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Let's say we take a triple integral and use it to find "3d angles", which use the idea of of taking a portion of a 3-d sphere, compared to the entire sphere.

If 2-d trig ratios can be used to find side-lengths of 2-d triangles, is it possible we can use "3-d trig ratios" to find "surface area sections" of 3-d right pyramids. A "right" 3-d angle is one-fourth of a sphere with a volume of 1 cubic centimeter.

So for finding surface areas, 3-d trigonometry could be used to solve the surface areas of right pyramids by only knowing one side of a surface area.

3-d Pythagorean theorem can be used to be applied to surface areas, or at least that's what I think?

So are there any questions or comments? Is this "possible", could this require help from several people?

Arbuja
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So indeed there do exist concepts called Solid angles and there is a version of the pythagorean theorem in 3d space (not only for line segment lengths BUT ALSO for entire areas of SIDES) which is extremely elegant.

That being the issue with attempting to create a trigonmetry for angles in 3d space is that that there are infinitely many such angles! Since a solid angle can have an arbitrary number of line segments connecting to it and therefore to gain the some control over 3d geometry as 2d geometry we need to essentially define

Sin for the 4 sided pyramid Sin for the 5 sided pyramid Sin for the 6 sided pyramid

Etc...

As far as I know no one had done this but it is definitely something interesting to look at.

And I would imagine you would traditionally consider various ratios of side lengths as well as face areas to generate your functions (why not try to prove some funky identities?)

Now if you generalized the concept of an 'Angle' as a collection of various pieces of information (how many lines are connecting to it, what are the angles between these lines, etc...) then you can define your new ocean of trig functions over which angles are valid arguments and which aren't and see if you can find some identities.

Good luck!

Sidharth Ghoshal
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    Hey, it's has been a long time, but I noticed a peculiar relation with the surface areas of a right triangular pyramid. The squares of each of the 3 surface areas is equal to the square of the largest surface area (I call it the "hypotenuse surface"). More work to come with that. Thanks! – Arbuja Jun 24 '14 at 23:27
  • Good Observation! See if you can prove it :) – Sidharth Ghoshal Jun 26 '14 at 00:44