Expressing $\sin(18^{°})$ in algebraic form from $z^5-1$ as starting point

Question

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.

Answer 1

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.

Answer 2

Start with the fact that $$\zeta=\cos(2\pi/5)+i\sin(2\pi/5)=\sin 18^{\circ}+i\cos 18^{\circ}$$ is a root of $z^5-1=0$ and obviously $\zeta\neq 1$ so that it is a root of $$z^4+z^3+z^2+z+1=0$$ Dividing this by $z^2$ and setting $y=z+z^{-1}$ we have $$y^2+y-1=0$$ On the other hand note that $$\zeta +\zeta^{-1}=2\sin 18^{\circ}$$ and hence the desired value of $\sin 18^{\circ}$ is $y/2$. From quadratic formula $y=(\sqrt{5}-1)/2 $ (the other root is negative) so that $$\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$$