Trisecting an angle - Page 3

xilman · 2005-12-03, 18:29

Quote:

Originally Posted by R.D. Silverman

I see nothing wrong with your doing so for your own education and
entertainment. In fact, I applaud such efforts.
...
(2) Caiming that the result will be "valuable information" is ludicrous beyond
words. This information is *useless*. And it is easily reconstructed.

Which information?

I agree that a list of angles would be useless.

On the other hand, a nicely laid out exposition of how the angles may be constructed with compass and unmarked straight edge would be quite educational for others to read and would be worth the effort of posting. I'd like to see a geometric construction of a 257-gon, for instance. I know it's possible but I've never seen such a construction. The analogous proof of constructibility of a 65537-gon may be too much of a challenge...

Don't you like seeing new and elegant proofs of a theorem that has been proved long before by other means?

It's possible that information about a "ready-reckoner" may also be useful to me if it leads me to a tool which I find useful and which is more convenient to use than, say bc(1) or Pari/gp.

Paul

alpertron · 2005-12-04, 01:08

Quote:

Originally Posted by mfgoode

I will try and tabulate the different angles starting from 3* angles up to 90* angles which can be constructed geometrically. Do you by chance have a ready reckoner for these constructions? It will be valuable information for the forum.

After constructing the 3 degree angle it is trivial to construct any multiple of it.

Maybe you want the formulas for sines and cosines of multiples of 3 degrees.

sin 0º = cos 90º = 0

sin 3º = cos 87º = $\sqrt{-\sqrt{5/128-\sqrt{5}/128}-\sqrt{15}/16-\sqrt{3}/16+1/2}$

sin 6º = cos 84º = $\sqrt{15/32-3*\sqrt{5}/32}-\sqrt{5}/8-1/8$

sin 9º = cos 81º = $-\sqrt{5/16-\sqrt{5}/16}+\sqrt{10}/8+\sqrt{2}/8$

sin 12º = cos 78º = $\sqrt{\sqrt{5}/32+5/32}-\sqrt{15}/8+\sqrt{3}/8$

sin 15º = cos 75º = $\sqrt{6}/4-\sqrt{2}/4$

sin 18º = cos 72º = $\sqrt{5}/4-1/4$

sin 21º = cos 69º = $\sqrt{-\sqrt{\sqrt{5}/128+5/128}-\sqrt{15}/16+\sqrt{3}/16+1/2}$

sin 24º = cos 66º = $-\sqrt{5/32-\sqrt{5}/32}+\sqrt{15}/8+\sqrt{3}/8$

sin 27º = cos 63º = $\sqrt{\sqrt{5}/16+5/16}-\sqrt{10}/8+\sqrt{2}/8$

sin 30º = cos 60º = $1/2$

sin 33º = cos 57º = $\sqrt{\sqrt{5/128-\sqrt{5}/128}-\sqrt{15}/16-\sqrt{3}/16+1/2}$

sin 36º = cos 54º = $\sqrt{5/8-\sqrt{5}/8}$

sin 39º = cos 51º = $\sqrt{-\sqrt{\sqrt{5}/128+5/128}+\sqrt{15}/16-\sqrt{3}/16$ +1/2}

sin 42º = cos 48º = $\sqrt{3*\sqrt{5}/32+15/32}-\sqrt{5}/8+1/8$

sin 45º = cos 45º = $\sqrt{2}/2$

sin 48º = cos 42º = $\sqrt{\sqrt{5}/32+5/32}+\sqrt{15}/8-\sqrt{3}/8$

sin 51º = cos 39º = $\sqrt{\sqrt{\sqrt{5}/128+5/128}-\sqrt{15}/16+\sqrt{3}/16+1/2}$

sin 54º = cos 36º = $\sqrt{5}/4+1/4$

sin 57º = cos 33º = $\sqrt{-\sqrt{5/128-\sqrt{5}/128}+\sqrt{15}/16+\sqrt{3}/16+1/2}$

sin 60º = cos 30º = $\sqrt{3}/2$

sin 63º = cos 27º = $\sqrt{\sqrt{5}/16+5/16}+\sqrt{10}/8-\sqrt{2}/8$

sin 66º = cos 24º = $\sqrt{15/32-3*\sqrt{5}/32}+\sqrt{5}/8+1/8$

sin 69º = cos 21º = $\sqrt{\sqrt{\sqrt{5}/128+5/128}+\sqrt{15}/16-\sqrt{3}/16+1/2}$

sin 72º = cos 18º = $\sqrt{\sqrt{5}/8+5/8}$

sin 75º = cos 15º = $\sqrt{6}/4+\sqrt{2}/4$

sin 78º = cos 12º = $\sqrt{3*\sqrt{5}/32+15/32}+\sqrt{5}/8-1/8$

sin 81º = cos 9º = $\sqrt{5/16-\sqrt{5}/16}+\sqrt{10}/8+\sqrt{2}/8$

sin 84º = cos 6º = $\sqrt{5/32-\sqrt{5}/32}+\sqrt{15}/8+\sqrt{3}/8$

sin 87º = cos 3º = $\sqrt{\sqrt{5/128-\sqrt{5}/128}+\sqrt{15}/16+\sqrt{3}/16+1/2}$

sin 90º = cos 0º = 1

mfgoode · 2005-12-04, 16:15

Quote:

Originally Posted by R.D. Silverman

I see nothing wrong with your doing so for your own education and
entertainment. In fact, I applaud such efforts.

However:

(1) Unless you restrict to angles that are an integral number of degrees, your
list will be infinite. (e.g. bisecting a 45 degree angle etc.)

]

Yes Bob, I agree that restricting the angles to integral numbers is a fine suggestion. Thanks for the tip..I appreciate your direct and straight from the shoulder pointers which are concise, precise and to the point

Quote Fatphil[I had to google for him (which lead immediately to wikipedia).Author of /The Trisectors/, it appears.
Thanks Bob - my Christmas pressy wish list has grown by one!
Phil ]

Following your example I checked on Dudley and found it interesting. I may also order one of his books

Quote Xilman [Don't you like seeing new and elegant proofs of a theorem that has been proved long before by other means?

It's possible that information about a "ready-reckoner" may also be useful to me if it leads me to a tool which I find useful and which is more convenient to use than, say bc(1) or Pari/gp.]

Thanks Paul for the encouragement I will start work on this ready reckoner.
The fact is that all constructable numbers are algebraic.

Quote Alpertron: [Maybe you want the formulas for sines and cosines of multiples of 3 degrees.

sin 0º = cos 90º = 0

sin 3º = cos 87º = ]

Thanks Alpertron for the table of sines and cosines.
Please give me an example say sin 3* angle as to how you arrive at the value
in the sq.rt.form?

I have come across a wonderful method to prove that a heptagon inscribed in a unit circle cannot be constructed by unmarked ruler and compass.
It is yours for the asking. Also you can come across it in "What is Mathematics?"
By Courant and Robbins
Please excuse my method of quoting as I still dont have the hang of it
Mally

alpertron · 2005-12-05, 17:47

Quote:

Originally Posted by mfgoode

]
Please give me an example say sin 3* angle as to how you arrive at the value
in the sq.rt.form?

In order to compute sin 3º we have to first compute sin 72º and cos 72º (equivalent to constructing the pentagon) and then sin 60º and cos 60º (equivalent to constructing the hexagon). Using the difference of angle formulas we can find sin (72º - 60º) = sin 12º and cos (72º - 60º) = cos 12º, and then using the formulas for halving angles twice we can find sin 3º and cos 3º.

First step: computing sin 72º and cos 72º:

We can start with the formula for multiple angles:

$cos\, 5t\, =\, (cos\,t)^5\, -\, 10 (cos\, t)^3\, (sin\, t)^2\, +\, 5 (cos\, t) (sin\, t)^4$ .

When t = 18º we get cos 5t = cos 90º = 0

Let c = cos t, s = sin t.

$\large 0 = c^5 - 10 c^3 s^2 + 5 cs^4$

c is not zero so we can divide both sides by c.

$\large 0 = c^4 - 10 c^2 s^2 + 5 s^4$

Since (cos t)^2 + (sin t)^2 = 1: (1)

$\large 0 = (1-s^2)^2 - 10 s^2 (1-s^2) + 5 s^4$

$\large 0 = 1 - 12s^2 + 16s^4$

This is a biquadratic equation that gives four solutions, one of them is sin 18º = cos 72º = $(\sqrt 5 - 1)/4$

From (1) we get the value of sin 72º.

The cos 60º and sin 60º are even easier to compute.

Then we use the following identities to compute sin 12º and cos 12º:

sin (a+b) = sin a cos b + cos a sin b
cos (a+b) = cos a cos b - sin a sin b

Let a = 72º and b = -60º, and of course, cos (-a) = cos a and sin (-a) = -sin a.

Finally use the formulas:

$\large \cos\left(\frac{x}{2}\right) = \pm\, \sqrt{\frac{1 + \cos(x)}{2}}$

$\large \sin\left(\frac{x}{2}\right) = \pm\, \sqrt{\frac{1 - \cos(x)}{2}}$

twice to find first sin 6º and cos 6º and then sin 3º and cos 3º

nibble4bits · 2005-12-12, 00:36

Note that you can only get an approximation but using the same techniques that allow for 3 degrees divided by any power of 2 could be used to create other angles inbetween. 3/2+3/4=1.5+0.75=2.25 -> That's 2.25 degrees from combining the trisection of two different harmonics of 3 degrees. I use the word 'harmonics' here in the radio frequency way - k/n or k*n are both harmonics for k=constant and n=nth harmonic. Actually, the same techniques I've seen here for going from 12 degrees to 3 degrees are also used to determine interference from two radio stations. There's a reason that FM stations tend to be near 3-digit primes.

Imagine frequencies A and B of a pure carrier wave (no signal).
Interference examples:
A+B
A-B
A/6+B
mA+nB; m and n = +/- harmonics, A and B = carrier frequency/wavelength.
Note that wavelength and frequency are inversely related using that simple little constant 'c' but I'm sure most of the people reading this know that! :P Sorry if my sense of humor is dry.

It can be argued that since the carriers are waves, then you can use trigonometry to determine their interactions. Of course it does, since we're talking circles=cycles. If A's wavelength is shorter than B, then from A's persepective B is always lagged behind. That is, for every A you will never see a complete copy of B since A ends too fast. A way to draw this, is to assume that 360 degrees is A, and B's angle is greater then 360 degrees since it's slower.

Does anyone have any diagrams explaining this in a visual way?

nibble4bits · 2005-12-12, 01:15

Here's one I made real quick. :)

mfgoode · 2005-12-13, 10:20

[QUOTE=alpertron]In order to compute sin 3º we have to first compute sin 72º and cos 72º (equivalent to constructing the pentagon) and then sin 60º and cos 60º (equivalent to constructing the hexagon). Using the difference of angle formulas we can find sin (72º - 60º) = sin 12º and cos (72º - 60º) = cos 12º, and then using the formulas for halving angles twice we can find sin 3º and cos 3º. ....etc./ Unquote.

Thanks Alpertron for the very valuable information. Im sorry I did not reply earlier.
Mally

mfgoode · 2005-12-13, 10:54

[QUOTE=nibble4bits]Note that you can only get an approximation but using the same techniques that allow for 3 degrees divided by any power of 2 could be used to create other angles in between. 3/2+3/4=1.5+0.75=2.25 -> That's 2.25 degrees from combining the trisection of two different harmonics of 3 degrees. I use the word 'harmonics' here in the radio frequency way - k/n or k*n are both harmonics for k=constant and n=nth harmonic. Actually, the same techniques I've seen here for going from 12 degrees to 3 degrees are also used to determine interference from two radio stations. There's a reason that FM stations tend to be near 3-digit primes. ...etc. Unquote/

Thank you Niddles4bits for your interest.

In simple language I understand you are saying that the process can be carried on indefinitely of halving a given angle.

Please read Silverman's post n. 21 point no. (1). Its useful to know the integral values of angles. Most angle values in cos and sin are approximate except for 0* and 2 pi and a few other particular trig signs like tan 45*.

The rest of your post though irrelevant is interesting. Its right up Alpertrons street and I leave it to him to answer.

Your point of FM stations tending to be near 3-digit primes is very interesting.
However the stations marked and which can be tuned into are mostly divisible by 3 on my FM radio.. Kindly clarify on this point

The thumb nail sketch resembles SHM of the beginning of a sinusoidal wave and if two waves are out of phase then interference can occur.

Regards the term 'c' its not as simple as it seems. I am labouring on a post for Xilman to get a thorough meaning of the term esp. in Relativity.
Mally

nibble4bits · 2005-12-13, 16:24

If I took ANY binary number and used the technique I showed you, then yes adding a series of halves+4ths+8ths+16ths+...(a_n)/(2^n) would generate approximations.
k/f(x) -> k=integer, f(x)=sum of (a_0)+(a_1)/2+...(a_n)/(2^n)
This formula should work. :)

I was thinking of stations such as 92.9 and 94.3 which are close to each other - the closer ones are the ones I'd worry about. 98.1, 98.5, and 98.9 are also deliberately arranged that way also.

I showed angles A and B in my diagram as coinciding every 8th (or 9th depending on point of view) cycle. In reallity they would take a lot longer period of time to even come close to alignment. I was lazy - it's was easy to draw. :P

For most people working with radio waves, c can be considered constant. If I'm making an antenna for 20 MHz then I know I need a length of twice the wavelength. "All else being equal" applies here - most people need to just simplify the process to make it effecient. Yes, I know c as a constant is misleading when you take account of nonuniform frames of reference.

<EDIT> Actually an even simpler formula is k*f(x)*2^-n where k and f(x) are integers for a constant and a sum, and n is the number of times to halve that constant.

alpertron · 2005-12-13, 19:24

By the way, I found that the polynomial

$65536X^{16}-262144X^{14}+430080X^{12}-372736X^{10}+182784X^8-50176X^6+7040X^4-384X^2+1$

has the roots sin 3°, sin 21°, sin 33°, sin 39°, sin 51°, sin 57°, sin 69°, sin 87° and their negatives.

nibble4bits · 2005-12-13, 21:45

In binary:
int(log(101010110100001010101011110101))=29 in decimal=11101

101010110100001010101011110101/(10^11101)=
101010110100001010101011110101/100000000000000000000000000000=
1.01010110100001010101011110101=~1.33797214366495609283447265625

To get the original string '101010110100001010101011110101' from '1.33797214366495609283447265625' is not too hard. The value of k can be constant=1 or whatever you need to estimate your solution. Too bad that any arbitary increase in binary digits will just end up making it slightly less wrong. :/ The point is not to make a table, but to make an algorythm that creates tables on the fly.

2005-12-12, 01:15	#28
nibble4bits Nov 2005 2·7·13 Posts	Here's one I made real quick. :) Attached Thumbnails

2005-12-13, 10:20	#29
mfgoode Bronze Medalist Jan 2004 Mumbai,India 4004₈ Posts	Trisecting an angle [QUOTE=alpertron]In order to compute sin 3º we have to first compute sin 72º and cos 72º (equivalent to constructing the pentagon) and then sin 60º and cos 60º (equivalent to constructing the hexagon). Using the difference of angle formulas we can find sin (72º - 60º) = sin 12º and cos (72º - 60º) = cos 12º, and then using the formulas for halving angles twice we can find sin 3º and cos 3º. ....etc./ Unquote. Thanks Alpertron for the very valuable information. Im sorry I did not reply earlier. Mally

2005-12-13, 10:54	#30
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2052₁₀ Posts	Trisecting an angle. [QUOTE=nibble4bits]Note that you can only get an approximation but using the same techniques that allow for 3 degrees divided by any power of 2 could be used to create other angles in between. 3/2+3/4=1.5+0.75=2.25 -> That's 2.25 degrees from combining the trisection of two different harmonics of 3 degrees. I use the word 'harmonics' here in the radio frequency way - k/n or kn are both harmonics for k=constant and n=nth harmonic. Actually, the same techniques I've seen here for going from 12 degrees to 3 degrees are also used to determine interference from two radio stations. There's a reason that FM stations tend to be near 3-digit primes. ...etc. Unquote/ Thank you Niddles4bits for your interest. In simple language I understand you are saying that the process can be carried on indefinitely of halving a given angle. Please read Silverman's post n. 21 point no. (1). Its useful to know the integral values of angles. Most angle values in cos and sin are approximate except for 0 and 2 pi and a few other particular trig signs like tan 45*. The rest of your post though irrelevant is interesting. Its right up Alpertrons street and I leave it to him to answer. Your point of FM stations tending to be near 3-digit primes is very interesting. However the stations marked and which can be tuned into are mostly divisible by 3 on my FM radio.. Kindly clarify on this point The thumb nail sketch resembles SHM of the beginning of a sinusoidal wave and if two waves are out of phase then interference can occur. Regards the term 'c' its not as simple as it seems. I am labouring on a post for Xilman to get a thorough meaning of the term esp. in Relativity. Mally

2005-12-13, 16:24	#31
nibble4bits Nov 2005 B6₁₆ Posts	If I took ANY binary number and used the technique I showed you, then yes adding a series of halves+4ths+8ths+16ths+...(a_n)/(2^n) would generate approximations. k/f(x) -> k=integer, f(x)=sum of (a_0)+(a_1)/2+...(a_n)/(2^n) This formula should work. :) I was thinking of stations such as 92.9 and 94.3 which are close to each other - the closer ones are the ones I'd worry about. 98.1, 98.5, and 98.9 are also deliberately arranged that way also. I showed angles A and B in my diagram as coinciding every 8th (or 9th depending on point of view) cycle. In reallity they would take a lot longer period of time to even come close to alignment. I was lazy - it's was easy to draw. :P For most people working with radio waves, c can be considered constant. If I'm making an antenna for 20 MHz then I know I need a length of twice the wavelength. "All else being equal" applies here - most people need to just simplify the process to make it effecient. Yes, I know c as a constant is misleading when you take account of nonuniform frames of reference. <EDIT> Actually an even simpler formula is kf(x)2^-n where k and f(x) are integers for a constant and a sum, and n is the number of times to halve that constant. Last fiddled with by nibble4bits on 2005-12-13 at 16:27

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2005-12-12, 00:36	#27
nibble4bits Nov 2005 2·7·13 Posts	Note that you can only get an approximation but using the same techniques that allow for 3 degrees divided by any power of 2 could be used to create other angles inbetween. 3/2+3/4=1.5+0.75=2.25 -> That's 2.25 degrees from combining the trisection of two different harmonics of 3 degrees. I use the word 'harmonics' here in the radio frequency way - k/n or k*n are both harmonics for k=constant and n=nth harmonic. Actually, the same techniques I've seen here for going from 12 degrees to 3 degrees are also used to determine interference from two radio stations. There's a reason that FM stations tend to be near 3-digit primes. Imagine frequencies A and B of a pure carrier wave (no signal). Interference examples: A+B A-B A/6+B mA+nB; m and n = +/- harmonics, A and B = carrier frequency/wavelength. Note that wavelength and frequency are inversely related using that simple little constant 'c' but I'm sure most of the people reading this know that! :P Sorry if my sense of humor is dry. It can be argued that since the carriers are waves, then you can use trigonometry to determine their interactions. Of course it does, since we're talking circles=cycles. If A's wavelength is shorter than B, then from A's persepective B is always lagged behind. That is, for every A you will never see a complete copy of B since A ends too fast. A way to draw this, is to assume that 360 degrees is A, and B's angle is greater then 360 degrees since it's slower. Does anyone have any diagrams explaining this in a visual way?

2005-12-13, 19:24	#32
alpertron Aug 2002 Buenos Aires, Argentina 2·683 Posts	By the way, I found that the polynomial $65536X^{16}-262144X^{14}+430080X^{12}-372736X^{10}+182784X^8-50176X^6+7040X^4-384X^2+1$ has the roots sin 3°, sin 21°, sin 33°, sin 39°, sin 51°, sin 57°, sin 69°, sin 87° and their negatives.

2005-12-13, 21:45	#33
nibble4bits Nov 2005 182₁₀ Posts	In binary: int(log(101010110100001010101011110101))=29 in decimal=11101 101010110100001010101011110101/(10^11101)= 101010110100001010101011110101/100000000000000000000000000000= 1.01010110100001010101011110101=~1.33797214366495609283447265625 To get the original string '101010110100001010101011110101' from '1.33797214366495609283447265625' is not too hard. The value of k can be constant=1 or whatever you need to estimate your solution. Too bad that any arbitary increase in binary digits will just end up making it slightly less wrong. :/ The point is not to make a table, but to make an algorythm that creates tables on the fly.