# 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.

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.

• There is no need to go to $z^{20}$. A minor trig identity helps us to deal with $z^5-1=0$ directly. See my answer. – Paramanand Singh Sep 21 '18 at 12:48
• @ParamanandSingh It's not about "need", it's about clarity of exposition. Students don't learn when you try to make them drink out of a fire hydrant. – B. Goddard Sep 21 '18 at 12:54
• Ok, I did have a look at your answer again. As you claim $a=\cos 18^{\circ}+i\sin 18^{\circ}$ satisfies $a^4+a^3+a^2+a+1=0$ and multiplying by $(a-1)$ we see that $a^5=1$, but $a^5=\cos 90^{\circ}+i\sin 90^{\circ} =i$, so your solution is wrong. – Paramanand Singh Sep 22 '18 at 15:31
• It's also surprising that people just upvote without checking correctness. And it's get accepted too. I doubt if OP really checked the details – Paramanand Singh Sep 22 '18 at 15:32
• @paramanandSingh Thanks for your contribution. Both of your answers helped me learn something. – Cro-Magnon Sep 23 '18 at 17:03

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 $$\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$$