New trigonometric identity

[Vinicius Claudino Ferraz]

https://pbs.twimg.com/media/El0Z7KvWoAUAqU5?format=png&name=900x900

\begin{align}
\tan^2 \biggl(\cfrac{\pi}{3} + \cfrac{1}{3} \text{ arctan } |x|\biggl) = 2 + 3x^2 + 2 \sqrt{x^2 + 1}\sqrt{9 x^2 + 1} \cos\biggl(\cfrac{1}{3} \text{ arctan } \cfrac{8|x|}{27 x^4 + 18 x^2 - 1}\biggl)
\end{align}

sketch of proof:

m = tan θ
v = tan 3θ = (3m - m³)/(1 - 3m²)
v/m =1/3 - 8/(9m² - 3)
x = m²
x³ + px + q = 0
Cardano

[William Elliot]

Post by Vinicius Claudino Ferraz
https://pbs.twimg.com/media/El0Z7KvWoAUAqU5?format=png&name=900x900
\begin{align}
\tan^2 \biggl(\cfrac{\pi}{3} + \cfrac{1}{3} \text{ arctan } |x|\biggl) = 2 + 3x^2 + 2 \sqrt{x^2 + 1}\sqrt{9 x^2 + 1} \cos\biggl(\cfrac{1}{3} \text{ arctan } \cfrac{8|x|}{27 x^4 + 18 x^2 - 1}\biggl)
\end{align}

Unreadable. Use plain text to state the theorem.

Post by Vinicius Claudino Ferraz
m = tan Ξ
v = tan 3Ξ = (3m - m³)/(1 - 3m²)
v/m =1/3 - 8/(9m² - 3)
x = m²
x³ + px + q = 0
Cardano

[sobriquet]

Post by Vinicius Claudino Ferraz
https://pbs.twimg.com/media/El0Z7KvWoAUAqU5?format=png&name=900x900
\begin{align}
\tan^2 \biggl(\cfrac{\pi}{3} + \cfrac{1}{3} \text{ arctan } |x|\biggl) = 2 + 3x^2 + 2 \sqrt{x^2 + 1}\sqrt{9 x^2 + 1} \cos\biggl(\cfrac{1}{3} \text{ arctan } \cfrac{8|x|}{27 x^4 + 18 x^2 - 1}\biggl)
\end{align}
m = tan θ
v = tan 3θ = (3m - m³)/(1 - 3m²)
v/m =1/3 - 8/(9m² - 3)
x = m²
x³ + px + q = 0
Cardano

It doesn't seem to hold. If you substitute 0 for x, the left side is 3 and the right side is 4.

Loading Image...

https://www.wolframalpha.com/input/?i=%28tan%28%28pi%2F3%29%2B%281%2F3%29arctan%28abs%280%29%29%29%29%5E2

https://www.wolframalpha.com/input/?i=2%2B3+0%5E2%2B2sqrt%280%5E2%2B1%29sqrt%289+0%5E2%2B1%29cos%28%281%2F3%29arctan%288%28abs%280%29%29%2F%2827+0%5E4%2B18+0%5E2-1%29%29%29

https://www.wolframalpha.com/input/?i=%28tan%28%28pi%2F3%29%2B%281%2F3%29arctan%28abs%28x%29%29%29%29%5E2%3D2%2B3x%5E2%2B2sqrt%28x%5E2%2B1%29sqrt%289x%5E2%2B1%29cos%28%281%2F3%29arctan%288%28abs%28x%29%29%2F%2827x%5E4%2B18x%5E2-1%29%29%29

[Vinicius Claudino Ferraz]

Replying to all

sobriquet tried zero and it did not hold. well, zero is as dangerous as pi/2... I did not define the domain hahaha

Try x = 1.

william elliot said it is unreadable.

For 1 the formula says that

tan² 75° = 5 + 4 sqrt(5) cos(1/3 arctan 2/11)

The angle arctan 2/11 is constructable and its third part is constructable with ruler and compass.

For other ecses, I've just posted that

tan²( pi/3 + 1/3 arctan |x| ) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( 1/3 ARCTAN 8|x|/(27x^4 + 18x² - 1) )

where ARCTAN is positive. if it were negative, add pi to it.

Now it holds for all x in the set of real numbers.

x = 0 ==> tan² 60° = 3 = 2 + 2 cos( 1/3 pi ) = 2 + 1

I think it will never hold for negatives and complex numbers because sqrt(x²) = |x| >= 0.

Sincerely,

Vinicius
the twenty twenty the November the third

[Vinicius Claudino Ferraz]

https://www.wolframalpha.com/input/?i=5+%2B+4+*+sqrt%285%29+*+cos%281%2F3+*+arctan%282%2F11%29%29+-+tan+%2875*pi%2F180%29%5E2

tan²( pi/3 + 1/3 arctan |x| ) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( 1/3 ARCTAN 8|x|/(27x^4 + 18x² - 1) )

where ARCTAN is positive. if it were not positive, add pi to it. So its image is (0, pi].

[Vinicius Claudino Ferraz]

List of constructible lengths

a + b, ab, a/b, sqrt(a)

List of constructible angles

3 * pi/180 * k ∈ {0, 1, 2, ..., 119, 120}
arctan b/a, given two constructible lengths
a/2, bissectrix

Now add to this Field eight cubic roots or trisections:

x = tan 0°
1) x = tan 9°
2) x = tan 18°
3) x = tan 27°
4) x = tan 36°
5) x = tan 45° yields (11 + 2i)^(1/3) ~ construction of the angle 1/3 arctan 2/11
6) x = tan 54°
7) x = tan 63°
8) x = tan 72°

[sobriquet]

Post by Vinicius Claudino Ferraz
Replying to all
sobriquet tried zero and it did not hold. well, zero is as dangerous as pi/2... I did not define the domain hahaha
Try x = 1.
william elliot said it is unreadable.
For 1 the formula says that
tan² 75° = 5 + 4 sqrt(5) cos(1/3 arctan 2/11)
The angle arctan 2/11 is constructable and its third part is constructable with ruler and compass.
For other ecses, I've just posted that
tan²( pi/3 + 1/3 arctan |x| ) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( 1/3 ARCTAN 8|x|/(27x^4 + 18x² - 1) )
where ARCTAN is positive. if it were negative, add pi to it.
Now it holds for all x in the set of real numbers.
x = 0 ==> tan² 60° = 3 = 2 + 2 cos( 1/3 pi ) = 2 + 1
I think it will never hold for negatives and complex numbers because sqrt(x²) = |x| >= 0.
Sincerely,
Vinicius
the twenty twenty the November the third

https://www.desmos.com/calculator/8fcyum02zf

It fails to hold for small numbers close to zero.

[Vinicius Claudino Ferraz]

Replying to sobriquet

Post by sobriquet
It fails to hold for small numbers close to zero.

That's because for small numbers you have to add pi to the arctan.

Put there this condition

p(x) = 27x^4 + 18x² - 1
f(x) = tan²( pi/3 + 1/3 arctan |x| )
g(x) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( 1/3 arctan 8|x|/p )
h(x) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( pi/3 + 1/3 arctan 8|x|/p )
q(x) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( 1/3 * pi/2 )

if p(x) > 0 then f(x) = g(x)
if p(x) < 0 then f(x) = h(x)
if p(x) = 0 then f(x) = q(x)

27x^4 + 18x² - 1 < 0
-0.227083 < x < 0.227083 ~ the angle is between 12° and 13°

https://www.wolframalpha.com/input/?i=27x%5E4+%2B+18x%5E2+-+1+%3C%3D+0

I'm worried
1) 8|x|/p = y ==> x = T(y)
2) there are 8 cubic roots which are constructible. now I want the tangent of 5t.
3) as the polynomial of sqrt(2)/2 is 2x² - 1 = 0 I want the polynomials of cos 3° k, the constructible angles,
compare the polynomial of t = arctan 2/11 with t/3.

First I thought as if it were as Q(2^(1/3)) = {a + b * 2^(1/3) + c * 4^(1/3)}
but no. Simply, the cubic root has disappeared and has become square root.

[[e-mail]]

What about a triangle with an infinitely small angle?
that looks like lines that are infinitely close to parallel.

[sobriquet]

Post by Vinicius Claudino Ferraz
Replying to sobriquet

Post by sobriquet
It fails to hold for small numbers close to zero.

That's because for small numbers you have to add pi to the arctan.

Ok, but that seems a bit like saying abs(x) = x holds although for
x < 0 you have to multiply x by -1 to make it hold.
So it's not entirely clear what it means for some equation to hold
if there is a range of cases where you have to tweak it a bit to
make it hold.

Post by Vinicius Claudino Ferraz
Put there this condition
p(x) = 27x^4 + 18x² - 1
f(x) = tan²( pi/3 + 1/3 arctan |x| )
g(x) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( 1/3 arctan 8|x|/p )
h(x) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( pi/3 + 1/3 arctan 8|x|/p )
q(x) = 2 + 3x² + 2 sqrt{x² + 1} sqrt{9x² + 1} cos( 1/3 * pi/2 )
if p(x) > 0 then f(x) = g(x)
if p(x) < 0 then f(x) = h(x)
if p(x) = 0 then f(x) = q(x)
27x^4 + 18x² - 1 < 0
-0.227083 < x < 0.227083 ~ the angle is between 12° and 13°
https://www.wolframalpha.com/input/?i=27x%5E4+%2B+18x%5E2+-+1+%3C%3D+0
I'm worried
1) 8|x|/p = y ==> x = T(y)
2) there are 8 cubic roots which are constructible. now I want the tangent of 5t.
3) as the polynomial of sqrt(2)/2 is 2x² - 1 = 0 I want the polynomials of cos 3° k, the constructible angles,
compare the polynomial of t = arctan 2/11 with t/3.

How do you associate numbers like sqrt(2)/2 with (roots of) polynomials
like 2x² - 1 = 0, given that we might as well associate them with other polynomials that share the same root (like 2^(3/2)x^3 - 1 = 0 )?

Post by Vinicius Claudino Ferraz
First I thought as if it were as Q(2^(1/3)) = {a + b * 2^(1/3) + c * 4^(1/3)}
but no. Simply, the cubic root has disappeared and has become square root.

[Vinicius Claudino Ferraz]

Post by sobriquet
Ok, but that seems a bit like saying abs(x) = x holds although for
x < 0 you have to multiply x by -1 to make it hold.
So it's not entirely clear what it means for some equation to hold
if there is a range of cases where you have to tweak it a bit to
make it hold.

So, let's just say the same language, guy. It holds only for x >=0, if you want to drop the abs(x). of course!

The cases are very common in mathematics, don't you like them?

We can first define F : ℝ → ℝ
x ↦ g(x), if and only if p(x) > 0
x ↦ h(x), if and only if p(x) < 0
x ↦ q(x), if and only if p(x) = 0

And second state the theorem: f(x) = F(x), for all x in ℝ.

Post by sobriquet
How do you associate numbers like sqrt(2)/2 with (roots of) polynomials
like 2x² - 1 = 0, given that we might as well associate them with other polynomials that share the same root (like 2^(3/2)x^3 - 1 = 0 )?

That's the chapter "minimal polynomial". It's calculable.
https://www.wolframalpha.com/input/?i=MinimalPolynomial%28cos%2836*pi%2F180%29%2Cx%29

Definition: the minimal polynomial of x₀ is p(x) if and only if p(x₀) = 0, deg p(x) is the minimum possible, p(x) has integer coefficients, and the leader coefficient is positive.

I've just got the vision:
x₀ = 1 ⇒ f(x₀) = 2/11
x₁ = 2/11 ⇒ f(x₁) = another constructible thing = x₂
This yields an INFINITE sequence.
Such that t = 1/3 arctan x_n is constructible with ruler and compass.

Sincerely,

Vinicius
20.20.Nov.04th

[Vinicius Claudino Ferraz]

Second ugly number
https://www.wolframalpha.com/input/?i=cos%281%2F3+*+arctan%28sqrt%285+%2B+2*sqrt%285%29%29%2F%28163+%2B+72*sqrt%285%29%29%29%29+-+1%2F4+sqrt%281%2F31+%28219+%2B+41+sqrt%285%29+%2B+sqrt%286+%283065+%2B+1189+sqrt%285%29%29%29%29%29

Now, let's take the simmetry with the inverse of the parameter:

x = 1/y for instance y = 2/11 => x = 11/2
f(x) = 8x/(27x^4 + 18x² - 1)
f(1/y) = 8/y (27/y^4 + 18/y^2 - 1)
f(1/y) = 8(27 + 18y^2 - y^4)/y^5

It was 1 unique sequence (x_n), now we have the second sequence y_n = 1/x_n .

[Vinicius Claudino Ferraz]

third ugly number
https://www.wolframalpha.com/input/?i=cos%281%2F3+*+arctan%285*sqrt%285+%2B+2*sqrt%285%29%29%2F%2890+%2B+41*sqrt%285%29%29%29%29+-+1%2F4+sqrt%281%2F11+%28119+-+5+sqrt%285%29+%2B+sqrt%2830+%2885+%2B+31+sqrt%285%29%29%29%29%29

fourth ugly number
https://www.wolframalpha.com/input/?i=cos%281%2F3+*+arctan%285*sqrt%285+%2B+2*sqrt%285%29%29%2F%2890+%2B+41*sqrt%285%29%29%29%29+-+1%2F4+sqrt%281%2F11+%28119+-+5+sqrt%285%29+%2B+sqrt%2830+%2885+%2B+31+sqrt%285%29%29%29%29%29

I've been thinking about x = z = r(cos t + i sen t)
and if 0 <= t < pi then abs(x) := z
else if pi <= t < 2 pi then abs(x) := -z (opposite of negative, for example)
because it came from sqrt(x^2). it "maybe" will hold in half plane.

Somebody please teach me. According to wikipedia,
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Extension_to_complex_plane

arctan(2i) = integral from 0 to 1 of dz/(1 + z^2) + integral from 1 to 2i of dz/(1 + z^2)

The first is arctan(1) = pi/4. And the second?

I forgot how to integrate f(z) dz from a to b.
Everything used to end up in Cauchy and number of holes.

Vinicius

[[e-mail]]

Post by Vinicius Claudino Ferraz
third ugly number
https://www.wolframalpha.com/input/?i=cos%281%2F3+*+arctan%285*sqrt%285+%2B+2*sqrt%285%29%29%2F%2890+%2B+41*sqrt%285%29%29%29%29+-+1%2F4+sqrt%281%2F11+%28119+-+5+sqrt%285%29+%2B+sqrt%2830+%2885+%2B+31+sqrt%285%29%29%29%29%29
fourth ugly number
https://www.wolframalpha.com/input/?i=cos%281%2F3+*+arctan%285*sqrt%285+%2B+2*sqrt%285%29%29%2F%2890+%2B+41*sqrt%285%29%29%29%29+-+1%2F4+sqrt%281%2F11+%28119+-+5+sqrt%285%29+%2B+sqrt%2830+%2885+%2B+31+sqrt%285%29%29%29%29%29
I've been thinking about x = z = r(cos t + i sen t)
and if 0 <= t < pi then abs(x) := z
else if pi <= t < 2 pi then abs(x) := -z (opposite of negative, for example)
because it came from sqrt(x^2). it "maybe" will hold in half plane.
Somebody please teach me. According to wikipedia,
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Extension_to_complex_plane
arctan(2i) = integral from 0 to 1 of dz/(1 + z^2) + integral from 1 to 2i of dz/(1 + z^2)
The first is arctan(1) = pi/4. And the second?
I forgot how to integrate f(z) dz from a to b.
Everything used to end up in Cauchy and number of holes.
Vinicius

Trig is your first polygon. That polygon is also QM...

[Vinicius Claudino Ferraz]

Post by ***@gmail.com
Trig is your first polygon. That polygon is also QM...

What the hell was this?

I know there is a straight line from (1,0) to (p,q).
y = q(x - 1)/(p - 1)
I want line integral from 1 to p + qi.

[[e-mail]]

Post by Vinicius Claudino Ferraz

Post by ***@gmail.com
Trig is your first polygon. That polygon is also QM...

What the hell was this?

There is a first polygon. When it is right angle it is both QM and Trig...

[Vinicius Claudino Ferraz]

Post by ***@gmail.com

Post by Vinicius Claudino Ferraz

Post by ***@gmail.com
Trig is your first polygon. That polygon is also QM...

What the hell was this?

There is a first polygon. When it is right angle it is both QM and Trig...

Among all polygons, who is the first?
Trig seems to qualify... Define that property, if it's meaningful.
QM for me is a segment. Where are the points Q and M? ; - P

At least we have countably many trisections. Luckily, it's not uncountable case.
Maybe we should "complete" that set by some way.

[Ross A. Finlayson]

Post by Vinicius Claudino Ferraz

Post by ***@gmail.com

Post by Vinicius Claudino Ferraz

Post by ***@gmail.com
Trig is your first polygon. That polygon is also QM...

What the hell was this?

There is a first polygon. When it is right angle it is both QM and Trig...

Among all polygons, who is the first?
Trig seems to qualify... Define that property, if it's meaningful.
QM for me is a segment. Where are the points Q and M? ; - P
At least we have countably many trisections. Luckily, it's not uncountable case.
Maybe we should "complete" that set by some way.

(I don't know it but it reminds me tri-lateral-ometry that basically
after sine/cosine as the orthogonal and for the Pythagorean, that
it is the un-hinged regular n-gon, or n-lateral, that unfolding the
endpoint sweeps sine then folding back up and unfolding again
sweeps the other half of the period of sine, about what are all
defined as the functions in n-lateral-ometry past the trilateral's,
that each regular n-gon defines an n-lateral that happens to
trace out first sine/cosine then other functions like them, each
with all the results what follow like for Fourier make for summing
the terms of for orthogonal functions, that all real functions have
a way to so be written as a series in powers of those. Also then
it reminds me of the quadratic sieve but I don't know. Here for
computing the residue, or for pull-back or along those lines in the
complex, here it's nice that you are finding these terms as what
are basically modular there, it seems there suffices enough modular
terms in the prime-counting, that you are finding these expressions,
of the variables that result in terms that under usual algebraic
manipulations: appear to drop out as either "dimensionless", or,
not just like multiplying by 1/1, i.e. still being algebraic.)

[Vinicius Claudino Ferraz]

Em terça-feira, 1 de dezembro de 2020 às 01:21:41 UTC-3, Ross A. Finlayson escreveu:
Fourier is by the way. An infinite sum of sines and cossines.
A kind of transform of functions.
But here we are dividing alpha /p and the result is a pretty: cos(p arccos(beta))

No link tonight. Everything is the same. Including the Spirits that make me owned LOL.
They are the same legion since 1982 when I reincarnated.

# Dividing quaternion by complex is not working. I need a better def powm(T, z)...

Below 4 + 6i + 10j + 37k ; p = 3 ; i^2 = j^2 = k^2 = -1

delta = [ (4.0 + 6.0j) (10.0 + 37.0j)]
[(-10.0 + 37.0j) (4.0 - 6.0j)]
m := delta*pi/180 = [(0.06981317008 + 0.1047197551j) (0.1745329252 + 0.6457718232j)]
[(-0.1745329252 + 0.6457718232j) (0.06981317008 - 0.1047197551j)]
a = [ (-0.8160820057 + 0.01219775137j) (0.02032958562 + 0.07521946678j)]
[(-0.02032958562 + 0.07521946678j) (-0.8160820057 - 0.01219775137j)]
b = [(0.2616975215 - 0.214297521j) (-0.3571625349 - 1.321501379j)]
[(0.3571625349 - 1.321501379j) (0.2616975215 + 0.214297521j)]
Re z = [(0.2572946101 - 0.153964933j) (-0.2566082216 - 0.94945042j)]
[ (0.2566082216 - 0.94945042j) (0.2572946101 + 0.153964933j)]
Im z = [(-0.9567507312 - 0.03246906898j) (-0.05411511496 - 0.2002259253j)]
[ (0.05411511496 - 0.2002259253j) (-0.9567507312 + 0.03246906898j)]
Mod z = [ (-0.186167613 + 0.04592389502j) (0.07653982503 + 0.2831973526j)]
[(-0.07653982503 + 0.2831973526j) (-0.186167613 - 0.04592389502j)]
cos Arg z = [(-2.796572984 + 0.1371635402j) (0.2286059003 + 0.845841831j)]
[(-0.2286059003 + 0.845841831j) (-2.796572984 - 0.1371635402j)] = [fraction] beta/gamma^p sometimes
sin Arg z = [(0.9426264608 + 0.4069351612j) (0.6782252686 + 2.509433494j)]
[(-0.6782252686 + 2.509433494j) (0.9426264608 - 0.4069351612j)]
tan Arg z = [(-0.03516496266 - 0.1472368196j) (-0.2453946994 - 0.9079603876j)]
[ (0.2453946994 - 0.9079603876j) (-0.03516496266 + 0.1472368196j)]
==> cos( [(3.0 + 0.0j) 0.0]
[ 0.0 (3.0 + 0.0j)] * arccos((tan^2(m)-a)/b)) = cos Arg z
delta = [ (124.0 + 6.0j) (10.0 + 37.0j)]
[(-10.0 + 37.0j) (124.0 - 6.0j)]
m := delta*pi/180 = [ (2.164208272 + 0.1047197551j) (0.1745329252 + 0.6457718232j)]
[(-0.1745329252 + 0.6457718232j) (2.164208272 - 0.1047197551j)]
a = [ (-0.8160820057 + 0.01219775137j) (0.02032958562 + 0.07521946678j)]
[(-0.02032958562 + 0.07521946678j) (-0.8160820057 - 0.01219775137j)]
b = [(0.2616975215 - 0.214297521j) (-0.3571625349 - 1.321501379j)]
[(0.3571625349 - 1.321501379j) (0.2616975215 + 0.214297521j)]
Re z = [(-0.2041383098 + 0.1007894302j) (0.1679823836 + 0.6215348195j)]
[(-0.1679823836 + 0.6215348195j) (-0.2041383098 - 0.1007894302j)]
Im z = [(-0.5890042647 - 0.02041656186j) (-0.03402760311 - 0.1259021314j)]
[ (0.03402760311 - 0.1259021314j) (-0.5890042647 + 0.02041656186j)]
Mod z = [ (-0.186167613 + 0.04592389502j) (0.07653982503 + 0.2831973526j)]
[(-0.07653982503 + 0.2831973526j) (-0.186167613 - 0.04592389502j)]
cos Arg z = [(1.884829414 - 0.07644037443j) (-0.1274006241 - 0.4713823091j)]
[(0.1274006241 - 0.4713823091j) (1.884829414 + 0.07644037443j)] = [fraction] beta/gamma^p sometimes
sin Arg z = [(0.5736246683 + 0.2511695785j) (0.4186159643 + 1.548879068j)]
[(-0.4186159643 + 1.548879068j) (0.5736246683 - 0.2511695785j)]
tan Arg z = [(0.07336041516 + 0.1362336953j) (0.2270561589 + 0.8401077879j)]
[(-0.2270561589 + 0.8401077879j) (0.07336041516 - 0.1362336953j)]
==> cos( [(3.0 + 0.0j) 0.0]
[ 0.0 (3.0 + 0.0j)] * arccos((tan^2(m)-a)/b)) = cos Arg z

#if: x = 2ad
# y = 2bd
# z = 2cd
# t = a**2 + b**2 + c**2 - d**2
# w = a**2 + b**2 + c**2 + d**2
#then: x^2 + y^2 + z^2 + t^2 == w^2

[Vinicius Claudino Ferraz]

Why do the children cry when we say that zero is equal to 2*pi rad?
Why do other children cry when we say that 20 degrees are undiscovered.
Why do the kids cry when we say that 2^(1/12) must exist, but nobody discovered yet.
Maybe an extraterrestrial... Some kind of life that lasts time enough to discover.
We are not alone at the universe. Maybe an ET lives more than Matusalem and his 10240 biblical years having sex.

[Vinicius Claudino Ferraz]

with 2 complexes, we make a quaternion, it could be 4x4
with 2 quaternions, we make an octonion 8x8
with 2 octonions, we make a sedenion 16x16
why is nobody interessed in 2 sedenions?
with them we make unnamed matrixes 32x32.
#badWay
https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction

[Vinicius Claudino Ferraz]

https://www.wolframalpha.com/input/?i=24.48378444127528097477087281354090828809026731570800802127910383060111

the input is x = 24.48378444127528097477087281354090828809026731570800802127910383060111
the fucking output is x = 13584 - 6064 sqrt(5)
I need to implement that in Python.
Fast.
How does wolfram alpha do that disgrace?

[Vinicius Claudino Ferraz]

https://www.wolframalpha.com/input/?i=cos%283*arccos%28%28tan%5E2%2812%2F1*pi%2F180%29-%2817-6*sqrt%285%29%29%29%2Fsqrt%281824-800*sqrt%285%29%29%29%29-%2813584-6064*sqrt%285%29%29%2F%28456-200*sqrt%285%29%29%5E1.5

Everything is intuitive here. 36 degrees / 3 = 12 degrees ~ sqrt(5) ~ golden number.

[Vinicius Claudino Ferraz]

https://www.wolframalpha.com/input/?i=tan%5E2%285*pi%2F180%29%3D23-12*sqrt%283%29%2B2*sqrt%28944%2B-544*sqrt%283%29%29*cos%282*pi%2F3%2B1%2F3*arccos%28%2840960%2B-23648*sqrt%283%29%29%2F%28944%2B-544*sqrt%283%29%29%5E1.5%29%29

Everything should be intuitive here too. 15 degrees / 3 = 5 degrees. This angle must exist.

https://www.wolframalpha.com/input/?i=cos%283*arccos%28%28tan%5E2%285%2F1*pi%2F180%29-%2823%2B-12*sqrt%283%29%29%29%2F%282*sqrt%28944%2B-544*sqrt%283%29%29%29%29%29-%2840960%2B-23648*sqrt%283%29%29%2F%28944%2B-544*sqrt%283%29%29%5E1.5

[Vinicius Claudino Ferraz]

48 lines is a little better than zero.
But it seems they are modulo 2. LOL

cos(3*arccos((tan^2(12/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(132/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(252/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(24/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(144/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(264/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(36/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(156/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(276/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(48/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(168/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(288/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(72/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(192/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(312/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(84/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(204/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(324/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(96/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(216/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(336/1*pi/180)-(17+6*sqrt(5)))/(2*sqrt(456+200*sqrt(5)))))-(13584+6064*sqrt(5))/(456+200*sqrt(5))^1.5
cos(3*arccos((tan^2(108/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(228/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(348/1*pi/180)-(17-6*sqrt(5)))/(2*sqrt(456-200*sqrt(5)))))-(13584-6064*sqrt(5))/(456-200*sqrt(5))^1.5
cos(3*arccos((tan^2(5/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(125/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(245/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(25/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(145/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(265/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(35/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(155/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(275/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(55/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(175/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(295/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(65/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(185/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(305/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(85/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(205/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(325/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(95/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(215/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(335/1*pi/180)-(23+12*sqrt(3)))/(2*sqrt(944+544*sqrt(3)))))-(40960+23648*sqrt(3))/(944+544*sqrt(3))^1.5
cos(3*arccos((tan^2(115/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(235/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5
cos(3*arccos((tan^2(355/1*pi/180)-(23-12*sqrt(3)))/(2*sqrt(944-544*sqrt(3)))))-(40960-23648*sqrt(3))/(944-544*sqrt(3))^1.5

Below the quaternion: 4 + 6i + 10j + 37k ; p = 3 ; i^2 = j^2 = k^2 = -1

delta = [ 4.0 -6.0 -10.0 -37.0]
[ 6.0 4.0 -37.0 10.0]
[10.0 37.0 4.0 -6.0]
[37.0 -10.0 6.0 4.0]
m := delta*pi/180 = [0.06981317008 -0.1047197551 -0.1745329252 -0.6457718232]
[ 0.1047197551 0.06981317008 -0.6457718232 0.1745329252]
[ 0.1745329252 0.6457718232 0.06981317008 -0.1047197551]
[ 0.6457718232 -0.1745329252 0.1047197551 0.06981317008]
a = [-0.8160820057 -0.01219775137 -0.02032958562 -0.07521946678]
[0.01219775137 -0.8160820057 -0.07521946678 0.02032958562]
[0.02032958562 0.07521946678 -0.8160820057 -0.01219775137]
[0.07521946678 -0.02032958562 0.01219775137 -0.8160820057]
b = [ 0.2616975215 0.214297521 0.3571625349 1.321501379]
[ -0.214297521 0.2616975215 1.321501379 -0.3571625349]
[-0.3571625349 -1.321501379 0.2616975215 0.214297521]
[ -1.321501379 0.3571625349 -0.214297521 0.2616975215]
Re z = [ 0.25729461 0.153964933 0.2566082216 0.9494504199]
[ -0.153964933 0.25729461 0.9494504199 -0.2566082216]
[-0.2566082216 -0.9494504199 0.25729461 0.153964933]
[-0.9494504199 0.2566082216 -0.153964933 0.25729461]
Im z = [ -0.9567507311 0.03246906896 0.05411511493 0.2002259252]
[-0.03246906896 -0.9567507311 0.2002259252 -0.05411511493]
[-0.05411511493 -0.2002259252 -0.9567507311 0.03246906896]
[ -0.2002259252 0.05411511493 -0.03246906896 -0.9567507311]
Mod z = [ -0.186167613 -0.04592389501 -0.07653982502 -0.2831973525]
[0.04592389501 -0.186167613 -0.2831973525 0.07653982502]
[0.07653982502 0.2831973525 -0.186167613 -0.04592389501]
[ 0.2831973525 -0.07653982502 0.04592389501 -0.186167613]
cos Arg z = [-2.796572984 -0.1371635402 -0.2286059004 -0.8458418315]
[0.1371635402 -2.796572984 -0.8458418315 0.2286059004]
[0.2286059004 0.8458418315 -2.796572984 -0.1371635402]
[0.8458418315 -0.2286059004 0.1371635402 -2.796572984] = [fraction] beta/gamma^p sometimes
sin Arg z = [0.9426264614 -0.4069351612 -0.6782252686 -2.509433494]
[0.4069351612 0.9426264614 -2.509433494 0.6782252686]
[0.6782252686 2.509433494 0.9426264614 -0.4069351612]
[ 2.509433494 -0.6782252686 0.4069351612 0.9426264614]
tan Arg z = [-0.03516496267 0.1472368196 0.2453946994 0.9079603876]
[ -0.1472368196 -0.03516496267 0.9079603876 -0.2453946994]
[ -0.2453946994 -0.9079603876 -0.03516496267 0.1472368196]
[ -0.9079603876 0.2453946994 -0.1472368196 -0.03516496267]
==> cos( [3.0 0.0 0.0 0.0]
[0.0 3.0 0.0 0.0]
[0.0 0.0 3.0 0.0]
[0.0 0.0 0.0 3.0] * arccos((tan^2(m)-a)/b)) = cos Arg z
delta = [124.0 -6.0 -10.0 -37.0]
[ 6.0 124.0 -37.0 10.0]
[ 10.0 37.0 124.0 -6.0]
[ 37.0 -10.0 6.0 124.0]
m := delta*pi/180 = [ 2.164208272 -0.1047197551 -0.1745329252 -0.6457718232]
[0.1047197551 2.164208272 -0.6457718232 0.1745329252]
[0.1745329252 0.6457718232 2.164208272 -0.1047197551]
[0.6457718232 -0.1745329252 0.1047197551 2.164208272]
a = [-0.8160820057 -0.01219775137 -0.02032958562 -0.07521946678]
[0.01219775137 -0.8160820057 -0.07521946678 0.02032958562]
[0.02032958562 0.07521946678 -0.8160820057 -0.01219775137]
[0.07521946678 -0.02032958562 0.01219775137 -0.8160820057]
b = [ 0.2616975215 0.214297521 0.3571625349 1.321501379]
[ -0.214297521 0.2616975215 1.321501379 -0.3571625349]
[-0.3571625349 -1.321501379 0.2616975215 0.214297521]
[ -1.321501379 0.3571625349 -0.214297521 0.2616975215]
Re z = [-0.2041383099 -0.1007894302 -0.1679823836 -0.6215348194]
[ 0.1007894302 -0.2041383099 -0.6215348194 0.1679823836]
[ 0.1679823836 0.6215348194 -0.2041383099 -0.1007894302]
[ 0.6215348194 -0.1679823836 0.1007894302 -0.2041383099]
Im z = [ -0.5890042646 0.02041656189 0.03402760317 0.1259021316]
[-0.02041656189 -0.5890042646 0.1259021316 -0.03402760317]
[-0.03402760317 -0.1259021316 -0.5890042646 0.02041656189]
[ -0.1259021316 0.03402760317 -0.02041656189 -0.5890042646]
Mod z = [ -0.186167613 -0.04592389501 -0.07653982502 -0.2831973525]
[0.04592389501 -0.186167613 -0.2831973525 0.07653982502]
[0.07653982502 0.2831973525 -0.186167613 -0.04592389501]
[ 0.2831973525 -0.07653982502 0.04592389501 -0.186167613]
cos Arg z = [ 1.884829414 0.07644037436 0.127400624 0.4713823087]
[-0.07644037436 1.884829414 0.4713823087 -0.127400624]
[ -0.127400624 -0.4713823087 1.884829414 0.07644037436]
[ -0.4713823087 0.127400624 -0.07644037436 1.884829414] = [fraction] beta/gamma^p sometimes
sin Arg z = [0.5736246678 -0.2511695786 -0.4186159644 -1.548879068]
[0.2511695786 0.5736246678 -1.548879068 0.4186159644]
[0.4186159644 1.548879068 0.5736246678 -0.2511695786]
[ 1.548879068 -0.4186159644 0.2511695786 0.5736246678]
tan Arg z = [0.07336041509 -0.1362336953 -0.2270561589 -0.8401077879]
[ 0.1362336953 0.07336041509 -0.8401077879 0.2270561589]
[ 0.2270561589 0.8401077879 0.07336041509 -0.1362336953]
[ 0.8401077879 -0.2270561589 0.1362336953 0.07336041509]
==> cos( [3.0 0.0 0.0 0.0]
[0.0 3.0 0.0 0.0]
[0.0 0.0 3.0 0.0]
[0.0 0.0 0.0 3.0] * arccos((tan^2(m)-a)/b)) = cos Arg z

Below the 'bi-quaternion': (3,4) + (6,8)i + (10,20)j + (37,39)k ; p = 3 ; i^2 = j^2 = k^2 = -1

delta = [ (3.0 + 4.0j) (-6.0 - 8.0j) (-10.0 - 20.0j) (-37.0 - 39.0j)]
[ (6.0 + 8.0j) (3.0 + 4.0j) (-37.0 - 39.0j) (10.0 + 20.0j)]
[(10.0 + 20.0j) (37.0 + 39.0j) (3.0 + 4.0j) (-6.0 - 8.0j)]
[(37.0 + 39.0j) (-10.0 - 20.0j) (6.0 + 8.0j) (3.0 + 4.0j)]
m := delta*pi/180 = [(0.05235987756 + 0.06981317008j) (-0.1047197551 - 0.1396263402j) (-0.1745329252 - 0.3490658504j) (-0.6457718232 - 0.6806784083j)]
[ (0.1047197551 + 0.1396263402j) (0.05235987756 + 0.06981317008j) (-0.6457718232 - 0.6806784083j) (0.1745329252 + 0.3490658504j)]
[ (0.1745329252 + 0.3490658504j) (0.6457718232 + 0.6806784083j) (0.05235987756 + 0.06981317008j) (-0.1047197551 - 0.1396263402j)]
[ (0.6457718232 + 0.6806784083j) (-0.1745329252 - 0.3490658504j) (0.1047197551 + 0.1396263402j) (0.05235987756 + 0.06981317008j)]
a = [ (-0.9823679967 + 0.2235804878j) (0.0162080435 - 0.01109890549j) (0.04009725789 - 0.01793437468j) (0.07966963086 - 0.06931714229j)]
[ (-0.0162080435 + 0.01109890549j) (-0.9823679967 + 0.2235804878j) (0.07966963086 - 0.06931714229j) (-0.04009725789 + 0.01793437468j)]
[(-0.04009725789 + 0.01793437468j) (-0.07966963086 + 0.06931714229j) (-0.9823679967 + 0.2235804878j) (0.0162080435 - 0.01109890549j)]
[(-0.07966963086 + 0.06931714229j) (0.04009725789 - 0.01793437468j) (-0.0162080435 + 0.01109890549j) (-0.9823679967 + 0.2235804878j)]
b = [ (1.0195121 - 1.129522723j) (-0.06875963585 - 0.004667902568j) (-0.1494040378 - 0.03777323999j) (-0.370070555 + 0.01770437451j)]
[(0.06875963585 + 0.004667902568j) (1.0195121 - 1.129522723j) (-0.370070555 + 0.01770437451j) (0.1494040378 + 0.03777323999j)]
[ (0.1494040378 + 0.03777323999j) (0.370070555 - 0.01770437451j) (1.0195121 - 1.129522723j) (-0.06875963585 - 0.004667902568j)]
[ (0.370070555 - 0.01770437451j) (-0.1494040378 - 0.03777323999j) (0.06875963585 + 0.004667902568j) (1.0195121 - 1.129522723j)]
Re z = [ (-0.3698273216 + 0.3013064752j) (-0.008724672152 - 0.05438227852j) (0.002558632624 - 0.1231308816j) (-0.08030676209 - 0.2849920705j)]
[(0.008724672152 + 0.05438227852j) (-0.3698273216 + 0.3013064752j) (-0.08030676209 - 0.2849920705j) (-0.002558632624 + 0.1231308816j)]
[(-0.002558632624 + 0.1231308816j) (0.08030676209 + 0.2849920705j) (-0.3698273216 + 0.3013064752j) (-0.008724672152 - 0.05438227852j)]
[ (0.08030676209 + 0.2849920705j) (0.002558632624 - 0.1231308816j) (0.008724672152 + 0.05438227852j) (-0.3698273216 + 0.3013064752j)]
Im z = [ (-0.5877099231 - 0.4384877699j) (0.04566808056 + 0.03927263299j) (0.08476072401 + 0.1046670248j) (0.2682165826 + 0.1814016501j)]
[(-0.04566808056 - 0.03927263299j) (-0.5877099231 - 0.4384877699j) (0.2682165826 + 0.1814016501j) (-0.08476072401 - 0.1046670248j)]
[ (-0.08476072401 - 0.1046670248j) (-0.2682165826 - 0.1814016501j) (-0.5877099231 - 0.4384877699j) (0.04566808056 + 0.03927263299j)]
[ (-0.2682165826 - 0.1814016501j) (0.08476072401 + 0.1046670248j) (-0.04566808056 - 0.03927263299j) (-0.5877099231 - 0.4384877699j)]
Mod z = [ (-0.416836924 - 0.190958951j) (0.003459564238 + 0.05988414779j) (-0.01634261779 + 0.1331289508j) (0.05560224466 + 0.3176364194j)]
[(-0.003459564238 - 0.05988414779j) (-0.416836924 - 0.190958951j) (0.05560224466 + 0.3176364194j) (0.01634261779 - 0.1331289508j)]
[ (0.01634261779 - 0.1331289508j) (-0.05560224466 - 0.3176364194j) (-0.416836924 - 0.190958951j) (0.003459564238 + 0.05988414779j)]
[ (-0.05560224466 - 0.3176364194j) (-0.01634261779 + 0.1331289508j) (-0.003459564238 - 0.05988414779j) (-0.416836924 - 0.190958951j)]
cos Arg z = [ (-0.3324620903 - 1.697245848j) (0.2425360862 - 0.04249407155j) (0.5505770182 + 0.003527476998j) (1.268796376 - 0.3772907155j)]
[ (-0.2425360862 + 0.04249407155j) (-0.3324620903 - 1.697245848j) (1.268796376 - 0.3772907155j) (-0.5505770182 - 0.003527476998j)]
[(-0.5505770182 - 0.003527476998j) (-1.268796376 + 0.3772907155j) (-0.3324620903 - 1.697245848j) (0.2425360862 - 0.04249407155j)]
[ (-1.268796376 + 0.3772907155j) (0.5505770182 + 0.003527476998j) (-0.2425360862 + 0.04249407155j) (-0.3324620903 - 1.697245848j)] = [fraction] beta/gamma^p sometimes
sin Arg z = [ (2.1443828 - 0.3224297845j) (0.04240410447 + 0.1917512319j) (0.01658853711 + 0.4388143521j) (0.3453236817 + 0.9976610339j)]
[(-0.04240410447 - 0.1917512319j) (2.1443828 - 0.3224297845j) (0.3453236817 + 0.9976610339j) (-0.01658853711 - 0.4388143521j)]
[(-0.01658853711 - 0.4388143521j) (-0.3453236817 - 0.9976610339j) (2.1443828 - 0.3224297845j) (0.04240410447 + 0.1917512319j)]
[ (-0.3453236817 - 0.9976610339j) (0.01658853711 + 0.4388143521j) (-0.04240410447 - 0.1917512319j) (2.1443828 - 0.3224297845j)]
tan Arg z = [ (0.07631256997 + 2.251756695j) (0.2072214583 - 0.001707164894j) (0.4565700741 + 0.0791328205j) (1.105504145 - 0.1375935647j)]
[(-0.2072214583 + 0.001707164894j) (0.07631256997 + 2.251756695j) (1.105504145 - 0.1375935647j) (-0.4565700741 - 0.0791328205j)]
[ (-0.4565700741 - 0.0791328205j) (-1.105504145 + 0.1375935647j) (0.07631256997 + 2.251756695j) (0.2072214583 - 0.001707164894j)]
[ (-1.105504145 + 0.1375935647j) (0.4565700741 + 0.0791328205j) (-0.2072214583 + 0.001707164894j) (0.07631256997 + 2.251756695j)]
==> cos( [3.0 0.0 0.0 0.0]
[0.0 3.0 0.0 0.0]
[0.0 0.0 3.0 0.0]
[0.0 0.0 0.0 3.0] * arccos((tan^2(m)-a)/b)) = cos Arg z

Delta(2):

v &= \cfrac{2m}{1 - m^2} \\
- vm^2 - 2m + v &= 0 \\
m^2 &= x \Rightarrow m = \sqrt{x} \\
v - vx &= 2 sqrt{x} \\
v^2(x^2 - 2x + 1) = 4x
v^2 x^2 -2 x(v^2 + 2) + v^2 = 0
Delta = 4 (v^2 + 2)^2 - 4 v^4 = 16(v^2 + 1)
x = [v^2 + 2 \pm 2 sqrt{v^2 + 1}]/v^2

a = 1 + 2v^2
s = pa
c = 1
r = Mod z = d/2 = 2v^2(1 + v^2)
sqrt(r) = v sqrt{2} sqrt{1 + v^2}

x &= s/2 + 2 \sqrt{r} \cos (t/2)
[v^2 + 2 \pm 2 sqrt{v^2 + 1}]/v^2 = 1 + 2v^2 + 2 sqrt{r} \cos (t/2)
[-v^4 + 1 - sqrt{v^2 + 1}]^2/v^4 = 2 v^2 (1 + v^2) \cos^2 (t/2)
cos Arg z = [-v^4 + 1 - E]^2/[v^6 * E^2] - 1

Delta(4):

w = f(v) como acima
x = f(sqrt(w))

E = sqrt{v^2 + 1}
w = (v^2 + 2 - 2 E)/v^2
x = [w + 2 - 2 sqrt{w + 1}]/w
x = 2v * [v - sqrt{2v^2 + 2 - 2 E}]/(v^2 + 2 - 2 E) + 1

a = 3 + 4v^2
c = 38 + 32 v^2
d = 6a^2 - c = 6(9 + 16v^4 + 24v^2) - 38 - 32v^2 = 96v^4 + 112 v^2 + 16
r = Mod z = (b/2)^4 = (d/p)^2 = d^2/16 = 2^4(6v^4 + 7v^2 + 1)^2

x = a + 2 sqrt[4](r) cos (t/4)
cos(4t) = 2cos^2(2t) - 1 = 2(2c^2 - 1)^2 - 1 = 2(4c^4 - 4c^2 + 1) - 1
2v * [v - sqrt{2v^2 + 2 - 2 E}]/(v^2 + 2 - 2 E) + 1 = 3 + 4v^2 + 2*2 sqrt{6v^4 + 7v^2 + 1} cos(t/4)
[v * [v - sqrt{2v^2 + 2 - 2 E}]/(v^2 + 2 - 2 E) - 1 - 2v^2]/[2 sqrt{6v^4 + 7v^2 + 1}] = c
cos Arg z = 8c^4 - 8c^2 + 1

Exercise. We have 2, 3 and 4. induction(p). "The world wants" cos Arg z as a function of p.

[Vinicius Claudino Ferraz]

no.
if a_1 ~ a_2 (mod p) ~ a_3 (mod q) are brothers,
he who constructs the angle a_1, he also constructs the angles (a_n).
and vice versa
he who does not construct the angle a_1, he does neither construct the angles (a_n).

it's like Q and R - Q. The constructible is densely greater then the unconstrutible.

https://drive.google.com/file/d/1XxcsllczeqMo0EgA46PoAXwis4--qJuM/view?usp=sharing

Vinicius Claudino Ferraz
2020/12/09th

[Vinicius Claudino Ferraz]

playlist

https://www.youtube.com/playlist?list=PLNaTqAGfaUsTLYx1HB6f4-soO8sUXQb2X

[www.claudino.webs.com /]

x^5 + rx + s = 0
This code in Python works: https://drive.google.com/file/d/1XxcsllczeqMo0EgA46PoAXwis4--qJuM/view?usp=sharing

See also this paper I have found. It seems it's an answer for n=5: https://arxiv.org/abs/math/0005026v1

I'm going to solve tan 5x as a function of tan x.

a = \tan 2t = \cfrac{2m}{1 - m^2} \\
b = \tan 3t = \cfrac{3m - m^3}{1 - 3m^2} \\
v = \tan 5t = \cfrac{a + b}{1 - ab} = \cfrac{2m(1 - 3m^2) + (3m - m^3)(1 - m^2)}{(1 - m^2)(1 - 3m^2) - 2m(3m - m^3)} \\
= \cfrac{m^5 - 10m^3 + 5m}{5m^4 - 10m^2 + 1} \\
m^5 - 5vm^4 - 10m^3 + 10vm^2 + 5m - v = 0
m^2 = x \Rightarrow m = \sqrt{x} \\
x^2\sqrt{x} - 5vx^2 - 10x\sqrt{x} + 10vx + 5\sqrt{x} - v = 0
x(x^2 - 10x + 5)^2 = v^2(5x^2 - 10x + 1)^2
x^5 + (- 20 - 25v^2)x^4 + (110 + 100v^2) x^3 + (- 100 - 110v^2) x^2 + (25 + 20v^2) x - v^2 = 0
EXERCISE: Tschirnhaus HERE.

I know the five roots of this quintic.
v = \tan \theta
x = \tan^2 ((|\theta| + k \pi)/5)

[Vinicius Claudino Ferraz]

Utility:
z = a + bi
|z| = 1
tau = b/a
cos(1/p arctan tau) = Re w
w = z^(1/p)

We almost have how to evaluate cos(1/3 arctan tau) = f(tau)
We almost have how to evaluate cos(1/p arctan tau) = f(tau, p)

1/p = n
z = cos t + i sin t
tau = tan t
w = z^n
Re w = cos nt = f(cos t)
Chebyshev, cos 2t, cos 3t, cos 4t, ...
https://en.wikipedia.org/wiki/Chebyshev_polynomials#Examples

Exercise: amplify Chebyshev to real p // complex p.
cos(1/p * t) = f(cos t)

[Vinicius Claudino Ferraz]

Registering: tan nt = p(tan t)
https://pbs.twimg.com/media/EsH_cxGW4AAz11w?format=jpg&name=medium

\begin{align*}
2 \le n \in \mathbb{N}, \forall \theta \in \text{GL}(4, \mathbb{C}) &\Rightarrow \sum_{i = 0}^n (-1)^i \binom{n}{i} (\tan \theta)^{n - i} (\tan n\theta)^{[1 - (-1)^i]/2} = 0
\end{align*}

out of charity, there is no salvation at all.

[sobriquet]

Post by Vinicius Claudino Ferraz
third ugly number
https://www.wolframalpha.com/input/?i=cos%281%2F3+*+arctan%285*sqrt%285+%2B+2*sqrt%285%29%29%2F%2890+%2B+41*sqrt%285%29%29%29%29+-+1%2F4+sqrt%281%2F11+%28119+-+5+sqrt%285%29+%2B+sqrt%2830+%2885+%2B+31+sqrt%285%29%29%29%29%29
fourth ugly number
https://www.wolframalpha.com/input/?i=cos%281%2F3+*+arctan%285*sqrt%285+%2B+2*sqrt%285%29%29%2F%2890+%2B+41*sqrt%285%29%29%29%29+-+1%2F4+sqrt%281%2F11+%28119+-+5+sqrt%285%29+%2B+sqrt%2830+%2885+%2B+31+sqrt%285%29%29%29%29%29

Links for 3rd and 4th one are identical?

Post by Vinicius Claudino Ferraz
I've been thinking about x = z = r(cos t + i sen t)
and if 0 <= t < pi then abs(x) := z
else if pi <= t < 2 pi then abs(x) := -z (opposite of negative, for example)
because it came from sqrt(x^2). it "maybe" will hold in half plane.
Somebody please teach me. According to wikipedia,
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Extension_to_complex_plane
arctan(2i) = integral from 0 to 1 of dz/(1 + z^2) + integral from 1 to 2i of dz/(1 + z^2)
The first is arctan(1) = pi/4. And the second?
I forgot how to integrate f(z) dz from a to b.
Everything used to end up in Cauchy and number of holes.
Vinicius

[Vinicius Claudino Ferraz]

Post by sobriquet
Links for 3rd and 4th one are identical?

Twin numbers. Try this: https://www.facebook.com/mathspirituality/posts/3719921324685949

Let me summarize my "paper", maybe somebody can help.

tan t = m
tan 2t = 2m/(1 - m^2)
tan 3t = v = \cfrac{3m - m^3}{1 - 3m^2} \\

We want m extracted as a function of m.
m^3 - 3vm^2 - 3m + v = 0 \\

Align 3 points as below, for trisecting some angles.

A = (a, 0) \\
B = (b, bm) \Rightarrow AB: y = bm\cdot\cfrac{x - a}{b - a} \\
C = \biggl( c, c \cdot \cfrac{3m - m^3}{1 - 3m^2} \biggl) \in AB \\
a = bc\cdot\cfrac{v - m}{cv - bm}\\
\cfrac{v}{m} = \cfrac{b}{c} \cdot \cfrac{c - a}{b - a)} = \cfrac{1}{3} - \cfrac{8}{9m^2 - 3} \\
m^2 = x \Rightarrow m = \sqrt{x} \\

x\sqrt{x} - 3vx - 3\sqrt{x} + v = 0 \\
v^2(3x - 1)^2 = x(x - 3)^2 \\
v^2(9x^2 - 6x + 1) = x^3 - 6x^2 + 9x \\
s = 6 + 9v^2 \\
q = 9 + 6v^2 \\
p = v^2 \\
x = y + s/3 \\
y^3 + Py + Q = 0 \\
P = q - s^2/3 = -3(9v^4 + 10v^2 + 1) \\
Q = -2/27 s^3 + q s/3 - p = -2(27v^6 + 45v^4 + 17v^2 - 1) \\
Delta = Q^2/4 + P^3/27 = - 64v^2(v^2 + 1)^2 \\
w = a + bi = r e^{it} \\
a = -Q/2 = (v^2 + 1) (27 v^4 + 18 v^2 - 1) \\
b = \sqrt{-\Delta} = 8|v|(v^2 + 1) < 0 \\
x = s/3 + 2 \sqrt[3]{r} \cos (t/3)

RENAME x TO tan²( pi/3 + 1/3 arctan |x| )
RENAME s/3 TO 2 + 3x²
RENAME \sqrt[3]{r} TO sqrt{x² + 1} sqrt{9x² + 1}
RENAME t TO arctan 8|x|/(27x^4 + 18x² - 1), BUT THE IMAGE IS IN (0, pi]

Replace x by tan 9k * pi/180.

HABEMUS UGLY NUMBERS
https://pbs.twimg.com/media/EmPUMiIXIAIeK3Z?format=jpg&name=medium

Let my tree be as follows:
a = 1 = tan 45°
aa = 2/11
ab = 11/2
aaa = 21296/5497
aab = 5497/21296
aba = 704/404003
abb = 404003/704
They are countable. Are they dense? Are they completeable?

Now we might go to fifth degree.

a = \tan 2t = \cfrac{2m}{1 - m^2} \\
b = \tan 3t = \cfrac{3m - m^3}{1 - 3m^2} \\
v = \tan 5t = \cfrac{a + b}{1 - ab} = \cfrac{2m(1 - 3m^2) + (3m - m^3)(1 - m^2)}{(1 - m^2)(1 - 3m^2) - 2m(3m - m^3)} \\
= \cfrac{m^5 - 10m^3 + 5m}{5m^4 - 10m^2 + 1} \\
m^5 - 5vm^4 - 10m^3 + 10vm^2 + 5m - v = 0
m^2 = x \Rightarrow m = \sqrt{x} \\
x^2\sqrt{x} - 5vx^2 - 10x\sqrt{x} + 10vx + 5\sqrt{x} - v = 0
x(x^2 - 10x + 5)^2 = v^2(5x^2 - 10x + 1)^2
x^5 + (- 20 - 25v^2)x^4 + (110 + 100v^2) x^3 + (- 100 - 110v^2) x^2 + (25 + 20v^2) x - v^2 = 0

I know the five roots of this quintic.
v = \tan \theta
x = \tan^2 ((|\theta| + k \pi)/5)

The question now is:
I want to suppose that it is solvable by radicals (then what?)
calculate its discriminant somehow, like Cardano's method,
and explicit x = f(v).

Somebody might go to seventh degree. LOL

sweatfully

Vinicius

[Vinicius Claudino Ferraz]

2² + 11² = 5³
10² + 55² = 5⁵
25² + 50² = 5⁵
38² + 41² = 5⁵
4² + 4² = 2⁵

The paper has a next page.

https://pbs.twimg.com/media/EmUMhr0XcAIFJPM?format=png&name=small

I used to know how-to solve by radicals. Why did I forget?

All we need is automation. Give me a faster processor! ; - D

procedure search for ugly numbers(degree: integer);
begin
if degree < 5 then exit;
if not is prime(degree) then exit;
...

Post by Vinicius Claudino Ferraz
sweatfully
Vinicius

[Vinicius Claudino Ferraz]

This is a verification in R++ Language that the trisection formula works for complex variables.

https://drive.google.com/file/d/1Oyf_OeTErjXKppyjUW551A76spVSfdBI/view?usp=sharing

Post by Vinicius Claudino Ferraz
sweatfully
Vinicius