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

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.

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

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.

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

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.

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;
>
>

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.

>>