2

Question

Let $\omega=\cos\dfrac{4\pi}{7}+i\sin\dfrac{4\pi}{7}$. Show that

$\omega-1=2\sin\dfrac{2\pi}{7}\left(\cos\dfrac{11\pi}{14}+i\sin\dfrac{11\pi}{14}\right)$.

My attempt

Observe that $\omega$ is a seventh root of unity. Label the roots $1, \nu, \nu^2,\ldots,\nu^6$. Then $\omega=\nu^2$.

We have $1+ \nu+ \nu^2+\ldots+\nu^6=0$ and so $1+ \nu+ \omega+\ldots+\nu^6=0$.

Then $\omega-1=-2-(\nu+\nu^6)-(\nu^3+\nu^4)-\nu^5$.

But $\nu+\nu^6=2\cos\dfrac{2\pi}{7}$ and $\nu^3+\nu^4=2\cos\dfrac{6\pi}{7}$

I do not know how to continue.

bibo_extreme
  • 560
  • 3
  • 14

2 Answers2

3

Using $$\cos(2\theta)-1=-2\sin^2\theta,\ \ \sin(2\theta)=2\sin\theta\cos\theta$$ $$-\sin\theta=\cos\left(\frac{\pi}{2}+\theta\right),\ \ \cos\theta=\sin\left(\frac{\pi}{2}+\theta\right)$$gives you$$\begin{align}\omega-1&=\cos\frac{4\pi}{7}+i\sin\frac{4\pi}{7}-1\\&=\cos\frac{4\pi}{7}-1+i\sin\frac{4\pi}{7}\\&=-2\sin^2\frac{2\pi}{7}+i\cdot 2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\\&=2\sin\frac{2\pi}{7}\left(-\sin\frac{2\pi}{7}+i\cos\frac{2\pi}{7}\right)\\&=2\sin\frac{2\pi}{7}\left(\cos\frac{11\pi}{14}+i\sin\frac{11\pi}{14}\right)\end{align}$$

mathlove
  • 139,939
1

I would do it differently. Recall identities

$$\cos 2x = 1 - 2\sin^2 x \\[1ex] \sin 2x = 2 \sin x \cos x$$

to obtain

$$\cos \frac{4 \pi}{7} = 1 - 2 \sin^2 \frac{2 \pi}{7} \\[1ex] \sin \frac{4 \pi}{7} = 2 \sin \frac{2 \pi}{7} \cos \frac{2 \pi}{7}$$

and therefore

$$\omega - 1 = 2 \sin \frac{2 \pi}{7} \left( - \sin \frac{2 \pi}{7} + i \cos \frac{2 \pi}{7} \right).$$

But

$$ \begin{align*} - \sin \frac{2 \pi}{7} + i \cos \frac{2 \pi}{7} & = i \left( \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7} \right) = \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \left( \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7} \right) \\[2ex] & = \cos \left( \frac{\pi}{2} + \frac{2 \pi}{7} \right) + i \sin \left( \frac{\pi}{2} + \frac{2 \pi}{7} \right) = \cos \frac{11 \pi}{14} + i \sin \frac{11 \pi}{14}. \end{align*}$$

Adayah
  • 10,468