## How to convert it to radicals?...

Asked by:

How to express sin(-(1/6)*Pi+(1/2)*arccos(1/3)) in radicals with Maple? The use of the applyrule command is not desired. Here is one of  my tries:
>convert(expand(sin(-(1/6)*Pi+(1/2)*arccos(1/3))), radical);
-(1/2)*cos((1/2)*arccos(1/3))+(1/2)*sqrt(3)*sin((1/2)*arccos(1/3))

## Simplification of radicals...

Asked by:

The expression  expr  the command  simplify  simplifies without any problems. Even certain automatic simplification is produced:

expr:=sqrt(4-sqrt(7))*sqrt(4+sqrt(7));

simplify(expr);

But if we slightly modify the expression, the simplification is not performed:

expr1:=sqrt(4-sqrt(6))*sqrt(4+sqrt(6));

simplify(expr1);

The last expression  expr1  succeed to simplify the only combination of commands:

expand(combine(expr1));

What is the reason for this strange behavior of  simplify?

## Another proof of stunningly beautiful identity

by:

I propose a different proof of this remarkable identity (see  http://www.mapleprimes.com/posts/144499-Stunningly-Beautiful-Identity-Proved ) in which  directly constructed a polynomial, whose root is the value of LHS, and this is expressed in radicals.

For the proof, we need three simple identities with cubic roots (a, b, c -any real numbers):

## Trigonometric functions in radicals

by: Maple

It is known that the trigonometric functions of an integer number of degrees may be expressed by radicals if the number of degrees is divisible by 3. Simple code finds all these values ​​in the range 0 to 90 degrees:

[sin(`0`^`o`)=0,`   cos`(`0`^`o`)=1,`   tan`(`0`^`o`)=0,`   cot`(`0`^`o`)=infinity];

for n from 3 to 87 by 3 do

[sin(n^`o`)=convert(sin(n*Pi/180),radical...

## solutions by radicals for polynomials ?...

Asked by:
`Usually Maple gives solutions in terms of radicals only up to degree=4(for example using RootOf + allvalues).Using 'irreduc' (to test first) and 'galois' (to check for the Galoisgroup, if degree <= 9) I have cases, where the result implies, that bytheory the roots can be given through radicals:Though the results are difficult to read in my case they are 'C(6)' or'C(8)', the cyclic groups of that order - thus abelian and IIRC thosegroups are solvable.`
 Page 1 of 1
﻿