is defined for 0 to pi/2 $\endgroup$ – Chen Guo Dec 6 '17 at 17:45
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    $\begingroup$ Since some users are unhappy about their answers being invalidated by the edit, I removed it. Please ask a new question referencing this one for context. $\endgroup$ – quid Dec 6 '17 at 19:03
  • $\begingroup$ On $(0,\pi/2)$, my answer still applies. $\endgroup$ – Simply Beautiful Art Dec 6 '17 at 20:06
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More general than gimusi's answer, notice that

$$\sin(x)=\cos(\pi/2-x)\\\cos(x)=\sin(\pi/2-x)$$

And so

$$f(\sin(x))+f(\cos(x))=f(\sin(\pi/2-x))+f(\cos(\pi/2-x))$$

But

$$\tan(x)\ne\tan(\pi/2-x)$$

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    $\begingroup$ nice generalization! $\endgroup$ – gimusi Dec 6 '17 at 17:29
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    $\begingroup$ So all we need to ensure is that the domain $D$ of $f$ is small enough never to contain both $x$ and $y$ when $x^2+y^2=1$ (except that $D$ may contain $\sqrt 2/2$) $\endgroup$ – Hagen von Eitzen Dec 6 '17 at 20:05
  • $\begingroup$ @HagenvonEitzen so you are proposing that whenever $f(\sin(x))$ exists, then $f(\cos(x))$ doesn't exist, and vice versa? I'm not sure of any reasonable interpretation of the question if that is he case you seek. $\endgroup$ – Simply Beautiful Art Dec 6 '17 at 20:27
  • $\begingroup$ @Hagen von Eitzen Yes exactly what i want is the one you said $\endgroup$ – Chen Guo Dec 7 '17 at 0:47
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    $\begingroup$ It would be interesting to check if there are other solutions, in the domain of R (or C) outside of [-1,1] (and x in C) $\endgroup$ – ypercubeᵀᴹ Dec 7 '17 at 8:13
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It is not possible, infact:

for $x=0: f(0)+f(1)=0$

for $x=\frac{\pi}{2}$: $f(1)+f(0)$ = RHS is not defined

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    $\begingroup$ @ChenGuo This answer shows that there is no solution: but even in general, "there must be some way" is not correct. For example, consider the equation $x = \sin x$. There is no way, with just algebraic or trigonometric steps, to solve for $x$. It has been proved to be impossible. The only thing we can do is approximate a numerical solution. $\endgroup$ – Ovi Dec 7 '17 at 2:08
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    $\begingroup$ @ChenGuo I don't see how a limit can help when commutativity is broken. For this to work, you'd have to redefine the $+$ operation, and then you might as make up whatever you want. $\endgroup$ – muru Dec 7 '17 at 5:22
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    $\begingroup$ @ChenGuo no they don't. But do tell us what $\sin(x+\epsilon)$ is supposed to be for real $x$ and infinitesimal $\epsilon$? $\endgroup$ – Simply Beautiful Art Dec 7 '17 at 11:46
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    $\begingroup$ +1) @gimusi God! How did you think? $\endgroup$ – освящение Dec 28 '17 at 14:32
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    $\begingroup$ @Abhishek Thanks I immediately saw it, despite Chen Guo never admited my the correctness point but he has considered correct the "generalization" by Simply Beautiful Art! The lesson is: simply answer often don't like so much to many mathematicians. $\endgroup$ – gimusi Dec 28 '17 at 14:55

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