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Using the ninth root of unity of 1, show that $cos\frac{π}{9}cos\frac{2π}{9}cos\frac{4π}{9}=\frac{1}{8}$.

Here is my solution.

Let $\omega=cos\theta+isin\theta$,

As $\omega^9=1$,

$$cos9\theta+isin9\theta=1$$ so $$9\theta=2nπ {(n\in Z)}$$
$$\theta=\frac{2nπ}{9} {(n\in Z)}$$ $$\theta=\frac{\pm2π}{9}, \frac{\pm4π}{9}, \frac{\pm6π}{9}, \frac{\pm8π}{9},$$ $$\omega=cis\frac{\pm2π}{9}, cis\frac{\pm4π}{9}, cis\frac{\pm6π}{9}, cis\frac{\pm8π}{9},$$ therefore $$\omega^9-1=0$$
$$(\omega-1)(1+\omega +\omega^2 +\omega^3+\omega^4+\omega^5+\omega^6+\omega^7+\omega^8)=0$$ and because $\omega$ is not real,
therefore $$1+\omega +\omega^2 +\omega^3+\omega^4+\omega^5+\omega^6+\omega^7+\omega^8=0$$ through factorisation,
$$(\omega - cis\frac{2π}{9})(\omega - cis\frac{-2π}{9})...(\omega - cis\frac{8π}{9})(\omega - cis\frac{-8π}{9})=1+\omega +\omega^2 +...+\omega^7+\omega^8$$

$$(\omega^2 - 2cos\frac{2π}{9}\omega +1)...(\omega^2 - 2cos\frac{8π}{9}\omega +1)=1+\omega +\omega^2 +...+\omega^7+\omega^8$$

divide both sides by $\omega^4$, $$(\omega - 2cos\frac{2π}{9} +\frac{1}{\omega})...(\omega - 2cos\frac{8π}{9} +\frac{1}{\omega})=\frac{1}{\omega^4}+\frac{1}{\omega^3}+... +\omega^3+\omega^4$$ Let $x=\omega+\frac{1}{\omega}$,
$$\omega^2+\frac{1}{\omega^2}=(\omega+\frac{1}{\omega})^2-2=x^2-2$$ similarly, $$\omega^3+\frac{1}{\omega^3}=x^3-3x$$ $$\omega^4+\frac{1}{\omega^4}=x^4-4(x^2-2)-6=x^4-4x^2+2$$

therefore, substituting in $$x^4+x^3-3x^2-2x+1=(x - 2cos\frac{2π}{9})...(x - 2cos\frac{8π}{9})$$ $$2^4cos\frac{2π}{9}cos\frac{4π}{9}cos\frac{6π}{9}cos\frac{8π}{9}=1$$ as $cos\frac{8π}{9}=-cos\frac{π}{9}$ and $cos\frac{6π}{9}=-\frac{1}{2}$ $$cos\frac{π}{9}cos\frac{2π}{9}cos\frac{4π}{9}=\frac{1}{8}$$

I believe that I might have over-complicated this question. Any other thoughts?

Winson
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1 Answers1

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Root of Unity Approach

Using $$ \omega=\cos\left(\frac\pi9\right)+i\sin\left(\frac\pi9\right) $$ we get $\omega^9=-1$ $$ \begin{align} \overbrace{\left(\omega-\omega^8\right)}^{2\cos\left(\frac\pi9\right)}\overbrace{\left(\omega^2-\omega^7\right)}^{2\cos\left(\frac{2\pi}9\right)}\overbrace{\left(\omega^4-\omega^5\right)}^{2\cos\left(\frac{4\pi}9\right)} &=\omega^7-\omega^8-\omega^{12}+\omega^{13}-\omega^{14}+\omega^{15}+\omega^{19}-\omega^{20}\\ &=\omega^7-\omega^8+\omega^3-\omega^4+\omega^5-\omega^6+\omega^1-\omega^2\\[3pt] &=1 \end{align} $$

Trigonometric Identity Approach

We can also use $\sin(2x)=2\sin(x)\cos(x)$, $$ \begin{align} \sin\left(\frac{\pi}9\right)\cos\left(\frac{\pi}9\right)\cos\left(\frac{2\pi}9\right)\cos\left(\frac{4\pi}9\right) &=\frac12\sin\left(\frac{2\pi}9\right)\cos\left(\frac{2\pi}9\right)\cos\left(\frac{4\pi}9\right)\\ &=\frac14\sin\left(\frac{4\pi}9\right)\cos\left(\frac{4\pi}9\right)\\ &=\frac18\sin\left(\frac{8\pi}9\right) \end{align} $$ Now divide by $\sin\left(\frac{\pi}9\right)=\sin\left(\frac{8\pi}9\right)$.

robjohn
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