Finding a new way to calculate one side of a triangle

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Discussion Overview

The discussion revolves around finding an alternative method to calculate one side of a triangle, specifically focusing on the integration of trigonometric functions. Participants explore the relationship between triangle sides and angles, particularly in the context of non-right triangles, and the use of elliptic integrals.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant seeks a method to compute d1 without using the law of cosines, expressing difficulty in integrating (cos a)^(1/2).
  • Another participant suggests that if r1 = r0 + x, then d1 could be related to r0 and r1 through the Pythagorean theorem, although they note that there are no right angles in the figure.
  • Some participants discuss the visualization of trigonometric relationships using a unit circle, but express skepticism about its utility in the absence of right angles.
  • A participant proposes looking into elliptic integrals as a potential solution for the integration challenge.
  • There is a discussion about the specific form of elliptic integrals, with one participant noting that their problem involves a circle traced from an off-center point rather than a constant radius.
  • Another participant provides a link to a resource for elliptic integrals and discusses approximations for small angles.
  • One participant mentions a more complex integral they are trying to solve, which Mathematica could not handle, and seeks advice on how to format LaTeX correctly in the forum.
  • There are suggestions to expand trigonometric expressions and utilize Euler's formula for simplification.

Areas of Agreement / Disagreement

Participants express a range of views on the methods to approach the problem, with no consensus reached on a definitive solution or method. The discussion remains unresolved regarding the best approach to compute d1 and the integration techniques involved.

Contextual Notes

Participants highlight the complexity of integrating (cos a)^(1/2) and the specific conditions of their geometric setup, which may affect the applicability of standard methods. There is also mention of limitations in Mathematica's ability to solve certain integrals.

nearc
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hopefully this is the right place for this question, the first part is a trig/geometry question but it is really a integration question:

i'm trying to find another way to compute d1 without using law of cosines because i don't know how to integrate (cos a)^.5, if someone knows how to do that please let me know.

in the figure provided I'm attempting to find d1 as a function of angle a, r0 and r1 are constants and angles t, p and lengths d0, x can be computed trivially.

i can use law of sines to find d1 if i can make b, m or c a function of a
 

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...r0 and r1 are constants...
If r1=r0+x, then d1 is related to r0 and r1 by pythagoras.
 
This diagram is a handy way to visualize the relationship between the trig functions and the angle using a unit circle: (credit: wikipedia)

500px-Circle-trig6.svg.png
 
Simon Bridge said:
If r1=r0+x, then d1 is related to r0 and r1 by pythagoras.

yes, r1=r0+x, but there are no right angles in the figure
 
SteamKing said:
This diagram is a handy way to visualize the relationship between the trig functions and the angle using a unit circle: (credit: wikipedia)

500px-Circle-trig6.svg.png

thanks i'll reflect on this, but i suspect that this would only be useful in the case where i have right angles
 
nearc said:
yes, r1=r0+x, but there are no right angles in the figure
Ah - so d1 is not a tangent ... it looks a lot like it is supposed to be one in the figure.

This is where I suspect you need to look for another way to set up the integral - or learn to use elliptic integrals.
 
Simon Bridge said:
Ah - so d1 is not a tangent ... it looks a lot like it is supposed to be one in the figure.

This is where I suspect you need to look for another way to set up the integral - or learn to use elliptic integrals.


can you give me some good lead for elliptic integrals? thansk
 
Simon Bridge said:
$$\int \cos^{\frac{1}{2}}\!(x)\; dx = 2E\big(\frac{x}{2}\big|2\big)$$

##E(x|m)##
Elliptic Integral of the seconds kind [properties]

thanks for your help, but I've looked through some basic elliptical integrals and I'm not finding anything like (cos x)^.5 the basic form is 1-(sin x )^2

also what I'm working with is not an ellipse it is a circle. the difference is that instead of a constant radius from the center tracing out the circle it is circle traced from a point off center and instead of a constant radius length the this segment length changes
 
  • #10
"elliptic integral" not "elliptical".
It is the name for a class of functions - they have application beyond elliptic curves.

How about:
http://integrals.wolfram.com/index.jsp?expr=sqrt(cos(x))&random=false

Um - note: if x is very small compared with r0, then cos(m)~1 and sin(m)~m and tan(m)~m.

In the diagram in post #3 - your triangles are AOD and ADE.
The exsec side is you side x.
AO=OD=r0
AD=d0
... like that.
 
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  • #11
thanks to simon I've figured out the simple elliptic however i now have a more difficult version:

$$\int_0^{2 \pi } \sqrt{\left(1-\frac{60}{61} \cos (x-179)\right) \left(1-\frac{60}{61} \cos (x-t)\right)} \, dx $$

using mathematica i can find the solution to

$$\int \sqrt{1-\frac{60}{61} \cos (x-t)} \, dx $$

but mathematica was not able to solve the harder version, any ideas?

also how do get the latex to show correctly?

thanks mark44
 
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  • #12
nearc said:
thanks to simon I've figured out the simple elliptic however i know have a more difficult version:

\int_0^{2 \pi } \sqrt{\left(1-\frac{60}{61} \cos (x-179)\right) \left(1-\frac{60}{61} \cos (x-t)\right)} \, dx

using mathematica i can find the solution to

\int \sqrt{1-\frac{60}{61} \cos (x-t)} \, dx

but mathematica was not able to solve the harder version, any ideas?

also how do get the latex to show correctly?
Add $$[/color] before and after your LaTeX. Here's what your stuff looks like with this addition.
$$\int_0^{2 \pi } \sqrt{\left(1-\frac{60}{61} \cos (x-179)\right) \left(1-\frac{60}{61} \cos (x-t)\right)} \, dx $$

$$\int \sqrt{1-\frac{60}{61} \cos (x-t)} \, dx
$$
 
  • #13
$$\left(1-\frac{60}{61} \cos (x-179)\right) \left(1-\frac{60}{61} \cos (x-t)\right)$$ ... hmmm? expand the brackets and try trig identities?
A quick path can be to expand the trig function by the Euler relations since manipulating exponentials is usually easier.
 
  • #14
thanks simon, are you referring to

$$e^{\text{ix}}=x \cos + i x \sin $$
 
  • #15
nearc said:
thanks simon, are you referring to

$$e^{\text{ix}}=x \cos + i x \sin $$
That does not make any sense!
You mean:
$$e^{\text{ix}}=\cos(x) + i \sin(x) $$... then yes
http://en.wikipedia.org/wiki/Euler's_formula
 
  • #16
yep that's it, mathematica likes to switch things?
 
  • #17
Dunno - never used mathematica.
 

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