8 ms·

As the math teacher at the first year of university once said: "i know two things: [cos(x)]^2 + [sin(x)]^2 = 1 i*i = -1 but can re-derive all the rest"

eru 6y ago
Good luck deriving group theory from that. :) (Though, yes, there's a bit of a connection with group theory, but not enough to help you.)
- dragontamer 6y ago
  Well, unit-circles over a finite field are a periodic "square". So you gotta redefine the norm from norm = sqrt(x^2 + y^2) into norm = max(x, y). From there, you get extension fields from real vs imaginary already. (Ex: you can form a new extension field from x + y*j, where x and y are complex numbers), which forms a new periodic cycle. I mean, deriving it all is hard because group theory is hard. I'm not sure if its because the tools "aren't there". Some super-AI or super-human probably can derive it all from those given facts.
  - eru 6y ago
    Yes, I know about the unit circle. But that's just a way to embed some of group theory. It doesn't actually help you much. (It's similar to how you can use eg set theory to construct the integers. It's possible, but doesn't actually help you prove anything about interesting about integers that you wouldn't have been able to prove without embedding them like this.) > I mean, deriving it all is hard because group theory is hard. I'm not sure if its because the tools "aren't there". Some super-AI or super-human probably can derive it all from those given facts. Yes, but that task wouldn't be made easier by this approach compared to starting from just the group theory axioms instead.
amelius 6y ago
From these two facts, you can come to the conclusion that sin(x) == 0 and cos(x) == 1, for any x. I.e., you need to know more.
- filmor 6y ago
  No, you can't. You can come to the conclusion that if there exits an x with sin(x) = 0, then cos(x) = 1. The expression as it stands is true for every x \in R without any additional assumptions.
  - amelius 6y ago
    Yes, I agree that was badly worded, but in any case you need more than those two equations to do anything meaningful with sin(x) and cos(x).
  - laszlokorte 6y ago
    but you could conclude that cos and sin are both jsut the constant function: x->sqrt(0.5)
    - filmor 6y ago
      Not at all, same reason.
lrmunoz 6y ago
My math teacher was in love with the Euler's identity written like this: 1+e^(i*pi) = 0 He said it relates the most important numbers and constants of the universe. He had it embroidered and framed on his office wall
- jacobolus 6y ago
  “Euler’s identity” is a flashy way of stating that a half-turn rotation in the plane is equivalent to a reflection across the center of rotation. This is a fact which can be easily understood by young children. The part that makes the expression flashy is that the rotation is written via its logarithm, which can be represented as the length of a circular arc (“angle measure”). It’s a useful fact to know, but is vastly overrated.
MayeulC 6y ago
Well, as far as I know, cosinus and sinus have relatively little to do with exponential and imaginary numbers, just like imaginary numbers have intrinsically little to do with the 2D plane. I just think of it as a convenient notation since those obey similar composition rules (that probably stem from their odd and even-ness, and taylor series decomposition from there). So you need at least that one, since it is by convention, and not really derivable[1]: exp(ix) = cos(x) + i*sin(x) I personally just use Euler's notation (using exp) as a tool to derive trig formulas, since I know power composition rules quite well already, and those tend to be more useful in general. [cos(x)]^2 + [sin(x)]^2 = 1 Is just the Pythagorean theorem on a unit circle, which also defines sin and cos :) [1] OK, you can look at the taylor series expansion, but you need to remember the derivatives for this.
- dragontamer 6y ago
  cos(x) and sin(x) are just the unit circles applied over the 2-dimesnional real + imaginary axis. Because x^2 + y^2 = 1 defines a unit circle, if you perform x^2 + i * y^2 = 1, you get a unit circle over the real/imaginary planes. Or to put it in a picture: https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Euler%27s_formula.svg/1024px-Euler%27s_formula.svg.png https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Eu... Remember the definition of sin / cos: https://upload.wikimedia.org/wikipedia/commons/b/bd/Sine_and_cosine_animation.gif https://upload.wikimedia.org/wikipedia/commons/b/bd/Sine_and... ------- Once you realize that "the imaginary axis" is just an arbitrary 2-dimension extension field (and that "imaginary" is a very bad name for it), it becomes way easier to see. ------- Now if you had a 3-dimension graph (real, imaginary, and time), and then you project the unit-circle over the real + time axis, you get a sin (or cos) respectively. ----------- Or to put it another way: there's a periodic nature of i: * i = i * i^2 = -1 * i^3 = -i * i^4 = 1 -- Cycle-length 4 * i^5 = i * i ^ x = i^(x mod 4) (In general) This period itself forms a circle, moving from real-to-imaginary and back. (e^i^x) = (e^(ix)) therefore is periodic as well. And the most natural periodic cycle is a circle (which is of course, just sin/cos). To make this "period 4" into a proper circle, multiply by pi/2. i^(pi/2)^x == period 3.141592.... or the "circumference" of the unit circle. e^i^(pi/2)^x == e^(i * pi/2 * x). Done.
- gmadsen 6y ago

eru 6y ago

Good luck deriving group theory from that. :) (Though, yes, there's a bit of a connection with group theory, but not enough to help you.)

dragontamer 6y ago
Well, unit-circles over a finite field are a periodic "square". So you gotta redefine the norm from norm = sqrt(x^2 + y^2) into norm = max(x, y). From there, you get extension fields from real vs imaginary already. (Ex: you can form a new extension field from x + y*j, where x and y are complex numbers), which forms a new periodic cycle. I mean, deriving it all is hard because group theory is hard. I'm not sure if its because the tools "aren't there". Some super-AI or super-human probably can derive it all from those given facts.
- eru 6y ago
  Yes, I know about the unit circle. But that's just a way to embed some of group theory. It doesn't actually help you much. (It's similar to how you can use eg set theory to construct the integers. It's possible, but doesn't actually help you prove anything about interesting about integers that you wouldn't have been able to prove without embedding them like this.) > I mean, deriving it all is hard because group theory is hard. I'm not sure if its because the tools "aren't there". Some super-AI or super-human probably can derive it all from those given facts. Yes, but that task wouldn't be made easier by this approach compared to starting from just the group theory axioms instead.

amelius 6y ago

From these two facts, you can come to the conclusion that sin(x) == 0 and cos(x) == 1, for any x. I.e., you need to know more.

filmor 6y ago
No, you can't. You can come to the conclusion that if there exits an x with sin(x) = 0, then cos(x) = 1. The expression as it stands is true for every x \in R without any additional assumptions.
- amelius 6y ago
  Yes, I agree that was badly worded, but in any case you need more than those two equations to do anything meaningful with sin(x) and cos(x).
- laszlokorte 6y ago
  but you could conclude that cos and sin are both jsut the constant function: x->sqrt(0.5)
  - filmor 6y ago
    Not at all, same reason.

lrmunoz 6y ago

My math teacher was in love with the Euler's identity written like this: 1+e^(i*pi) = 0 He said it relates the most important numbers and constants of the universe. He had it embroidered and framed on his office wall

jacobolus 6y ago
“Euler’s identity” is a flashy way of stating that a half-turn rotation in the plane is equivalent to a reflection across the center of rotation. This is a fact which can be easily understood by young children. The part that makes the expression flashy is that the rotation is written via its logarithm, which can be represented as the length of a circular arc (“angle measure”). It’s a useful fact to know, but is vastly overrated.

MayeulC 6y ago

Well, as far as I know, cosinus and sinus have relatively little to do with exponential and imaginary numbers, just like imaginary numbers have intrinsically little to do with the 2D plane. I just think of it as a convenient notation since those obey similar composition rules (that probably stem from their odd and even-ness, and taylor series decomposition from there). So you need at least that one, since it is by convention, and not really derivable[1]: exp(ix) = cos(x) + i*sin(x) I personally just use Euler's notation (using exp) as a tool to derive trig formulas, since I know power composition rules quite well already, and those tend to be more useful in general. [cos(x)]^2 + [sin(x)]^2 = 1 Is just the Pythagorean theorem on a unit circle, which also defines sin and cos :) [1] OK, you can look at the taylor series expansion, but you need to remember the derivatives for this.

dragontamer 6y ago
cos(x) and sin(x) are just the unit circles applied over the 2-dimesnional real + imaginary axis. Because x^2 + y^2 = 1 defines a unit circle, if you perform x^2 + i * y^2 = 1, you get a unit circle over the real/imaginary planes. Or to put it in a picture: https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Euler%27s_formula.svg/1024px-Euler%27s_formula.svg.png https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Eu... Remember the definition of sin / cos: https://upload.wikimedia.org/wikipedia/commons/b/bd/Sine_and_cosine_animation.gif https://upload.wikimedia.org/wikipedia/commons/b/bd/Sine_and... ------- Once you realize that "the imaginary axis" is just an arbitrary 2-dimension extension field (and that "imaginary" is a very bad name for it), it becomes way easier to see. ------- Now if you had a 3-dimension graph (real, imaginary, and time), and then you project the unit-circle over the real + time axis, you get a sin (or cos) respectively. ----------- Or to put it another way: there's a periodic nature of i: * i = i * i^2 = -1 * i^3 = -i * i^4 = 1 -- Cycle-length 4 * i^5 = i * i ^ x = i^(x mod 4) (In general) This period itself forms a circle, moving from real-to-imaginary and back. (e^i^x) = (e^(ix)) therefore is periodic as well. And the most natural periodic cycle is a circle (which is of course, just sin/cos). To make this "period 4" into a proper circle, multiply by pi/2. i^(pi/2)^x == period 3.141592.... or the "circumference" of the unit circle. e^i^(pi/2)^x == e^(i * pi/2 * x). Done.
gmadsen 6y ago