Roots of unity question. 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.
 A: 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$$\begin{align}\omega-1&=\cos\frac{4\pi}{7}+i\sin\frac{4\pi}{7}-1\\&=\cos\frac{4\pi}{7}-1+i\sin\frac{4\pi}{7}\\&=-2\sin^2\frac{2\pi}{7}+i\cdot 2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\\&=2\sin\frac{2\pi}{7}\left(-\sin\frac{2\pi}{7}+i\cos\frac{2\pi}{7}\right)\\&=2\sin\frac{2\pi}{7}\left(\cos\frac{11\pi}{14}+i\sin\frac{11\pi}{14}\right)\end{align}$$
A: 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 
$$
\begin{align*}
- \sin \frac{2 \pi}{7} + i \cos \frac{2 \pi}{7} & = i \left( \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7} \right) = \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \left( \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7} \right) \\[2ex]
& = \cos \left( \frac{\pi}{2} + \frac{2 \pi}{7} \right) + i \sin \left( \frac{\pi}{2} + \frac{2 \pi}{7} \right) = \cos \frac{11 \pi}{14} + i \sin \frac{11 \pi}{14}.
\end{align*}$$
