How to use Euler formula to prove the following conclusion

Question

I only know how to prove it use Fourier series. $$\ln(\sin{x})=-\ln2-\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}$$

Answer 1

Hint: $$ \log (\sin x) = \Re \log (\sin x) = \Re \log \left( {\frac{{e^{ix} - e^{ - ix} }}{{2i}}} \right) = - \log 2 + \Re \log (1 - e^{ - 2ix} ). $$

Answer 2

$$\sum_{k=1}^\infty\dfrac{cos(2k x)}k$$

$=$ the real part of $$\sum_{k=1}^\infty\dfrac{(e^{2ix})^k}k=-\ln(1-e^{2ix})$$

Now $\ln(1-e^{2ix})$

$\equiv\ln(-1)+\ln(e^{ix})+\ln(e^{ix}-e^{-ix})\pmod{2\pi i}$

$\equiv i\pi+ix+\ln(2i\sin x)\pmod{2\pi i}$

$\equiv i\pi+ix+\ln2+\ln(\sin x)+\ln i$

whose real part is $\ln2+\ln(\sin x)$