Polygons inscribed in a circle: Let there be an equilateral triangle inscribed in a circle and the measure of the sides...

[prince mike]

Polygons inscribed in a circle of a diameter of 1 unit:

Let there be an equilateral triangle inscribed in a circle and the measure of the sides of the triangle is a measure of an angle of sin of 60°, as the measures of the angles of the triangles decrease by a multiple of 2 the sides of the triangles increase by a multiple of 2. From 3 sides to 6 sides to 12 sides to 24 etc... The sides of the polygons are a measure of angles with isosceles triangles in each polygons and in Isosceles triangle you only need one angle to find the lenght of the sides of the triangles since the sides correspond to the measure of an angle.

The following statements hold true for isosceles triangles:

$\ cos C=1-(\ (cos A)^2 + \ (cos B)^2)$

$\ sin C=1-(\ (sin A)^2 + \ (sin B)^2)$

I posted it in Wikipedia and it was an original search and

 

[prince mike]

Polygons inscribed in a circle of a diameter of 1 unit:

Let there be an equilateral triangle inscribed in a circle and the measure of the sides of the triangle is a measure of an angle of sin of 60°, as the measures of the angles of the triangles decrease by a multiple of 2 the sides of the triangles increase by a multiple of 2. From 3 sides to 6 sides to 12 sides to 24 etc... The sides of the polygons are a measure of angles with isosceles triangles in each polygons and in Isosceles triangle you only need one angle to find the lenght of the sides of the triangles since the sides correspond to the measure of an angle.

The following statements hold true for isosceles triangles:

[math]\ cos C=1-(\ (cos A)^2 + \ (cos B)^2)[/math]
[math]\ sin C=1-(\ (sin A)^2 + \ (sin B)^2)[/math]
I posted it in Wikipedia and it was an original research.

 

[The Highlander]

I posted it in Wikipedia and it was an original research.

Why have you started two threads saying essentially the same thing and what exactly is your question?

Is this what you are looking for?

 

[The Highlander]

I posted it in Wikipedia and it was an original search and

Why have you started two threads saying essentially the same thing and what exactly is your question?

Is this what you are looking for?

 

[Dr.Peterson]

Let there be an equilateral triangle inscribed in a circle and the measure of the sides of the triangle is a measure of an angle of sin of 60°, as the measures of the angles of the triangles decrease by a multiple of 2 the sides of the triangles increase by a multiple of 2. From 3 sides to 6 sides to 12 sides to 24 etc... The sides of the polygons are a measure of angles with isosceles triangles in each polygons and in Isosceles triangle you only need one angle to find the lenght of the sides of the triangles since the sides correspond to the measure of an angle.

This sounds like a distorted version of Archimedes calculation of pi; it's not true that the (interior) angles of a regular polygon are halved when the number of sides is doubled, if that's what you are saying here.

The following statements hold true for isosceles triangles:

[math]\ cos C=1-(\ (cos A)^2 + \ (cos B)^2)[/math]
[math]\ sin C=1-(\ (sin A)^2 + \ (sin B)^2)[/math]

If A and B are the equal base angles, then the first is almost correct, but the second is wrong.

Please show why you think they are true -- whether it's a proof or an example.

I posted it in Wikipedia and it was an original research.

There's a reason they don't accept "original research". They want everything to be correct, as demonstrated in published sources.

 

[stapel]

Why have you started two threads saying essentially the same thing and what exactly is your question?

Threads merged.

 

[prince mike]

This sounds like a distorted version of Archimedes calculation of pi; it's not true that the (interior) angles of a regular polygon are halved when the number of sides is doubled, if that's what you are saying here.


If A and B are the equal base angles, then the first is almost correct, but the second is wrong.

Please show why you think they are true -- whether it's a proof or an example.


There's a reason they don't accept "original research". They want everything to be correct, as demonstrated in published sources.

[math]\ sin \frac{60}{2^x}(3)(2^x)=\pi[/math]
[math]\ cos C=1-(\ (cos A)^2 + \ (cos B)^2)[/math]
This let's you divide each angle by 2,staring from [math]\ sin 60=\sqrt{ 0.75}[/math] and after you find the angle ,you multiply by the number of sides for each gons. It's the same as saying for cos C
for example:

1×1×2=2
2×2×2=8
3×3×2=18
4×4×2=32
5×5×2=50

and you subtract from 100 instead of 1.

[math]\ sin C=1-(\ (sin A)^2 + \ (sin B)^2)[/math]

 

[Dr.Peterson]

[math]\ sin \frac{60}{2^x}(3)(2^x)=\pi[/math]
[math]\ cos C=1-(\ (cos A)^2 + \ (cos B)^2)[/math]
This let's you divide each angle by 2,staring from [math]\ sin 60=\sqrt{ 0.75}[/math] and after you find the angle ,you multiply by the number of sides for each gons. It's the same as saying for cos C
for example:

1×1×2=2
2×2×2=8
3×3×2=18
4×4×2=32
5×5×2=50

and you subtract from 100 instead of 1.

[math]\ sin C=1-(\ (sin A)^2 + \ (sin B)^2)[/math]

This is not quite coherent.

The first line ought at least to have a limit in it, shouldn't it?

The second line is correct, but I asked how you justified it, and you haven't done that. (It is mostly a double angle formula; and since you are assuming A and B are equal, it seems inappropriate to distinguish them.)

I have no idea what the five multiplications are there for, or what you want to subtract from 100.

And, again, the last line is simply not true. Please explain why you think it is true.

 

[prince mike]

[math]\ sin \frac{60}{2^x}(3)(2^x)=\pi[/math]
[math]\ cos C=1-(\ (cos A)^2 + \ (cos B)^2)[/math]
This let's you divide each angle by 2,staring from [math]\ sin 60=\sqrt{ 0.75}[/math] and after you find the angle ,you multiply by the number of sides for each gons. It's the same as saying for cos C
for example:

1×1×2=2
2×2×2=8
3×3×2=18
4×4×2=32
5×5×2=50

and you subtract from 100 instead of 1.

[math]\ sin C=1-(\ (sin A)^2 + \ (sin B)^2)[/math]

This is not quite coherent.

The first line ought at least to have a limit in it, shouldn't it?

The second line is correct, but I asked how you justified it, and you haven't done that. (It is mostly a double angle formula; and since you are assuming A and B are equal, it seems inappropriate to distinguish them.)

I have no idea what the five multiplications are there for, or what you want to subtract from 100.

And, again, the last line is simply not true. Please explain why you think it is true.

The last line gives you the opposite of cos C and for cos C is only true for isosceles triangles where 2 angles A and B are of the same, and in each polygons the sides corresponds to the measure of sin 60°,sin 30°,sin 15°,etc...
If you draw a triangle of 60° in a circle of a diameter of 1 unit the measure of the side of the triangle is[math]\sqrt 0.75[/math] and from the side there is an opening of an angle of sin 30° which is for the hexagon and the side lenght is of 0.5 and so one and in each side there is an isosceles triangle. I hope it's clear

 

[Dr.Peterson]

The last line gives you the opposite of cos C and for cos C is only true for isosceles triangles where 2 angles A and B are of the same, and in each polygons the sides corresponds to the measure of sin 60°,sin 30°,sin 15°,etc...

No, you are just making a claim, not supporting it. (What does "opposite of cos C" even mean??)

But then, I don't see that you've actually used your claimed identities in your work, so maybe it doesn't matter, except that you aren't giving me any reason to think you are thinking clearly.

If you draw a triangle of 60° in a circle of a diameter of 1 unit the measure of the side of the triangle is[math]\sqrt {0.75}[/math] and from the side there is an opening of an angle of sin 30° which is for the hexagon and the side lenght is of 0.5 and so one and in each side there is an isosceles triangle. I hope it's clear

I'm not sure whether your lack of clarity is due to not knowing English, or not understanding how to reason in math. Some of what you say almost makes sense, but then it doesn't. Try breaking up your sentences and defining what you are talking about (what hexagon? what opening? what isosceles triangles?).

 

[prince mike]

No, you are just making a claim, not supporting it. (What does "opposite of cos C" even mean??)

But then, I don't see that you've actually used your claimed identities in your work, so maybe it doesn't matter, except that you aren't giving me any reason to think you are thinking clearly.


I'm not sure whether your lack of clarity is due to not knowing English, or not understanding how to reason in math. Some of what you say almost makes sense, but then it doesn't. Try breaking up your sentences and defining what you are talking about (what hexagon? what opening? what isosceles triangles?).

 

[Dr.Peterson]

I can make no sense of this at all. You haven't made any attempt yet to justify your claims, irrelevant as they are to your work in approximating pi, which would work if you actually carried it out.

You claim that
[math]\cos C=1-((\cos A)^2 + (\cos B)^2)[/math][math]\sin C=1-((\sin A)^2 + (\sin B)^2)[/math]Let's just take one example. Consider the isosceles triangle with base angles A=B=70^o and apex angle C=40^o.

Then
[math]\cos C=\cos40^\circ=0.7660444[/math][math]1-((\cos A)^2 + (\cos B)^2)=1-((\cos70^\circ)^2 + (\cos70^\circ)^2)=0.7660444[/math]These agree; that doesn't prove you're right, but you could show it using a double-angle identity.


[math]\sin C=\sin40^\circ=0.6427876[/math][math]1-((\sin A)^2 + (\sin B)^2)=1-((\sin70^\circ)^2 + (\sin70^\circ)^2)=-0.7660444[/math]These don't agree.

Now I'm finished with you, unless you interact appropriately in the future.

 

[prince mike]

1)The apex of an equilateral triangle which is called also Isosceles triangle inscribed with in the circle of a diameter of 1 unit is 60° and the base facing the apex is sin of 60°. 180° divided by 60° is equal to 3 as well, which are the 3 sides of the triangle.
2)For an hexagon with 6 equal angles of 120° and one of the angle is sin of 120° which is the apex facing the base and the side of the triangle with vertex of 30° and 30°. 120°+30°+30°=180° and 180÷120=6 representing 6 equal sides.
3)next is a polygon with 12 equal sides and 180°÷12=15°. 15°+15°+150°=180°.In this case the apex is 150° and sin of 150° is the base of the triangle which is the side of the hexagon equal to 0.5 unit of lenght.
4)next level is a polygon with 24 equal sides. 180°÷24=7.5°
7.5°+7.5°+165°=180° and sin of 165° is the apex with sin of 15° being the base and the side of the polygon with 12 sides.

[math]\sin B\cos C+\sin C\cos B=\sin A[/math]
[math]\sin B\sin C-\cos C\cos B=\cos A[/math]
[math]\sin A\cos C+\sin C\cos A=\sin B[/math]
[math]\sin A\sin C-\cos C\cos A=\cos B[/math]
[math]\sin A\cos B+\sin B\cos A=\sin C[/math]
[math]\sin A\sin B-\cos A\cos B=\cos C[/math]
Length of the sides of the triangles:

[math]b\times \cos C+c\times \cos B=a[/math]
[math]a\times \cos C+c\times \cos A=b[/math]
[math]a\times \cos B+b\times \cos A=c[/math]

 

[prince mike]

1)The apex of an equilateral triangle which is called also Isosceles triangle inscribed with in the circle of a diameter of 1 unit is 60° and the base facing the apex is sin of 60°. 180° divided by 60° is equal to 3 as well, which are the 3 sides of the triangle.
2)For an hexagon with 6 equal angles of 120° and one of the angle is sin of 120° which is the apex facing the base and the side of the triangle with vertex of 30° and 30°. 120°+30°+30°=180° and 180÷120=6 representing 6 equal sides.
3)next is a polygon with 12 equal sides and 180°÷12=15°. 15°+15°+150°=180°.In this case the apex is 150° and sin of 150° is the base of the triangle which is the side of the hexagon equal to 0.5 unit of lenght.
4)next level is a polygon with 24 equal sides. 180°÷24=7.5°
7.5°+7.5°+165°=180° and sin of 165° is the apex with sin of 15° being the base and the side of the polygon with 12 sides.

[math]\sin B\cos C+\sin C\cos B=\sin A[/math]
[math]\sin B\sin C-\cos C\cos B=\cos A[/math]
[math]\sin A\cos C+\sin C\cos A=\sin B[/math]
[math]\sin A\sin C-\cos C\cos A=\cos B[/math]
[math]\sin A\cos B+\sin B\cos A=\sin C[/math]
[math]\sin A\sin B-\cos A\cos B=\cos C[/math]
Length of the sides of the triangles:

[math]b\times \cos C+c\times \cos B=a[/math]
[math]a\times \cos C+c\times \cos A=b[/math]
[math]a\times \cos B+b\times \cos A=c[/math]

Correction

1)The apex of an equilateral triangle which is called also Isosceles triangle inscribed with in the circle of a diameter of 1 unit is 60° and the base facing the apex is sin of 60°. 180° divided by 60° is equal to 3 as well, which are the 3 sides of the triangle.
2)For an hexagon with 6 equal angles of 120° and one of the angle is sin of 120° which is the apex facing the base and the side the equilateral triangle with the hexagon vertex having angles of 30° and 30° and apex of 120°. 120°+30°+30°=180° and 180÷6=30° representing 6 equal sides.
3)next is a polygon with 12 equal sides and 180°÷12=15°. 15°+15°+150°=180°.In this case the apex is 150° and sin of 150° is the base of the triangle which is the side of the hexagon equal to 0.5 unit of lenght.
4)next level is a polygon with 24 equal sides. 180°÷24=7.5°
7.5°+7.5°+165°=180° and sin of 165° is the apex with sin of 15° being the base and the side of the polygon with 12 sides.

[math]\sin B\cos C+\sin C\cos B=\sin A[/math]
[math]\sin B\sin C-\cos C\cos B=\cos A[/math]
[math]\sin A\cos C+\sin C\cos A=\sin B[/math]
[math]\sin A\sin C-\cos C\cos A=\cos B[/math]
[math]\sin A\cos B+\sin B\cos A=\sin C[/math]
[math]\sin A\sin B-\cos A\cos B=\cos C[/math]
Length of the sides of the triangles:

[math]b\times \cos C+c\times \cos B=a[/math]
[math]a\times \cos C+c\times \cos A=b[/math]
[math]a\times \cos B+b\times \cos A=c[/math]