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Question Number 70913 by TawaTawa last updated on 09/Oct/19

Commented by mathmax by abdo last updated on 10/Oct/19

$w e h a v e \frac{z^{2017} - 1}{z - 1} = \prod_{k = 1}^{2016} (z - e^{\frac{i 2 k π}{2017}})$ $z = - 1 \Rightarrow 1 = \prod_{k = 1}^{2016} (- 1 - e^{\frac{i 2 k π}{2017}}) = \prod_{k = 1}^{2016} (1 + e^{\frac{i 2 k π}{2017}})$ $= \prod_{k = 1}^{2016} (1 + c o s (\frac{2 k π}{2017}) + i s i n (\frac{2 k π}{2017}))$ $= \prod_{k = 1}^{2016} (2 {c o s}^{2} (\frac{k π}{2017}) + 2 i c o s (\frac{k π}{2017}) s i n (\frac{k π}{2017}))$ $= 2^{2016} \prod_{k =}^{2016} c o s (\frac{k π}{2017}) \prod_{k = 1}^{2016} e^{\frac{i k π}{2017}}$ $= 2^{2016} \prod_{k = 1}^{2016} c o s (\frac{k π}{2017}) e^{\frac{i π}{2017} \sum_{k = 1}^{2016} k}$ $= 2^{2016} \prod_{k = 1}^{2016} c o s (\frac{k π}{2017}) e^{\frac{i π}{2017} \frac{2016.2017}{2}}$ $= i^{2016} . 2^{2016} \prod_{k = 1}^{2016} c o s (\frac{k π}{2017}) \Rightarrow \prod_{k = 1}^{2016} c o s (\frac{k π}{2017}) = \frac{1}{2^{2016}} \Rightarrow$ $\prod_{k = 1}^{2016} \frac{1}{c o s (\frac{k π}{2017})} = 2^{2016}$
Commented by mathmax by abdo last updated on 10/Oct/19

$l e t p (z) = z^{2017} - 1 r o o t s o f p (z)$ $p (z) = 0 \Rightarrow z^{2017} = e^{i 2 k π} \Rightarrow z_{k} = e^{\frac{i 2 π k}{2017}} a n d 0 ⩽ k ⩽ 2026$ $p (z) = \prod_{k = 0}^{2016} (z - z_{k}) = (z - 1) \prod_{k = 1}^{2016} (z - e^{\frac{i 2 k π}{2017}}) \Rightarrow$ $\frac{z^{207} - 1}{z - 1} = \prod_{k = 1}^{2016} (z - e^{\frac{i 2 k π}{2017}}) l e t p a s s e t o l i m i t (x \to 1) \Rightarrow$ $2017 = \prod_{k = 1}^{2016} (1 - c o s (\frac{2 k π}{2017}) - i s i n (\frac{2 k π}{2017}))$ $= \prod_{k = 1}^{2016} (2 {s i n}^{2} (\frac{k π}{2017}) - 2 i s i n (\frac{k π}{2017}) c o s (\frac{k π}{2017}))$ $= {(- 2 i)}^{2016} \prod_{k = 1}^{2016} s i n (\frac{k π}{2017}) \prod_{k = 1}^{2016} e^{i \frac{k π}{2017}}$ $= {(- 2 i)}^{2016} e^{\frac{i π}{2017} \sum_{k = 1}^{2016} k} \prod_{k = 1}^{2016} s i n (\frac{k π}{2017})$ $= {(- 2 i)}^{2016} e^{\frac{i π}{2017} \frac{(2016) (2017)}{2}} \prod_{k = 1}^{2016} s i n (\frac{k π}{2017})$ $= {(- 2 i)}^{2016} {(i)}^{2016} \prod_{k = 1}^{2016} s i n (\frac{k π}{2017})$ $= 2^{2016} \prod_{k = 1}^{2016} s i n (\frac{k π}{2017}) \Rightarrow \prod_{k = 1}^{2016} s i n (\frac{k π}{2017}) = \frac{2017}{2^{2016}}$ $r e s t t o c a l c u l a t e \prod_{k = 1}^{2016} c o s (\frac{k π}{2017}) . . . . b e c o n t i n u e d . . . .$
Commented by TawaTawa last updated on 10/Oct/19

$God bless you sir$
Commented by mathmax by abdo last updated on 10/Oct/19

$y o u a r e w e l c o m e .$
	Answered by mind is power last updated on 10/Oct/19
	$Π_(n=1) ^(2016) ((e^(inx) +e^(−inx) )/2)=(1/2^(2016) ).Π_(n=1) ^(2016) (e^(2inx) +1).(1/(Π_(n=1) ^(2016) e^(inx) )) x=(π/(2017)) Z^(2017) −1=0⇒Z^(2017) =e^(2ikπ) Z=e^(i((2kπ)/(2017))) ,k∈{0,1,,2016} Z^(2017) −1=(z−1).Π_(k=1) ^(k=2016) (z−e^((2ikπ)/(2017)) ) ⇒Π_(k=1) ^(2016) (−1−e^((2kπ)/(2017)) )=(((−1)^(2017) −1)/(−2)) ⇒Π_(n=1) ^(2016) (−1)(1+e^(2inx) )=1 ⇒Π_(n=1) ^(2016) (1+e^(2inx) )=1 ⇒∐_(n=1) ^(2016) cos(nx)=(1/2^(2016) ).1.(1/e^(ixΣ_(n=1) ^(2016) n) ) Σ_1 ^(2016) n=1008.2017 e^(ixΣn) =e^(i(π/(2017)) .2017.1008) =e^(i1008π) =1 ⇒∐_(n=1) ^(2016) cos(((nπ)/(2017)))=(1/2^(2016) )⇒Π_(n=1) ^(2016) sec(((nπ)/(2017)))=2^(2016)$
	$\underset{n = 1}{\prod^{2016}} \frac{e^{i n x} + e^{- i n x}}{2} = \frac{1}{2^{2016}} . \underset{n = 1}{\prod^{2016}} (e^{2 i n x} + 1) . \frac{1}{\underset{n = 1}{\prod^{2016}} e^{i n x}} x = \frac{π}{2017}$ $Z^{2017} - 1 = 0 \Rightarrow Z^{2017} = e^{2 i k π}$ $Z = e^{i \frac{2 k π}{2017}}, k \in {0, 1,, 2016}$ $Z^{2017} - 1 = (z - 1) . \underset{k = 1}{\prod^{k = 2016}} (z - e^{\frac{2 i k π}{2017}})$ $\Rightarrow \underset{k = 1}{\prod^{2016}} (- 1 - e^{\frac{2 k π}{2017}}) = \frac{{(- 1)}^{2017} - 1}{- 2}$ $\Rightarrow \underset{n = 1}{\prod^{2016}} (- 1) (1 + e^{2 i n x}) = 1$ $\Rightarrow \underset{n = 1}{\prod^{2016}} (1 + e^{2 i n x}) = 1$ $\Rightarrow \underset{n = 1}{\overset{2016}{∐}} c o s (n x) = \frac{1}{2^{2016}} . 1 . \frac{1}{e^{i x \underset{n = 1}{\sum^{2016}} n}}$ $\underset{1}{\sum^{2016}} n = 1008.2017$ $e^{i x Σ n} = e^{i \frac{π}{2017} . 2017.1008} = e^{i 1008 π} = 1$ $\Rightarrow \underset{n = 1}{\overset{2016}{∐}} c o s (\frac{n π}{2017}) = \frac{1}{2^{2016}} \Rightarrow \underset{n = 1}{\prod^{2016}} s e c (\frac{n π}{2017}) = 2^{2016}$
	Commented by TawaTawa last updated on 10/Oct/19

	$God bless you sir$
	Commented by mind is power last updated on 10/Oct/19

	$y^{'} r e w e l c o m$
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