Talk:Sine of 72 Degrees

I don't believe that particular nested radical has a simple surd form --Ybab321 (talk) 01:20, 18 September 2014 (UTC)


 * Odd had a naive belief that *all* such expressions could be rendered as simple surds. May need research. --prime mover (talk) 05:02, 18 September 2014 (UTC)


 * $\sqrt{\sqrt5}$ itself cannot be simplified, acting as a counter-example. --kc_kennylau (talk) 12:41, 18 September 2014 (UTC)


 * Doesn't necessarily follow that $\sqrt {10 + 2 \sqrt 5}$ can't be simplified. After all, $\sqrt {8 - 4 \sqrt 3}$ simplifies to $\sqrt 6 - \sqrt 2$ -- see Sine of 15 Degrees. --prime mover (talk) 13:04, 18 September 2014 (UTC)


 * My point still holds: I believe it should be possible to express it in a form $a + \sqrt b + \sqrt c + \sqrt[4] d$ sort of thing, so as to convert the nestedness of the roots. --prime mover (talk) 13:07, 18 September 2014 (UTC)

The transformation of $\sqrt {a + \sqrt b}$ to $\sqrt c + \sqrt d$ is done with the relationship $c = \frac {a + \sqrt {a^2 - b}} 2,\ d = \frac {a - \sqrt {a^2 - b}} 2$, that is, it is required that $a^2-b$ is a perfect square.

With $a=8, b=48$, it follows nicely that $a^2-b = 16$, but with $a=10, b=20$, it follows that $a^2-b = 80$; so the transformation will only get as far as $\sqrt {5 + \sqrt {20}} + \sqrt {5 - \sqrt {20}}$.

Applying the transformation to $\sqrt {5 + \sqrt {20}}$ gives $\frac 1 {\sqrt 2} \left({\sqrt {5 + \sqrt 5} + \sqrt {5 - \sqrt 5}}\right)$, applying the transformation to $\sqrt {5 + \sqrt 5}$ gives $\frac 1 {\sqrt 2} \left({\sqrt {5 + \sqrt {20}} + \sqrt {5 - \sqrt {20}}}\right)$, the transformation will continue to cycle between these two expressions.

Maybe there's hope for a $a + \sqrt b + \sqrt c + \sqrt[4] d$ sort of thing, but it's certainly not trivial. --Ybab321 (talk) 18:59, 18 September 2014 (UTC)


 * Are you in a position to be able to put a distillation of the above into a page on ? --prime mover (talk) 19:37, 18 September 2014 (UTC)