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Question Number 118899 by bobhans last updated on 20/Oct/20

\sin (\frac{π}{14}) \sin (\frac{2 π}{14}) \sin (\frac{3 π}{14}) \dots \sin (\frac{6 π}{14}) = ?

Answered by TANMAY PANACEA last updated on 20/Oct/20

t a k i n g h e l p f r o m S . L L o n e y T r i g o n o m e t r y

2^{\frac{n - 1}{2}} s i n (\frac{2 π}{2 n}) s i n (\frac{4 π}{2 n}) s i n (\frac{6 π}{2 n}) \dots s i n (\frac{n - 2}{2 n}) π = \sqrt{n}

p = s i n (\frac{2 π}{28}) s i n (\frac{4 π}{28}) s i n (\frac{6 π}{28}) s i n (\frac{8 π}{28}) s i n (\frac{10 π}{28}) s i n (\frac{12 π}{28})

s o c o m p a r i n g

2 n = 28 \to n = 14

r e q u i r e d a n s w e r = \frac{\sqrt{n}}{2^{\frac{n - 1}{2}}}

= \frac{\sqrt{14}}{2^{\frac{13}{2}}} = \frac{\sqrt{7} \times \sqrt{2}}{2^{6} \times \sqrt{2}} = \frac{\sqrt{7}}{2^{6}} A n s w e r

Commented by TANMAY PANACEA last updated on 20/Oct/20

Answered by mindispower last updated on 20/Oct/20

l e t z^{14} - 1 = 0

\Rightarrow z^{14} - 1 = \underset{k = 0}{\prod^{13}} (X - e^{\frac{2 i k π}{14}}) \Rightarrow \underset{k = 1}{\prod^{13}} (1 - e^{\frac{2 i k π}{14}}) = 14

(1 - e^{\frac{2 i k π}{14}}) = e^{\frac{i k π}{14}} (e^{\frac{- i k π}{14}} - e^{\frac{i k π}{14}})

= - 2 i s i n (\frac{k π}{14}) e^{i k \frac{π}{14}}

\Rightarrow \underset{k = 1}{\prod^{13}} (- 2 i s i n (\frac{k π}{14})) . e^{i \frac{π}{14} Σ k}) = 14

\Rightarrow {(- 2 i)}^{13} \underset{k = 1}{\prod^{6}} s i n (\frac{k π}{14}) . \underset{k = 7}{\prod^{13}} s i n (\frac{k π}{14}) e^{i \frac{π}{14} (7.13)} = 14

\Rightarrow {(- 2)}^{13} . i . \underset{k = 1}{\prod^{6}} s i n (\frac{k π}{14}) . \underset{k = 1}{\prod^{7}} s i n (\frac{(14 - k) π}{14}) e^{\frac{1}{2} 13 i π} = 14

s i n (π - x) = s i n (x), s i n (\frac{7 π}{14}) = s i n (\frac{π}{2}) = 1, e^{\frac{13 i π}{2}} = i

\Rightarrow {(2)}^{13} {[\underset{k = 1}{\prod^{6}} s i n (\frac{k π}{14})]}^{2} = 14

\Rightarrow \underset{k = 1}{\prod^{6}} s i n (\frac{k π}{14}) = \frac{\sqrt{7}}{2^{6}}

Commented by TANMAY PANACEA last updated on 20/Oct/20

e x c e l l e n t s i f

Commented by TANMAY PANACEA last updated on 20/Oct/20

s i r

Commented by mindispower last updated on 20/Oct/20

t h a n k y o u s i r