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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.

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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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  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh Sep 21 '18 at 12:48
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    $\begingroup$ @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. $\endgroup$ – B. Goddard Sep 21 '18 at 12:54
  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh Sep 22 '18 at 15:31
  • $\begingroup$ 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 $\endgroup$ – Paramanand Singh Sep 22 '18 at 15:32
  • $\begingroup$ @paramanandSingh Thanks for your contribution. Both of your answers helped me learn something. $\endgroup$ – Cro-Magnon Sep 23 '18 at 17:03
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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 $$\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$$

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