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Cino Hilliard
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Jun0603, 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((3sqrt(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.


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Cino
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Jun0903, 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.cuttheknot.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


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Graham C
Member since Feb503

Jun1003, 08:35 AM (EST) 

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((3x)/2) = sqrt((3x)/8) = sqrt((62x)/16) = (1/4)*sqrt(x^22x+1) = (1/4)*(x1) (or 1/4*(1x)) = (sqrt(5)1)/4 (or (1sqrt(5))/4) = B (or B) I don't know anything about Maple. 

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Cino
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Jun1103, 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 = (Phi1)/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((3x)/2) > = sqrt((3x)/8) > = sqrt((62x)/16) > = (1/4)*sqrt(x^22x+1) > = (1/4)*(x1) (or 1/4*(1x)) > = (sqrt(5)1)/4 (or (1sqrt(5))/4) > = B (or B) > Here is is what I was looking for without strict sqrt sign handling. sin18 = 1/2sqrt((3sqrt(5))/2) (3sqrt(5))/2 = 3/2 + sqrt(5)/2 = (62sqrt(5))/4 sin18 = 1/4sqrt(62sqrt(5)) 6  2sqrt(5) = 5  2sqrt(5)+ 1. Factoring this, = (sqrt(5)  1)^2. Taking roots sqrt(62sqrt(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(75sin72) >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.parigphome.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 

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Cino Hilliard
guest

Jun1503, 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/(1x), 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 (y1)(y5)= 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) = (Phi1)/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 (3sqrt(5))/2 = (Phi1)^2 (5+sqrt(5))/2 = Phi+2 (5sqrt(5))/2 = (Phi1)^2+1 It appears (3sqrt(5))/2 = (Phi1)^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 

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Cino Hilliard
guest

Jun1803, 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. 6sqrt(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(6sqrt(20)) with sqrt(5)1 yielding a simplification of the origional nested radical for the sin18 saga.


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Graham C
Member since Feb503

Jun1903, 05:08 PM (EST) 

10. "RE: sin(18) degrees"
In response to message #9

>>How can you possibly do that? A perfect square has to be an >>integer. 6sqrt(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(6sqrt(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; > >


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Cino Hilliard
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Jun2103, 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. 6sqrt(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(6sqrt(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. >>


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