#How can I prove that?
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DarkBlood
limber mauve
Well, it's not.
graceful galleon
Well, it's not.
oh fuck
limber mauve
Like, why would you even think it is? The two sides of the equation don't even have the same domain.
graceful galleon
Yes but idk i was solving some problem and fund this
I thought it was weird too
limber mauve
What problem were you solving?
Because, like, -1/sqrt(1 - x^2) is the derivative of arccos(x), but the right side of that equation isn't a derivative. It's not even a limit.
graceful galleon
Yes, i was trying to prove that the arccos function had a central symetry (ofc he has) so ther's my beginning of answer
limber mauve
By "central symmetry", I assume you're talking about the rotational symmetry about its midpoint?
graceful galleon
yes
limber mauve
Well, the derivative is part of that, because you need to know whether the function is increasing or decreasing on its domain.
The derivative is negative on the domain, so it's decreasing.
Which means that what you want to prove is arccos(0 - x) - arccos(0) = arccos(0) - arccos(0 + x).
For all x.
...in [-1, 1].
graceful galleon
And how do I prove that?
limber mauve
Well, doing the algebra, it boils down to arccos(-x) + arccos(x) = pi.
So then arccos(x) = pi - arccos(-x), which means x = cos(pi - arccos(-x)) = sin(arccos(-x)), and just... y'know, solve that algebraically I guess?
Geometricallly, I mean.
graceful galleon
Well I think no
limber mauve
To be perfectly honest, I'm kinda half paying attention. I haven't been sleeping great.
graceful galleon
Oh go to sleep then
limber mauve
No, I woke up already today.
Anyway.
The, like, core point I was making is that what you want to prove is that if you go right from the center by x units, the graph goes down by the same amount that it goes up if you go left from the center x units.
graceful galleon
Oh I see that
limber mauve
Like, if you let f(x) = arccos(x) - arccos(0), you want to prove f is odd.
graceful galleon
so arccos(-x) -arccos(0)= -arccos(x)-arccos(0)
limber mauve
so arccos(-x) -arccos(0)= -arccos(x)-arccos(0)
...zero, not theta.
graceful galleon
Shit
limber mauve
...wait, I see where I made a mistake.
graceful galleon
arccos 0 is pi/2 so arccos(-x) -pi/2= -arccos(x)-pi/2
limber mauve
arccos 0 is pi/2 so arccos(-x) -pi/2= -arccos(x)-pi/2
...well, no.
It's what I said from the beginning. arccos(-x) - pi/2 = pi/2 - arccos(x).
You're not distributing the negative.
limber mauve
Well, it's like I said, arccos(x) + arccos(-x) = pi.
I made a mistake when I said cos(pi - arccos(-x)) = sin(arccos(-x)), because I was confusing pi with pi/2.
graceful galleon
With that, how can i prove there's a rotational symetry
limber mauve
Well, it's like I said, `arccos(x) + arccos(-x) = pi`.
...because this is literally just an algebraic transformation of the equation we had before.
graceful galleon
Okay, thank you very much !
limber mauve
Do you... know what to do from here?