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Question Number 201033 by Mingma last updated on 28/Nov/23

	Answered by Frix last updated on 28/Nov/23

	$\sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{9 π}{14} = x$ $0 < x < 1$ $Using trigonometric formulas \Rightarrow$ $\frac{1}{4} (1 - \cos \frac{π}{7} + \sin \frac{3 π}{14} - \sin \frac{π}{14}) = x ★$ $\cos \frac{π}{7} - \sin \frac{3 π}{14} + \sin \frac{π}{14} = 1 - 4 x$ ${(\cos \frac{π}{7} - \sin \frac{3 π}{14} + \sin \frac{π}{14})}^{3} = {(1 - 4 x)}^{3}$ $Expanding & using trig . formulas \Rightarrow$ $- \frac{15}{4} + \frac{31}{4} (\cos \frac{π}{7} - \sin \frac{3 π}{14} + \sin \frac{π}{14}) = {(1 - 4 x)}^{3}$ $Using ★ \Rightarrow$ $x^{3} - \frac{3 x^{2}}{4} - \frac{19 x}{64} + \frac{3}{64} = 0$ $x = \frac{1}{8}$ $(x = - \frac{3}{8} \land x = 1 not valid)$
	Commented by Mingma last updated on 28/Nov/23
	Nice solution!
	Answered by som(math1967) last updated on 28/Nov/23

	$l e t \frac{π}{14} = θ \Rightarrow 14 θ = π$ $s i n θ s i n 3 θ s i n 9 θ$ $= \frac{1}{2 c o s θ} \times 2 s i n θ c o s θ s i n 3 θ s i n 9 θ$ $= \frac{1}{2 c o s θ} \times s i n 2 θ s i n 3 θ s i n 9 θ$ $= \frac{1}{4 c o s θ} \times 2 s i n 2 θ s i n 3 θ s i n 9 θ$ $= \frac{1}{4 c o s θ} \times (c o s θ - c o s 5 θ) s i n 9 θ$ $= \frac{1}{8 c o s θ} \times [2 s i n 9 θ c o s θ - 2 s i n 9 θ c o s 5 θ)$ $= \frac{1}{8 c o s θ} \times (s i n 10 θ + s i n 8 θ - s i n 14 θ - s i n 4 θ)$ $= \frac{1}{8 c o s θ} [s i n (π - 4 θ) + s i n (π - 6 θ) - 0 - s i n 4 θ]$ $= \frac{1}{8 c o s θ} \times s i n 6 θ$ $= \frac{1}{8 c o s θ} \times c o s θ = \frac{1}{8}$ $[π = 14 θ \Rightarrow \frac{π}{2} = 7 θ \Rightarrow \frac{π}{2} - θ = 6 θ$ $∴ s i n (\frac{π}{2} - θ) = s i n 6 θ \Rightarrow c o s θ = s i n 6 θ]$
	Commented by Mingma last updated on 28/Nov/23
	Nice solution!
	Answered by Sutrisno last updated on 29/Nov/23

	$= \frac{2 . c o s (\frac{π}{14}) s i n (\frac{π}{14}) s i n (\frac{7 π}{14} - \frac{4 π}{14}) s i n (\frac{7 π}{14} + \frac{2 π}{14})}{2 c o s (\frac{π}{14})}$ $= \frac{s i n (\frac{2 π}{14}) c o s (\frac{4 π}{14}) c o s (\frac{2 π}{14})}{2 c o s (\frac{π}{14})}$ $= \frac{2 . s i n (\frac{2 π}{14}) c o s (\frac{2 π}{14}) c o s (\frac{4 π}{14})}{4 c o s (\frac{π}{14})}$ $= \frac{2 s i n (\frac{4 π}{14}) c o s (\frac{4 π}{14})}{8 s i n (\frac{7 π}{14} + \frac{π}{14})}$ $= \frac{s i n (\frac{8 π}{14})}{8 s i n (\frac{8 π}{14})}$ $= \frac{1}{8}$
	Answered by MathematicalUser2357 last updated on 20/Jan/24

	$\sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{9 π}{14}$ $= \cos (\frac{π}{2} - \frac{π}{14}) \cos (\frac{π}{2} - \frac{3 π}{14}) \cos (\frac{π}{2} - \frac{9 π}{14})$ $= \frac{1}{2} (\cos (\frac{π}{2} - \frac{9 π}{14}) \cdot [\cos {(\frac{π}{2} - \frac{π}{14}) + (\frac{π}{2} - \frac{3 π}{14})} + \cos {(\frac{π}{2} - \frac{π}{14}) - (\frac{π}{2} - \frac{3 π}{14})}])$ $= \frac{1}{2} (\cos (\frac{π}{2} - \frac{9 π}{14}) \cos {(\frac{π}{2} - \frac{π}{14}) + (\frac{π}{2} - \frac{3 π}{14})} + \cos (\frac{π}{2} - \frac{9 π}{14}) \cos {(\frac{π}{2} - \frac{π}{14}) - (\frac{π}{2} - \frac{3 π}{14})})$ $= \frac{1}{2} {\frac{1}{2} (\cos [(\frac{π}{2} - \frac{9 π}{14}) + {(\frac{π}{2} - \frac{π}{14}) + (\frac{π}{2} - \frac{3 π}{14})}] + \cos [(\frac{π}{2} - \frac{π}{14}) - {(\frac{π}{2} - \frac{π}{14}) + (\frac{π}{2} - \frac{3 π}{14})}]) + \frac{1}{2} (\cos [(\frac{π}{2} - \frac{9 π}{14}) + {(\frac{π}{2} - \frac{π}{14}) - (\frac{π}{2} - \frac{3 π}{14})}] + \cos [(\frac{π}{2} - \frac{9 π}{14}) - {(\frac{π}{2} - \frac{π}{14}) - (\frac{π}{2} - \frac{3 π}{14})}])}$ $= \frac{1}{2} [\frac{1}{2} {\cos \frac{4 π}{7} + \cos (- \frac{2 π}{7})} + \frac{1}{2} {\cos (- \frac{2 π}{7}) + 1}]$ $= \frac{1}{4} {\cos \frac{4 π}{7} + \cos (- \frac{2 π}{7}) + \cos (- \frac{2 π}{7}) + 1}$
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