Finding a new way to calculate one side of a triangle

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In summary, the current method for calculating the side of a triangle is using the Pythagorean theorem, but a new method is needed due to its limitations. This new method involves using trigonometric functions and is based on the laws of sines and cosines. Its benefits include versatility, accuracy, and applicability to real-world scenarios.
  • #1
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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  • #2
...r0 and r1 are constants...
If r1=r0+x, then d1 is related to r0 and r1 by pythagoras.
 
  • #3
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
 
  • #4
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
 
  • #5
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
 
  • #6
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.
 
  • #7
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
 
  • #9
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 $$ before and after your LaTeX. Here's what your stuff looks like with this addition.
$$\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.
 

1. What is the current method for calculating the side of a triangle?

The current method for calculating the side of a triangle is by using the Pythagorean theorem, which states that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

2. Why is there a need for a new method of calculating the side of a triangle?

While the Pythagorean theorem is a reliable and widely-used method, it can be complicated and time-consuming for certain types of triangles, such as those with non-integer side lengths or those with angles other than 90 degrees. Therefore, there is a need for a more efficient and accurate method for calculating triangle sides.

3. How does a new method of calculating the side of a triangle work?

The new method involves using trigonometric functions, such as sine, cosine, and tangent, to determine the length of the missing side. This method is based on the relationship between the sides and angles of a triangle, known as the law of sines and the law of cosines.

4. What are the benefits of using a new method for calculating triangle sides?

The new method offers more versatility and accuracy, as it can be used for all types of triangles and does not require the use of square roots or approximation. It also allows for easier calculations and can be applied to more complex geometric problems.

5. Can the new method be used in real-world applications?

Yes, the new method can be applied in various fields, such as engineering, architecture, and surveying. It can be used to accurately measure distances and angles in real-world scenarios, making it a useful tool for solving practical problems.

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