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?
 A: $\sin \theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$
$\begin{align*}\sin 1^\circ & = \sin\left(\dfrac{\pi}{180} \right) \\ & = \dfrac{e^{i\tfrac{\pi}{180}}-e^{-i\tfrac{\pi}{180}}}{2i} \\ & =\dfrac{\left(e^{i\tfrac{\pi}{6}}\right)^{1/30}-\left(e^{-i\tfrac{\pi}{6}}\right)^{1/30}}{2i} \\ & = \dfrac{\sqrt[30]{\cos \left( \dfrac{\pi}{6} \right) + i \sin \left( \dfrac{\pi}{6} \right)} - \sqrt[30]{\cos\left( \dfrac{\pi}{6} \right) - i \sin \left( \dfrac{\pi}{6} \right) } }{2i} \\ & = \dfrac{\sqrt[30]{\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}} - \sqrt[30]{\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}} }{2i}\end{align*}$
A: Just a comment. Also note that, since
$$\sqrt[^\pm3]{i}=\frac{\sqrt{3}}{2}\pm\frac{i}{2}$$
This was achieved using Euler's Identity or more specifically,
$$i^x=\cos\left(\frac{\pi x}{2}\right)+i\sin\left(\frac{\pi x}{2}\right)$$
We can even simply that to,
$$\sin(1^°)=\frac{1}{2i}\left(\sqrt[^{90}]{i}-\sqrt[^{90}]{\frac{1}{i}}\right)$$
$$\sin(1^°)=\frac{1}{2i}\left(\sqrt[^{90}]{i}-\sqrt[^{90}]{-i}\right)$$
Simplying this any further would just make this trivial.
Also note that in the above question,
$$\sin\left(\frac{29\pi}{60}\right)=\sin(87°)=\frac{1}{4}\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2\sqrt{5}}}$$
$$\cos\left(\frac{29\pi}{60}\right)=\cos(87°)=\frac{1}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}$$
You can get this by using the usual identity for $\sin(90-3)$
And as far as I can see,
$$\frac{1}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}=\frac{\left(1+\sqrt{3}\right)\sqrt{5+\sqrt{5}}}{8}+\frac{\left(\sqrt{5}-1\right)\left(\sqrt{3}-1\right)}{8\sqrt{2}}$$
I don't think we could denest it just like that, because there'e simply too many surds to take care of.
A: I have figured out how to find my answer using De Moivre's formula, not that the method in particular is of great importance but it is slightly alternative to InterstellarProbe's use of the definition of sine (which helped me figure this out).
De Moivre's formula is $$(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta, \qquad n\in\mathbb{Z}.$$
In my case, we let $\theta=1^{\circ}$. Then we can choose $n$ such that $\cos n\theta$ and $\sin n\theta$ are values that we know exactly. Let's choose $n=30$ because $\cos30^{\circ}=\frac{\sqrt{3}}{2}$ and $\sin30^{\circ}=\frac{1}{2}$.
Then \begin{align}(\cos1^{\circ}+i\sin1^{\circ})^{30}&=\frac{\sqrt{3}}{2}+\frac{i}{2}\\\Longrightarrow\qquad\cos1^{\circ}+i\sin1^{\circ}&=\sqrt[30]{\frac{\sqrt{3}}{2}+\frac{i}{2}}.\tag{1}\label{1}\end{align}
The trick is now to choose $\theta=-1^{\circ}$:\begin{align}(\cos1^{\circ}-i\sin1^{\circ})^{30}&=\frac{\sqrt{3}}{2}-\frac{i}{2}\\\Longrightarrow\qquad\cos1^{\circ}-i\sin1^{\circ}&=\sqrt[30]{\frac{\sqrt{3}}{2}-\frac{i}{2}}.\tag{2}\label{2}\end{align}
Then we subtract \eqref{2} from \eqref{1}:\begin{align}2i\sin1^{\circ}&=\sqrt[30]{\frac{\sqrt{3}}{2}+\frac{i}{2}}-\sqrt[30]{\frac{\sqrt{3}}{2}-\frac{i}{2}}\\\Longleftrightarrow\qquad\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}}.\end{align}
We could have also chosen $n=45$ or $n=60$, say, and gotten a different but equivalent result. In fact, we may use this method to find the exact values of $\sin\theta$, $\cos\theta$, $\tan\theta$, $\sec\theta$, $\csc\theta$, and $\cot\theta$ for all $\theta\in\mathbb{Q}$ in terms of radicals, without having to solve quintic or higher-order equations.
