/sci/ - Science & Math

>>12620268
Triangles do not have side numbers, triangles have side [math]\bold{lengths}[/math]. Lengths are non-negative reals only.

>>12620268
Incorrect metric. A metric defines how you measure lengths in a space and squaring complex numbers isn't how it's done. You multiply the number with it's conjugate (a+bi => a - bi) so i*(-i)=1, you get sqrt(2).

[math]a = i[/math]
[math]b = 1[/math]
[math]c = 0[/math]
So:

[math]A = \frac{ab}{2} = \frac{i}{2}[/math]
According to heron's formula, which is [math]\sqrt{s(s-a)(s-b)(s-c)}[/math] the area ends up [math]\frac{i}{2}[/math] again

[math]a = b\cos\gamma+\sqrt{c^2-b^2\sin^2\gamma}[/math]
then
[math]i = 1 * 0 + \sqrt{0 - 1 * 1}[/math]
[math]i = \sqrt{-1}[/math]

[math]c = \sqrt{a^2 + b^2 - 2ab\cos\gamma}[/math]
then
[math]0 = \sqrt{-1 + 1 - 0}[/math]
[math]0 = 0[/math]

by definition
[math]\tan\alpha = \frac{a}{b}[/math]
[math]\tan\alpha = i[/math]

but also
[math]\tan\alpha = \frac{a\sin\gamma}{b-a\cos\gamma}[/math]
then
[math]\tan\alpha = \frac{i * 1}{1 - i * 0}[/math]
[math]\tan\alpha = i[/math]

There are a couple of divisions by zero but since this is schizo general they're not illegal just undefined.
Also the division by zero loses commutativity and associativity.

>>12620268
This sort of triangle actually occurs in special relativity; the zero-length hypotenuse is what you call a "light-like interval" or "null interval" and it represents how something moving at the speed of light doesn't experience any time.

>>12621507
god physicists are so fucking stupid

>>12621519
Its just a function, not a direct observable. Spacetime distance is not space distance.

>>12620268
Pythagorean theorem only applies on the euclidean metric, which is defined for real numbers. You can construct triangles with imaginary lengths, but Pythagorean theorem is no longer valid and now you need to use something like a Reimann metric to find the hypotenuse.

>>12620268

>>12621519
Physicist here. That guy is an idiot. That triangle doesn't exist in SR. We define a Minkowski Metric. Then you can construct a triangle with legs of values t, and r. The spacetime interval is given by s^2 = t^2 - r^2. Diagrammatically, it's a meaningless triangle used by brainlets who don't understand math.

>>12621535
>That guy is an idiot.
>repeats exactly what I said in different words

>>12620492
[math]\tan(ix) \to i[/math] as [math]x \to \infty[/math]

>>12620268
>can /sci/ explain this shit?
not every quadratic form is positive definite

>>12621535
Brainlet. Let (t, r) be (0, 0), (0, 1), and (1, 0) for three vertices of a triangle and compute s for all three sides.

>>12621535
what the fuck are you smoking, that guy is 100% right and you haven't said anything different than him

I was writing an elaborate response when realized I was just being jebaited by sum dumbass.

Lemme just leave here a quate from Gauss about the subject and an image, both from Wikipedia.

"If one formerly contemplated this subject from a false point of view and therefore found a mysterious darkness, this is in large part attributable to clumsy terminology. Had one not called +1, −1, √−1 positive, negative, or imaginary (or even impossible) units, but instead, say, direct, inverse, or lateral units, then there could scarcely have been talk of such darkness." ~ Gauss (1831)

Lastly fuck you OP.

>>12620492
let [math]r[/math] be the radius of the inscribed circle, [math]r = \sqrt{\frac{1}{s}(s-a)(s-b)(s-c)}[/math], where [math]r = \frac{1}{2}+\frac{i}{2}[/math]
and [math]s[/math] the semi-perimeter [math]s = \frac{a+b+c}{2}[/math], where [math]s = \frac{1}{2}+\frac{i}{2}[/math]

From the law of cotangents we have
[math]\cot \frac{\alpha}{2} = \frac{s - a}{r}[/math]

So we get
[math]\cot \frac{\alpha}{2} = \frac{\frac{1}{2}+\frac{i}{2} - i}{\frac{1}{2}+\frac{i}{2}}[/math]
[math]\cot \frac{\alpha}{2} = -i[/math]

Extending the law further:

[math]\cot \frac{\beta}{2} = \frac{\frac{1}{2}+\frac{i}{2} - 1}{\frac{1}{2}+\frac{i}{2}}[/math]
[math]\cot \frac{\beta}{2} = i[/math]

[math]\cot \frac{\gamma}{2} = \frac{\frac{1}{2}+\frac{i}{2} - 0}{\frac{1}{2}+\frac{i}{2}}[/math]
[math]\cot \frac{\gamma}{2} = 1[/math]

We can then conclude with the last identity which is
[math]\frac{\cot \frac{\alpha}{2}}{s - a} = \frac{\cot \frac{\beta}{2}}{s - b} = \frac{\cot \frac{\gamma}{2}}{s - c} = \frac{1}{r}[/math]
which becomes
[math]\frac{-i}{s - a} = \frac{i}{s - b} = \frac{1}{s - c} = \frac{1}{\frac{1}{2}+\frac{i}{2}}[/math]
which is proven to be true here

https://www.wolframalpha.com/input/?i=%28-i%2F%28s-a%29%29%3D%28i%2F%28s-b%29%29%3D%281%2F%28s-c%29%29%3D%281%2F%281%2F2%2Bi%2F2%29%29+where+a%3Di%2C+b%3D1%2C+c%3D0%2C+s%3D%281%2F2%2Bi%2F2%29

As you know
[math]\sin(ix) = i \sinh(x)[/math]
[math]\cos(ix) = \cosh(x)[/math]
[math]\tan(ix) = i \tanh(x)[/math]
[math]\sin(x+iy) = \sin(x) \cosh(y) + i \cos(x) \sinh(y)[/math]
[math]\cos(x+iy) = \cos(x) \cosh(y) - i \sin(x) \sinh(y)[/math]
[math]\tan(x+iy) = \displaystyle \frac{\tan(x) + i \tanh(y)}{1 - i \tan(x) \tanh(y)}[/math]

cos is finite for finite inputs, but it approaches complex infinity as the imaginary part of the angle approaches [math]\pm \infty[/math], with which direction it goes in the complex plane dependent on the real part of the angle. Similarly for sin.

[math]\tan(x + iy) \to i[/math] as [math]y \to \infty[/math] and [math]\tan(x + iy) \to -i[/math] as [math]y \to -\infty[/math].

We can't assign finite angle values to the two non-right angles of this triangle, but we do approach its trigonometric ratios when the angle opposite the side of length i has an imaginary part tending towards positive infinity and the angle opposite the side of length 1 has an imaginary part tending towards negative infinity.