Finding the exact value of $\sin 30°$ using golden ratio $\sin \left(k\cdot 30^\circ\right) = \frac{\sqrt{2}}{4}\;\sqrt{\;4\;\pm_1\;\sqrt{\phi\,(a\phi+b\overline{\phi})}\;\pm_2\;\sqrt{\overline{\phi}\,(c\phi+d\overline{\phi})}\;}$
I'm currently trying to solve this for $a = 4, b = 0, c = 0, d = 4$ and $\pm_1 = -$, $\pm_2 = +$ 
where $\phi := \dfrac{\sqrt 5+ 1}{2}$ and $\overline{\phi} := \phi^{-1} = \frac{1}{2}(\sqrt{5}-1)$
There's an outer square root, which made me confused. Hence I couldn't begin to solve it.
Regards!
 A: Note that when $b=c=0$ the nested square roots simplify down to $4\phi^2$ and $4\bar\phi^2$ so the expression simplifies to $$\frac{\sqrt{2}}{4}\sqrt{4-2\phi +2\bar\phi}$$And so the $\sqrt{5}$ terms cancel out.
A: To simplify things, start with a definition. Let (with $\,a,b,c,d\,$ being $\frac14$ of original formula)
 $$\, F(a,b,c,d,e,f) := 2 
+ \, e \sqrt{\phi\,(a\phi + b\overline{\phi})} 
+ \, f \sqrt{\overline{\phi}\,(c\phi + d\overline{\phi})}.\,$$
The values of $\, 4\sin(k \cdot 30^\circ)^2 \,$ are $\,
 \{ 0, 1, 3, 4 \}. \,$ To get
these values from $\, F(a,b,c,d,e,f) \,$ we choose appropriate
parameter values where $\, a,b,c,d > 0 \,$ are integers and
$\, e,f \in \{-1,+1\}. \,$ Notice that if $\, n \,$ is a value then
$\, 4-n \,$ is a value by changing sign of $\, e \,$ and $\, f. \,$
Therefore, we only need to find the values $\, 0 \,$ and $\, 1. \,$
Note that
 $\, (s \phi + t)^2 = \phi\,(a\phi + b\overline{\phi}) \,$
where $\, a = s(s + 2t),\, b = t(t - 2s) \,$
and the corresponding identity
 $\, (u \overline{\phi}+v)^2 = \overline{\phi}\,(c\phi+d\overline{\phi}) \,$
where $\, d = u(u - 2v),\, c = v(v + 2u). \,$
Now  $\, \phi - \overline{\phi} = 1 \,$ and
$$\, F(a,b,c,d,e,f) = 2 + e(s\phi + t) + f(u\overline{\phi} + v) =
 2 + f v + e t - f u + \phi(e s + f u) \,$$
so use $\, f = 1,\, e = -1,\, u = 1, t = v = 0,\, s = ufe, \,$ and $\, -1 = f v + e t - f u \,$
to get $\, \sin(30^\circ). \,$
Your choice of
 $\, a = 1, b = 0, c = 0, d = 1, e = -1, f = +1 \,$
is one of an infinity of choices.
A: By direct substitution,
$$\frac{\sqrt{2}}{4}\;\sqrt{\;4\;\pm_1\;\sqrt{\phi\,(a\phi+b\overline{\phi})}\;\pm_2\;\sqrt{\overline{\phi}\,(c\phi+d\overline{\phi})}\;}=\\
\frac{\sqrt{2}}{4}\;\sqrt{\;4\;-\;\sqrt{\phi\,(4\phi)}\;+\;\sqrt{\overline{\phi}\,(4\overline{\phi})}\;}=\\
\frac{\sqrt{2}}{4}\;\sqrt{\;4\;-\;2\phi\;+\;2\overline{\phi}\,\;}.$$
And from the definitions of $\phi,\overline\phi$,
$$=\frac{\sqrt{2}}{4}\;\sqrt{\;4\;-2\,\;}=\frac12.$$
