<s>[INST] Simplifying nested radicals with higher-order radicals I've seen that $$\sin1^{\circ}=\frac{1}{2i}\sqrt[3]{\frac{1}{4}\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2\sqrt{5}}}+\frac{i}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}}-\frac{1}{2i}\sqrt[3]{\frac{1}{4}\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2\sqrt{5}}}-\frac{i}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}}.$$
But then someone was able to simplify this neat, but long, expression with higher-order radicals, and they said they used De Moivre's theorem: $$\sin1^{\circ}=\frac{1}{2i}\sqrt[30]{\frac{\sqrt{3}}{2}+\frac{i}{2}}-\frac{1}{2i}\sqrt[30]{\frac{\sqrt{3}}{2}-\frac{i}{2}}.$$
I have been looking at this for a while now, and I cannot see how they were able to successfully do this. I am very impressed by the result and would like to use a similar technique to simplify nested radicals in the future.
Edit: It seems like the person who originally used De Moivre's theorem did not use it to directly simplify the longer radical expression, but rather found $\sin1^{\circ}$ by the method I figured out in my answer to this question. I do think there is limited value to writing the exact value of, say, $\sin1^{\circ}$ out, but which way do you think is better, the longer combination of square and cube roots, or the compact thirtieth-root?
 [/INST] $\sin \theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$
$\begin{align*}\sin 1^\circ & = \sin\left(\dfrac{\pi}{180} \right) \\ & = \dfrac{e^{i\tfrac{\pi}{180}}-e^{-i\tfrac{\pi}{180}}}{2i} \\ & =\dfrac{\left(e^{i\tfrac{\pi}{6}}\right)^{1/30}-\left(e^{-i\tfrac{\pi}{6}}\right)^{1/30}}{2i} \\ & = \dfrac{\sqrt[30]{\cos \left( \dfrac{\pi}{6} \right) + i \sin \left( \dfrac{\pi}{6} \right)} - \sqrt[30]{\cos\left( \dfrac{\pi}{6} \right) - i \sin \left( \dfrac{\pi}{6} \right) } }{2i} \\ & = \dfrac{\sqrt[30]{\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}} - \sqrt[30]{\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}} }{2i}\end{align*}$
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