# Is there a function satisfying the following equation: $f(\sin x)+f(\cos x) = \frac{\tan x}{2}$?

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 $x+(1-x^2)^{1/2}$

• @Benjamin Moss: Thanks for editing – Chen Guo Dec 6 '17 at 17:09
• 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. – gimusi Dec 6 '17 at 17:34
• @gimusi i already told you my attempt, you should have given other suggested solve. It is defined for 0 to pi/2 – Chen Guo Dec 6 '17 at 17:45
• 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. – quid Dec 6 '17 at 19:03
• On $(0,\pi/2)$, my answer still applies. – Simply Beautiful Art Dec 6 '17 at 20:06

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

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

• nice generalization! – gimusi Dec 6 '17 at 17:29
• 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 $\sqrt 2/2$) – Hagen von Eitzen Dec 6 '17 at 20:05
• @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. – Simply Beautiful Art Dec 6 '17 at 20:27
• @Hagen von Eitzen Yes exactly what i want is the one you said – Chen Guo Dec 7 '17 at 0:47
• 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) – ypercubeᵀᴹ Dec 7 '17 at 8:13

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

• @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. – Ovi Dec 7 '17 at 2:08
• @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. – muru Dec 7 '17 at 5:22
• @ChenGuo no they don't. But do tell us what $\sin(x+\epsilon)$ is supposed to be for real $x$ and infinitesimal $\epsilon$? – Simply Beautiful Art Dec 7 '17 at 11:46
• +1) @gimusi God! How did you think? – освящение Dec 28 '17 at 14:32
• @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. – gimusi Dec 28 '17 at 14:55