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Question Number 91465 by Zainal Arifin last updated on 30/Apr/20

Commented by Prithwish Sen 1 last updated on 01/May/20

8.((2.sin((2π)/(15)).cos((2π)/(15)))/(sin((2π)/(15)))) .cos((4π)/(15))cos((8π)/(15))cos((14π)/(15)) =2.((sin((16π)/(15)))/(sin((2π)/(15)))).cos((14π)/(15))=2.((sin(π+(π/(15))))/(sin((2π)/(15)))).cos(π−(π/(15))) =2.((sin(π/(15)))/(2.sin(π/(15)).cos(π/(15)))) . cos(π/(15)) = 1

$$\mathrm{8}.\frac{\mathrm{2}.\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{15}}}{\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{15}}}\:.\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{15}}\mathrm{cos}\frac{\mathrm{8}\pi}{\mathrm{15}}\mathrm{cos}\frac{\mathrm{14}\pi}{\mathrm{15}} \\ $$$$=\mathrm{2}.\frac{\mathrm{sin}\frac{\mathrm{16}\pi}{\mathrm{15}}}{\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{15}}}.\mathrm{cos}\frac{\mathrm{14}\pi}{\mathrm{15}}=\mathrm{2}.\frac{\mathrm{sin}\left(\pi+\frac{\pi}{\mathrm{15}}\right)}{\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{15}}}.\mathrm{cos}\left(\pi−\frac{\pi}{\mathrm{15}}\right) \\ $$$$=\mathrm{2}.\frac{\mathrm{sin}\frac{\pi}{\mathrm{15}}}{\mathrm{2}.\mathrm{sin}\frac{\pi}{\mathrm{15}}.\mathrm{cos}\frac{\pi}{\mathrm{15}}}\:.\:\mathrm{cos}\frac{\pi}{\mathrm{15}}\:=\:\mathrm{1}\: \\ $$

Answered by john santu last updated on 01/May/20

let (π/(15)) = w , cos ((14π)/(15)) = cos(π−(π/(15))) = −cos (π/(15)) ⇒ ((−16cos w cos 2w cos 4w cos 8w)/(sin w)) × sin w = ((−8 sin 2w cos 2w cos 4w cos 8w)/(sin w)) = ((−4sin 4w cos 4w cos 8w)/(sin w)) = ((−2sin 8w cos 8w)/(sin w)) = ((−sin 16w)/(sin w)) sin (((16π)/(15))) = sin (π+(π/(15))) = −sin ((π/(15))) ∴ ((sin ((π/(15))))/(sin ((π/(15))))) = 1

$${let}\:\frac{\pi}{\mathrm{15}}\:=\:{w}\:,\:\mathrm{cos}\:\frac{\mathrm{14}\pi}{\mathrm{15}}\:=\:\mathrm{cos}\left(\pi−\frac{\pi}{\mathrm{15}}\right)\: \\ $$$$=\:−\mathrm{cos}\:\frac{\pi}{\mathrm{15}} \\ $$$$\Rightarrow\:\frac{−\mathrm{16cos}\:{w}\:\mathrm{cos}\:\mathrm{2}{w}\:\mathrm{cos}\:\mathrm{4}{w}\:\mathrm{cos}\:\mathrm{8}{w}}{\mathrm{sin}\:{w}}\:×\:\mathrm{sin}\:{w}\:= \\ $$$$\frac{−\mathrm{8}\:\mathrm{sin}\:\mathrm{2}{w}\:\mathrm{cos}\:\mathrm{2}{w}\:\mathrm{cos}\:\mathrm{4}{w}\:\mathrm{cos}\:\mathrm{8}{w}}{\mathrm{sin}\:{w}}\:= \\ $$$$\frac{−\mathrm{4sin}\:\mathrm{4}{w}\:\mathrm{cos}\:\mathrm{4}{w}\:\mathrm{cos}\:\mathrm{8}{w}}{\mathrm{sin}\:{w}}\:= \\ $$$$\frac{−\mathrm{2sin}\:\mathrm{8}{w}\:\mathrm{cos}\:\mathrm{8}{w}}{\mathrm{sin}\:{w}}\:=\:\frac{−\mathrm{sin}\:\mathrm{16}{w}}{\mathrm{sin}\:{w}} \\ $$$$\mathrm{sin}\:\left(\frac{\mathrm{16}\pi}{\mathrm{15}}\right)\:=\:\mathrm{sin}\:\left(\pi+\frac{\pi}{\mathrm{15}}\right)\:=\:−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{15}}\right) \\ $$$$\therefore\:\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{15}}\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{15}}\right)}\:=\:\mathrm{1} \\ $$

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