The full question goes: If $\omega$ is the solution to $z^7 = 1$ with the least positive argument, determine the argument of $1 + \omega$. Express your answer in terms of $\pi$.
And the mark scheme goes: $\omega$ has argument $2\pi/7$ and $1 + \omega$ has argument $\phi$, then $$\tan\phi = \frac{\sin\frac{2\pi}7}{1+\cos\frac{2\pi}7} = \frac{2\sin\frac\pi7\cos\frac\pi7}{2\cos^2\frac\pi7} = \tan\frac\pi7$$ therefore $\phi = \frac\pi7$.
Could someone please explain how tan came into the picture and why the expression is $ \tan\phi = \frac{\sin\frac{2\pi}7}{1+\cos\frac{2\pi}7}$.
Assume $e^{2\pi i/7}+1 = |e^{2\pi i/7}+1|e^{i\phi}$. We have $$\cos\phi + i\sin\phi = \frac{\left(1+\cos\frac{2\pi}7\right) + i\sin\frac{2\pi}7}{\sqrt{\left(1+\cos\frac{2\pi}7\right)^2 + \sin^2\frac{2\pi}7}}= \frac{\left(1+\cos\frac{2\pi}7\right) + i\sin\frac{2\pi}7}{\sqrt{2\left(1+\cos\frac{2\pi}7\right)}}$$
so $\cos\phi = \frac{1+\cos\frac{2\pi}7}{\sqrt{2\left(1+\cos\frac{2\pi}7\right)}}$ and $\sin\phi = \frac{\sin\frac{2\pi}7}{\sqrt{2\left(1+\cos\frac{2\pi}7\right)}}$.
Therefore
$$\tan\phi = \frac{\sin\phi}{\cos\phi} = \frac{\sin\frac{2\pi}7}{1+\cos\frac{2\pi}7} = \frac{2\sin\frac\pi7\cos\frac\pi7}{2\cos^2\frac\pi7} = \frac{\sin\frac\pi7}{\cos\frac\pi7}=\tan\frac\pi7$$ so $\phi = \frac\pi7$ because $e^{2\pi i/7}+1$ is in the first quadrant.