<s>[INST] Expressing $\sin(18^{°})$ in algebraic form from $z^5-1$ as starting point I am trying to express $\sin(18°)$ in algebraic form using only complex numbers. I know that when I factor $z^5-1.$ I get an expansion that looks like: $(z-1)(z^4+z^3+z^2+z^1+1)$ The exercise then says substitute for $z+\frac{1}{z}$ in the 'long factor' and then somehow derive $\sin(18°)$. I just have no idea how to I am supposed to do this. But knowing how, could teach me something about complex numbers. 
 [/INST] I guess you realize that $18^\circ$ is $1/20$ of a circle, so that $a=\sin 18^\circ+i\cos 18^\circ$ is a $5$ root of unity.  So $a$ must satisfy $z^{5}=1$.  
The polynomial $z^{5}-1$ factors and a bit of thought gets you to the point where you are, that $a$ must satisfy $z^4+z^3+z^2+z+1=0$.  So if you can find its roots, then one of the real parts will be $\sin 18^\circ.$
Divide the above equation by $z^2.$
$$z^2+z+1+\frac{1}{z}+\frac{1}{z^2}=0.$$
Let $w=z+\frac{1}{z}$ and note the the 2nd and 4th terms equal $w$.  Also note that $w^2= z^2+2+\frac{1}{z^2}$, so by adding and subtracting $1$ we make the above 
$$w^2+w-1 = 0.$$
Solve this to get 
$$w=\frac{-1\pm\sqrt{5}}{2}.$$
For each of these two solutions solve the quadratic equations
$$z+\frac{1}{z} = \frac{-1\pm\sqrt{5}}{2}.$$
Figure out which one is in the 1st quadrant and take its imaginary part.  
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