Finding a new way to calculate one side of a triangle

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The discussion revolves around finding an alternative method to calculate one side of a triangle (d1) without using the law of cosines, particularly focusing on integrating the function (cos a)^(1/2). Participants explore the relationship between the sides and angles, noting that while r1 equals r0 plus x, the absence of right angles complicates the calculations. Suggestions include using elliptic integrals, with one user sharing a specific integral formula and seeking further assistance on a more complex version. The conversation also touches on LaTeX formatting for mathematical expressions and methods for simplifying trigonometric 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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