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 $\prod\limits_{k=1}^{6}{\sin
\frac{k\pi }{7}=\frac{7}{{{2}^{6}}}}$
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