Proof for trigonometric identity using complex numbers and series

Exercise:

Show that, $\prod\limits_{k=1}^{n-1}{\sin \frac{k\pi }{n}=\frac{n}{{{2}^{n-1}}}\,\,\,,\,\,\forall \,\,n\ge 2}$

Solution: we know that, $\sin \theta =\frac{{{e}^{i\theta }}-{{e}^{-i\theta }}}{2i}=\frac{1}{2i}\left( {{e}^{i\theta }}-{{e}^{-i\theta }} \right)=\frac{1}{2i}{{e}^{i\theta }}\left( 1-{{e}^{-2i\theta }} \right)$

So $\prod\limits_{k=1}^{n-1}{\sin \frac{k\pi }{n}}=\prod\limits_{k=1}^{n-1}{\left( \frac{1}{2i}{{e}^{i\frac{k\pi }{n}}}\left( 1-{{e}^{-i\frac{2k\pi }{n}}} \right) \right)}=\prod\limits_{k=1}^{n-1}{\frac{1}{2i}\prod\limits_{k=1}^{n-1}{{{e}^{i\frac{k\pi }{n}}}}\prod\limits_{k=1}^{n-1}{\left( 1-{{e}^{-i\frac{2k\pi }{n}}} \right)}}$

But $\prod\limits_{k=1}^{n-1}{\frac{1}{2i}=\underbrace{\frac{1}{2i}\times \frac{1}{2i}\times ...\times \frac{1}{2i}}_{\left( n-1 \right)-times}}={{\left( \frac{1}{2i} \right)}^{n-1}}=\frac{1}{{{2}^{n-1}}}\times \frac{1}{{{i}^{n-1}}}$  also

$\prod\limits_{k=1}^{n-1}{{{e}^{i\frac{k\pi }{n}}}}={{e}^{i\frac{\pi }{n}}}\times {{e}^{i\frac{2\pi }{n}}}\times ....\times {{e}^{i\frac{\left( n-1 \right)\pi }{n}}}={{e}^{i\frac{\pi }{n}\left( 1+2+...+\left( n-1 \right) \right)}}$

But $1+2+3+...+n-1$ is athematic series with common difference is 1

so $1+2+3+...+n-1=\frac{n\left( n-1 \right)}{2}$

thus $\prod\limits_{k=1}^{n-1}{{{e}^{i\frac{k\pi }{n}}}={{e}^{i\frac{\pi }{n}\left( \frac{n}{2}\left( n-1 \right) \right)}}={{e}^{i\frac{\pi }{2}\left( n-1 \right)}}={{\left( {{e}^{i\frac{\pi }{2}}} \right)}^{n-1}}={{i}^{n-1}}}$

so $\prod\limits_{k=1}^{n-1}{\sin \frac{k\pi }{n}=\frac{1}{{{2}^{n-1}}}\times \frac{1}{{{i}^{n-1}}}\times {{i}^{n-1}}\times \prod\limits_{k=1}^{n-1}{\left( 1-{{e}^{-i\frac{2k\pi }{n}}} \right)}}=\frac{1}{{{2}^{n-1}}}\prod\limits_{k=1}^{n-1}{\left( 1-{{e}^{-i\frac{2k\pi }{n}}} \right)}$

Let $\xi ={{e}^{-i\frac{2\pi }{n}}}$ so $\prod\limits_{k=1}^{n-1}{\left( 1-{{e}^{-i\frac{2k\pi }{n}}} \right)=\prod\limits_{k=1}^{n-1}{\left( 1-{{\xi }^{k}} \right)}}=\left( 1-\xi  \right)\left( 1-{{\xi }^{2}} \right)\left( 1-{{\xi }^{3}} \right)...\left( 1-{{\xi }^{n-1}} \right)$

Notice that $\sum\limits_{k=0}^{n-1}{{{x}^{k}}}=1+x+{{x}^{2}}+...+{{x}^{n-1}}=\frac{{{x}^{n}}-1}{x-1}\Leftrightarrow \left( x-1 \right)\sum\limits_{k=0}^{n-1}{{{x}^{k}}={{x}^{n}}-1}$

And ${{x}^{n}}-1=\prod\limits_{k=0}^{n-1}{\left( x-{{\xi }^{k}} \right)}\Leftrightarrow \left( x-1 \right)\sum\limits_{k=0}^{n}{{{x}^{k}}=\prod\limits_{k=0}^{n-1}{\left( x-{{\xi }^{k}} \right)}}$

$\Leftrightarrow \left( x-1 \right)\sum\limits_{k=0}^{n}{{{x}^{k}}=\left( x-{{\xi }^{0}} \right)\prod\limits_{k=1}^{n-1}{\left( x-{{\xi }^{k}} \right)=\left( x-1 \right)\prod\limits_{k=1}^{n-1}{\left( x-{{\xi }^{k}} \right)}}}\Leftrightarrow \prod\limits_{k=1}^{n-1}{\left( x-{{\xi }^{k}} \right)=\sum\limits_{k=0}^{n}{{{x}^{k}}}}$

Put $x=1\Leftrightarrow \prod\limits_{k=1}^{n-1}{\left( 1-{{\xi }^{k}} \right)=\sum\limits_{k=0}^{n}{{{1}^{k}}=\underbrace{1+1+...+1}_{n-summand}}=1\times n=n}$

So $\prod\limits_{k=1}^{n-1}{\sin \frac{k\pi }{n}=\frac{n}{{{2}^{n-1}}}\,\,\,,\,\,\,n\ge 2}$ as result $\prod\limits_{k=1}^{6}{\sin \frac{k\pi }{7}=\frac{7}{{{2}^{6}}}}$