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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 ofseveral angles I had no notes as to how in that book. In bold inkI had sin(18) = 1/2sqrt((3-sqrt(5))/2) Then in pencil, dated 5 years later, which I think I did usinga 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. ThanksCinoPs: If you do you should get a fee.

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

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alexb
Charter Member
985 posts
Jun-06-03, 12:56 PM (EST)    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 ofhttps://www.cut-the-knot.com/ctk/NecessaryAndSufficient.shtmlthere's a reference to T. J. Osler's article from the Monthly where you can find several similar (and more general) identities. Cino guest
Jun-09-03, 08:34 PM (EST)

2. "RE: sin(18) degrees"
In response to message #1 Graham C
Member since Feb-5-03
Jun-10-03, 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((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. 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 alongor 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))/4sin18 = 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?Orsin75 = (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 9is 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 maplestyle nd 4.234 sec for mine. Pari can be downloaded atMy gui calculator which bundles pari and does not require learningpari to start is athttps://groups.msn.com/BC2LCC/page.msnw?fc_a=0&fc_p=%2FCalcpariHit 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 doingbig numbers without a lot of preliminary study which I hope willcome later by whetting the appitite of the curious. This is a Betaand has work to be done. suggestions are welcome.Cheers and roebuck,Cino 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 Graham C
Member since Feb-5-03
Jun-11-03, 09:16 AM (EST)    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. 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 placesPhi = (sqrt(5)+1)/2 \\set Phi floor(Phi^1000/sqrt(5)+.5) \\compute it 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875Test it? fibonacci(1000) \\Pari built in function 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875Also 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)/2Repetitive quantities in the construction if sin(3x) angles in mynote 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+1It appears (3-sqrt(5))/2 = (Phi-1)^2 is the only one for which we can eliminate the primary radical. Time to sign off.. Cino Graham C
Member since Feb-5-03
Jun-17-03, 08:35 AM (EST)    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.... 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. Iwill call it a "perfect" square because when you take the sqrt ofboth sides you can replace sqrt(6-sqrt(20)) with sqrt(5)-1 yieldinga simplification of the origional nested radical for the sin18 saga. Graham C
Member since Feb-5-03
Jun-19-03, 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. 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)^2aren't hard to find - if c = 25 try 25,81; 36,400; 18,4; 20,16;>> 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 belowSimplifying Nested RadicalsThe Basic IdeaThe basic idea is to simplify an expression of the form sqrt(c+sqrt(d)) such assqrt(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=4abThen by the same author in the section on Theorems Involving Perfect SquaresA "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. >>

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 four6-2sqrt(5)=6-2sqrt(5) Square both sides

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