Wolfram alpha vs me. How did they get from what I had to what they had?
#is this an identity I’m not aware of with arctan?
lone star
meager swanBOT
lone star
It’s the last step I don’t get
If it was log instead of tan-1 that would make sense to me cos u can bring the power to the front but I didn’t know u could do that with tan-1
real gyro
New discovery club
I also got to know this now
On the internet
Fresh discovery for me too because of you
lone star
Is there a nice explanation for why that’s true
real gyro
Wait
lone star
Cos this doesn’t work and I don’t see how it’s just off by a constant cos that is not the same graph
real gyro
Hmm, I sure did see someone do it like this on the internet
Idk why
But see, I also do not have a correct explanation for this
lone star
Ah I see
real gyro
Still, u can treat it like a property
lone star
Well I’m not convinced it will hold for all powers
lone star
And as the Desmos thing shows it doesn’t really seem true even with the ^-1
real gyro
Maybe it does not work for cos inverse
lone star
So it’s a bit confusing
real gyro
Cause cos returns positive value
For negative argument
We do need some advanced mathematicians for this problem
lone star
Yeah
Idk how to get advanced mathematicians to come here though 😂
It’s a huge server it’s hard to get noticed lol
waxen raft
Does that come from the fact that tan-1(x) + tan-1(1/x) = pi/2 ?
or -pi/2 but a constant anyway
real gyro
@lone star it's not a trigonometric identity it is just re-substituting $u=\mathrm{csec}, x$ in $\tan^{-1} u = \tan^{-1}(\mathrm{csec},x) $
or do you ask how one got $\int\frac{1}{1+u^2}\ \mathrm{d}u = \tan^{-1}u$ ?
real gyro
(if you ask the latter, try the substitution $u=\tan v$ and use the identity $1 + \tan^2 = 1/\cos^2$ that can easily obtained from $\cos^2 + \sin^2 = 1$ when dividing by $\cos^2$)
reef pikeBOT
Landau08
lone star
Does that come from the fact that tan-1(x) + tan-1(1/x) = pi/2 ?
Ohhhh that’ll be it!
lone star
<@803706369994915910> it's not a trigonometric identity it is just re-substituti...
Thanks but it’s okay I understand that I was confused about getting from the -tan^-1(cscx) to the tan^-1(sinx)
lone star
Ohhhh that’ll be it!
Because I put some numbers in and it was off by 1.5ish each time
lone star
Does that come from the fact that tan-1(x) + tan-1(1/x) = pi/2 ?
So why is this true
Yeah
Found a nice explanation for why
real gyro
sorry, I first misunderstood the question,; above geometric proof is nice
lone star
No worries :)
waxen raft
So why is this true
You can simply derive and prove the derivative is 0, that is enough for your initial problem, you don't care about the value of the constant
lone star
.solved