CTK Exchange
Front Page
Movie shortcuts
Personal info
Awards
Reciprocal links
Terms of use
Privacy Policy

Interactive Activities

Cut The Knot!
MSET99 Talk
Games & Puzzles
Arithmetic/Algebra
Geometry
Probability
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
My Logo
Math Poll
Other Math sit's
Guest book
News sit's

Recommend this site

Manifesto: what CTK is about |Store| Search CTK Buying a book is a commitment to learning Table of content Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

CTK Exchange

Subject: "sin(18) degrees"     Previous Topic | Next Topic
Printer-friendly copy     Email this topic to a friend    
Conferences The CTK Exchange High school Topic #241
Reading Topic #241
Cino Hilliard
guest
Jun-06-03, 12:48 PM (EST)
 
"sin(18) degrees"
 
   Hi,

Recently I found an old note book where I had derived the sin of
several angles I had no notes as to how in that book. In bold ink
I had

sin(18) = 1/2sqrt((3-sqrt(5))/2)

Then in pencil, dated 5 years later, which I think I did using
a scientific calculator

sin(18) = (sqrt(5)-1)/4 Amazing simplification!

Can anyone show the steps to get the simplification from the first?
I can't figure it out with algebra.

Thanks
Cino
Ps: If you do you should get a fee.



  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top

  Subject     Author     Message Date     ID  
sin(18) degrees Cino Hilliard Jun-06-03 TOP
  RE: sin(18) degrees alexb Jun-06-03 1
     RE: sin(18) degrees Cino Jun-09-03 2
         RE: sin(18) degrees Graham C Jun-10-03 3
             RE: sin(18) degrees Cino Jun-11-03 5
         RE: sin(18) degrees Tofique Fatehi Jun-10-03 4
             RE: sin(18) degrees Graham C Jun-11-03 6
                 RE: sin(18) degrees Cino Hilliard Jun-15-03 7
                     RE: sin(18) degrees Graham C Jun-17-03 8
                         RE: sin(18) degrees Cino Hilliard Jun-18-03 9
                             RE: sin(18) degrees Graham C Jun-19-03 10
                                 RE: sin(18) degrees Cino Hilliard Jun-21-03 12
  RE: sin(18) degrees anonymous Jun-20-03 11

Conferences | Forums | Topics | Previous Topic | Next Topic
alexb
Charter Member
985 posts
Jun-06-03, 12:56 PM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
1. "RE: sin(18) degrees"
In response to message #0
 
   Well, just write A = B, where A is the first number and B is the second. Squaring gives you (6 - 2sqrt(5)) on both sides.

> Ps: If you do you should get a fee.

What are we going to do next?

P.S. At the bottom of

https://www.cut-the-knot.com/ctk/NecessaryAndSufficient.shtml

there's a reference to T. J. Osler's article from the Monthly where you can find several similar (and more general) identities.


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Cino
guest
Jun-09-03, 08:34 PM (EST)
 
2. "RE: sin(18) degrees"
In response to message #1
 
   >Well, just write A = B, where A is the first number and B is
>the second. Squaring gives you (6 - 2sqrt(5)) on both sides.
A priori please!I will ask it another way.

When I plug z = 1/2*sqrt((3 - sqrt(5))/2) into Maple it gives
results below. Question. How does it do it?
is it a lookup? Please show how to convert step by step please.


Maple V
...........1/2......1/2 1/2
1/4 I (- 3 plus 5) * 2

Maple 8
.......... .....1/2
- 1/4 plus 5 / 4

Obviously, Maple V dosen't know how to get the simplification
I derived by astute observation using an HP calculator in 1978.
There is no doubt in my mind that many of the early discoveries
were made similarly based on prior knowledge. While not entirely
on this subject, It is tiresome to continually read the credits
bestowed to certain lofty individuals for implementing certain
mathematical truths when in fact millions before and after
discovered the same truths. Besides, how many of the discoveries
were made by their underlings - slaves,wives or what ever and credit
sycophantically given to their "master." Yes, how much did Gauss and
Euler mulct? Fermat? I doubt much. He was an amature.

>> Ps: If you do you should get a fee.
Is this bad grammer? If you do solve this problem you should get a
1.61803398874989484820458683436563811772030917980576286213544862270526
0462818902449707207204189391137...
Another one of my bad jokes.
>
>What are we going to do next?
Arthur C. Clark once said "I want a computer that will tell ME what
to do next."

>
>P.S. At the bottom of
>
>https://www.cut-the-knot.com/ctk/NecessaryAndSufficient.shtml

Yes! I felt the same joy im making my discovery. It appeared almost
to be a free lunch.

>
>there's a reference to T. J. Osler's article from the
>Monthly where you can find several similar (and more
>general) identities.
google search on sin(18),sin(1) also interesting.
Anyone interested in a super number theory calculator for windows
take a look

https://groups.msn.com/BC2LCC/page.msnw?fc_p=%2FPari&fc_a=0

read the calcpari.readme file for Description.

Have fun in the facinating world of numbers.

Cino


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Graham C
Member since Feb-5-03
Jun-10-03, 08:35 AM (EST)
Click to EMail Graham%20C Click to send private message to Graham%20C Click to view user profileClick to add this user to your buddy list  
3. "RE: sin(18) degrees"
In response to message #2
 
   >>Well, just write A = B, where A is the first number and B is
>>the second. Squaring gives you (6 - 2sqrt(5)) on both sides.
>A priori please!

If you don't like Alex's way of getting from A to B, let x=sqrt(5), and try:

A = (1/2)*sqrt((3-x)/2)
= sqrt((3-x)/8)
= sqrt((6-2x)/16)
= (1/4)*sqrt(x^2-2x+1)
= (1/4)*(x-1) (or 1/4*(1-x))
= (sqrt(5)-1)/4 (or (1-sqrt(5))/4)
= B (or -B)

I don't know anything about Maple.


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Cino
guest
Jun-11-03, 09:16 AM (EST)
 
5. "RE: sin(18) degrees"
In response to message #3
 
   >>>Well, just write A = B, where A is the first number and B is
>>>the second. Squaring gives you (6 - 2sqrt(5)) on both sides.
>>A priori please!
>
>If you don't like Alex's way of getting from A to B, let
>x=sqrt(5), and try:

Obviously, if I assume A=B knowing that A = (Phi-1)/2 all along
or with prior knowledge, naturally I can confirm that truth by some
calculation. I had already (pseudo confirmed) it to 12 places on a calculator 1n 1978.

>
>A = (1/2)*sqrt((3-x)/2)
> = sqrt((3-x)/8)
> = sqrt((6-2x)/16)
> = (1/4)*sqrt(x^2-2x+1)
> = (1/4)*(x-1) (or 1/4*(1-x))
> = (sqrt(5)-1)/4 (or (1-sqrt(5))/4)
> = B (or -B)

>
Here is is what I was looking for without strict
sqrt sign handling.

sin18 = 1/2sqrt((3-sqrt(5))/2)
(3-sqrt(5))/2 = 3/2 + sqrt(5)/2 = (6-2sqrt(5))/4
sin18 = 1/4sqrt(6-2sqrt(5))
6 - 2sqrt(5) = 5 - 2sqrt(5)+ 1. Factoring this,
= (sqrt(5) - 1)^2. Taking roots
sqrt(6-2sqrt(5))= sqrt(5) - 1 thus
sin18 = 1/4(sqrt(5)-1)
Same as yours, with a clarifying step plus I am a student who could
not learn algebra because I left all my x's in Texas.
Sin15 = sqrt(2)/4*(sqrt(3)-1) anyone?
Or
sin75 = (sqrt(6)+sqrt(2))/4 easy.
sin72 = 1/2*sqrt((5_sqrt(5))/2)
Sin3 = sine(75-sin72)

>I don't know anything about Maple.
Maple is a symbolic mathematics package similar to Mathematica and
others. Another great calculator is Pari which is a number theory
calculator with limited symbolics but has a very easy C like script
that you can use to perform operations with the NT functions. For
speed in calculation go with Pari. For conversion such as sin(18) go
with Maple. There was a free trial version on maple 8 but now maple 9
is out and I don't see anything about a trial version.

I guess it is a matter of taste on what we mean by simplify. Maple
gives -1/4+sqrt(5)/4. I like (sqrt(5)-1)/4 since Maple's result
requires 5 Arithmetic operations and mine 3. However, do not be
deceived; Maple style computes 1000000 calculations in 45.672 sec vs
49.844 sec using my "simplification." However, Pari, a free number
theory calculator, performs 1000000 calcs in 4.422 sec for maple
style nd 4.234 sec for mine. Pari can be downloaded at

https://www.parigp-home.de/

My gui calculator which bundles pari and does not require learning
pari to start is at

https://groups.msn.com/BC2LCC/page.msnw?fc_a=0&fc_p=%2FCalcpari

Hit the keys on a blank or Clr display to see what the keys do.
Not all functions rely on pari. But it is a good start to doing
big numbers without a lot of preliminary study which I hope will
come later by whetting the appitite of the curious. This is a Beta
and has work to be done. suggestions are welcome.

Cheers and roebuck,

Cino


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Tofique Fatehi
guest
Jun-10-03, 08:54 AM (EST)
 
4. "RE: sin(18) degrees"
In response to message #2
 
   Isn't this the Golden Ratio? And it is connected to the Fibonacci Series. Just FYI


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Graham C
Member since Feb-5-03
Jun-11-03, 09:16 AM (EST)
Click to EMail Graham%20C Click to send private message to Graham%20C Click to view user profileClick to add this user to your buddy list  
6. "RE: sin(18) degrees"
In response to message #4
 
   >Isn't this the Golden Ratio? And it is connected to the
>Fibonacci Series. Just FYI

Actually it is half the golden ratio, defined as 1/x = x/(1-x), which gives x^2 plus x - 1 = 0, and x = (sqrt(5)-1)/2.

Interesting though.


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Cino Hilliard
guest
Jun-15-03, 04:56 AM (EST)
 
7. "RE: sin(18) degrees"
In response to message #6
 
   >>Isn't this the Golden Ratio? And it is connected to the
>>Fibonacci Series. Just FYI
Yes. Very much so. You can remember the golden ratio easily by
stating it.
"The lesser is to the greater as the greater is to the whole."

The nth Fibonacci number F(n) can be calculated with the integer
part of (Phi^n)/sqrt(5)+.5. Eg.,using Pari,
\p 1000 \\set precision to 1000 decimal places
Phi = (sqrt(5)+1)/2 \\set Phi
floor(Phi^1000/sqrt(5)+.5) \\compute it 4346655768693745643568852767504062580256466051737178040248172908953655
5417949051890403879840079255169295922593080322634775209689623239873322
471161642996440906533187938298969649928516003704476137795166849228875
Test it
? fibonacci(1000) \\Pari built in function
4346655768693745643568852767504062580256466051737178040248172908953655
5417949051890403879840079255169295922593080322634775209689623239873322
471161642996440906533187938298969649928516003704476137795166849228875

Also on the simplification - yet another approach.

>
>Actually it is half the golden ratio, defined as 1/x =
>x/(1-x), which gives x^2 plus x - 1 = 0, and x =
>(sqrt(5)-1)/2.
>
>Interesting though.


Can sin(18) = 1/2sqrt((3 - sqrt(5))/2) be simplified? If Yes then
3/2 -sqrt(5)/2 must be a perfect square to remove the primary
radical. Then let, 3/2 - sqrt(5)/2 = (6 - sqrt(20))/4 so
sin(18) = 1/4sqrt(6 - sqrt(20)). Is 6 - sqrt(20) a perfect square?
Assume it is. Then
(1) sqrt(6 - sqrt(20)) = sqrt(x) - sqrt(y) for some x,y
so 6 - sqrt(20) = x + y - 2sqrt(xy) where
(2) 6 = x+y
sqrt(20) = 2sqrt(xy)
20 = 4xy
x = 5/y
substituting into (2) and
solving for y we have y^2 - 6y + 5 = 0
factoring (y-1)(y-5)= 0
so y = 1,y = 5 substituting y = 1 in (1) we have
sqrt(6 - sqrt(20)) = sqrt(5) - sqrt(1) and
sin(18) = 1/4sqrt(6 - sqrt(20))= 1/4(sqrt(5) - 1) = (Phi-1)/2
Repetitive quantities in the construction if sin(3x) angles in my
note book are.

sqrt(2)
sqrt(3)
sqrt(5)
(1+sqrt(5))/2 = Phi
(3-sqrt(5))/2 = (Phi-1)^2
(5+sqrt(5))/2 = Phi+2
(5-sqrt(5))/2 = (Phi-1)^2+1

It appears (3-sqrt(5))/2 = (Phi-1)^2 is the only one for which we can eliminate the primary radical.

References
https://mcraefamily.com/MathHelp/BasicSimplifying.htm
https://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibFormula.html

Time to sign off..
Cino


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Graham C
Member since Feb-5-03
Jun-17-03, 08:35 AM (EST)
Click to EMail Graham%20C Click to send private message to Graham%20C Click to view user profileClick to add this user to your buddy list  
8. "RE: sin(18) degrees"
In response to message #7
 
   >
>Can sin(18) = 1/2sqrt((3 - sqrt(5))/2) be simplified? If Yes
>then
>3/2 -sqrt(5)/2 must be a perfect square to remove the
>primary
>radical. Then let, 3/2 - sqrt(5)/2 = (6 - sqrt(20))/4 so
>sin(18) = 1/4sqrt(6 - sqrt(20)). Is 6 - sqrt(20) a perfect
>square?
>Assume it is.

How can you possibly do that? A perfect square has to be an integer. 6-sqrt(20) is not an integer. You can calculate it. There's no point in 'assuming' it equals anything other than what it does equal which is 1.52786....


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Cino Hilliard
guest
Jun-18-03, 06:36 PM (EST)
 
9. "RE: sin(18) degrees"
In response to message #8
 
   >>
>>Can sin(18) = 1/2sqrt((3 - sqrt(5))/2) be simplified? If Yes
>>then
>>3/2 -sqrt(5)/2 must be a perfect square to remove the
>>primary
>>radical. Then let, 3/2 - sqrt(5)/2 = (6 - sqrt(20))/4 so
>>sin(18) = 1/4sqrt(6 - sqrt(20)). Is 6 - sqrt(20) a perfect
>>square?
>>Assume it is.
>
>How can you possibly do that? A perfect square has to be an
>integer. 6-sqrt(20) is not an integer. You can calculate it.
>There's no point in 'assuming' it equals anything other than
>what it does equal which is 1.52786....

6 - sqrt(20) = (sqrt(5) - 1)^2 = 1.52786...

If you dont want to call it perfect square, that is your choice. I
will call it a "perfect" square because when you take the sqrt of
both sides you can replace sqrt(6-sqrt(20)) with sqrt(5)-1 yielding
a simplification of the origional nested radical for the sin18 saga.



  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Graham C
Member since Feb-5-03
Jun-19-03, 05:08 PM (EST)
Click to EMail Graham%20C Click to send private message to Graham%20C Click to view user profileClick to add this user to your buddy list  
10. "RE: sin(18) degrees"
In response to message #9
 
   >>How can you possibly do that? A perfect square has to be an
>>integer. 6-sqrt(20) is not an integer. You can calculate it.
>>There's no point in 'assuming' it equals anything other than
>>what it does equal which is 1.52786....
>
>6 - sqrt(20) = (sqrt(5) - 1)^2 = 1.52786...
>
>If you dont want to call it perfect square, that is your
>choice. I will call it a "perfect" square because when you take the
>sqrt of both sides you can replace sqrt(6-sqrt(20)) with sqrt(5)-1
>yielding a simplification of the origional nested radical for the
>sin18 saga.
>
I accept it does that and I generally accept people's right to define the terms they use. But it's confusing to use a term the rest of the world (not just me) uses to mean something else - i.e. the squares of the integers: 1,4,9,16,25.........

Incidentally sets of integers a,b,c that satisfy
a - sqrt(b) = (srt(c)-1)^2
aren't hard to find - if c = 25 try 25,81; 36,400; 18,4; 20,16;
>
>


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
Cino Hilliard
guest
Jun-21-03, 04:32 PM (EST)
 
12. "RE: sin(18) degrees"
In response to message #10
 
   >>>How can you possibly do that? A perfect square has to be an
>>>integer. 6-sqrt(20) is not an integer. You can calculate it.
>>>There's no point in 'assuming' it equals anything other than
>>>what it does equal which is 1.52786....
>>
>>6 - sqrt(20) = (sqrt(5) - 1)^2 = 1.52786...
>>
>>If you dont want to call it perfect square, that is your
>>choice. I will call it a "perfect" square because when you take the
>>sqrt of both sides you can replace sqrt(6-sqrt(20)) with sqrt(5)-1
>>yielding a simplification of the origional nested radical for the
>>sin18 saga.
>>
>I accept it does that and I generally accept people's right
>to define the terms they use. But it's confusing to use a
>term the rest of the world (not just me) uses to mean
That is stretching it a bit.
>something else - i.e. the squares of the integers:
>1,4,9,16,25.........
What do you call 1/4,1/9,1/16,1/25...4/25,9,25,x^2/y^2...a^2+2ab+b^2?
In fact rational numbers have to, of necessity, be included as
possible perfect squares. This is true because every whole number w
is a rational number namely w/1.

I just quoted the author in the link I gave. A snip is below

Simplifying Nested Radicals
The Basic Idea
The basic idea is to simplify an expression of the form sqrt(c+sqrt
(d)) such as

sqrt(9/32+sqrt(45)/32)

Let x = sqrt(c+sqrt(d))

Perhaps c+sqrt(d) is a perfect square, in which case this can be
simplified.

If it's a perfect square, then it has to be the square of an
expression of the form sqrt(a)+sqrt(b).

So c+sqrt(d) = a+b + 2sqrt(ab)

That means c = a+b, and d=4ab

Then by the same author in the section on
Theorems Involving Perfect Squares

A "perfect square" is a number that can be expressed as k^2, where k
is an integer. So he forgot about the rationals and then ascribed
perfect to irrationals. Oh well.

>
>Incidentally sets of integers a,b,c that satisfy
> a - sqrt(b) = (srt(c)-1)^2
>aren't hard to find - if c = 25 try 25,81; 36,400; 18,4;
>20,16;
>>
Interesting. b and c are rational perfect squares.

>>


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top
anonymous
guest
Jun-20-03, 05:08 PM (EST)
 
11. "RE: sin(18) degrees"
In response to message #0
 
   You can see that (sqrt(5)-1)/4=1/2sqrt((3-sqrt(5))/2) from:

sqrt(5)-1=2sqrt((3-sqrt(5))/2) Multiply both sides by four
6-2sqrt(5)=6-2sqrt(5) Square both sides


  Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top

Conferences | Forums | Topics | Previous Topic | Next Topic

You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.

|Front page| |Contents|

Copyright © 1996-2018 Alexander Bogomolny

67083803