cos ((π/(17)))×cos (((2π)/(17)))×cos (((4π)/(17)))×cos (((8π)/(17))) =


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Question Number 86716 by john santu last updated on 30/Mar/20

$\cos (\frac{π}{17}) \times \cos (\frac{2 π}{17}) \times \cos (\frac{4 π}{17}) \times \cos (\frac{8 π}{17}) =$
Commented by jagoll last updated on 30/Mar/20

$\frac{\cos x \cos 2 x \cos 4 x \cos 8 x}{2 \sin x} \times 2 \sin x$ $\frac{\sin 2 x \cos 2 x \cos 4 x \cos 8 x}{2 \sin x} =$ $\frac{\sin 4 x \cos 4 x \cos 8 x}{4 \sin x} =$ $\frac{\sin 8 x \cos 8 x}{8 \sin x} = \frac{\sin 16 x}{16 \sin x} = \frac{\sin \frac{π}{17}}{16 \sin \frac{π}{17}}$ $= \frac{1}{16}$ $[\sin \frac{π}{17} = \sin (π - \frac{π}{17}) = \sin (\frac{16 π}{17})]$
Commented by mathmax by abdo last updated on 30/Mar/20

$l e t A = c o s (\frac{π}{17}) c o s (\frac{2 π}{17}) c o s (\frac{4 π}{17}) c o s (\frac{8 π}{17}) w e h a v e$ $s i n (\frac{π}{17}) A = \frac{1}{2} s i n (\frac{2 π}{17}) c o s (\frac{2 π}{17}) c o s (\frac{4 π}{17}) c o s (\frac{8 π}{17})$ $= \frac{1}{4} s i n (\frac{4 π}{17}) c o s (\frac{4 π}{17}) c o s (\frac{8 π}{17}) = \frac{1}{8} s i n (\frac{8 π}{17}) c o s (\frac{8 π}{17})$ $= \frac{1}{16} s i n (\frac{16 π}{17}) = \frac{1}{16} s i n (π - \frac{π}{17}) = \frac{1}{16} s i n (\frac{π}{17}) \Rightarrow$ $A = \frac{1}{16}$
	Answered by redmiiuser last updated on 30/Mar/20

	$\frac{1}{16}$
	Answered by TANMAY PANACEA. last updated on 30/Mar/20

	$p = c o s a c o s 2 a c o s 4 a c o s 8 a [a = \frac{π}{17}]$ $2 p s i n a = (2 s i n a c o s a) c o s 2 a c o s 4 a c o s 8 a$ $= s i n 2 a c o s 2 a c o s 4 a c o s 8 a$ $2^{2} p s i n a = (2 s i n 2 a c o s 2 a) c o s 4 a c o s 8 a$ $2^{2} p s i n a = s i n 4 a c o s 4 a c o s 8 a$ $2^{3} p s i n a = (2 s i n 4 a c o s 4 a) c o s 8 a$ $2^{3} p s i n a = s i n 8 a c o s 8 a$ $2^{4} p s i n a = (2 s i n 8 a c o s 8 a)$ $2^{4} p s i n a = s i n 16 a$ $16 \times p \times s i n a = s i n (17 a - a)$ $p = \frac{1}{16 s i n a} \times s i n (π - a)$ $p = \frac{s i n a}{16 s i n a} = \frac{1}{16}$
	Commented by peter frank last updated on 31/Mar/20

	$t h a n k y o u b o t h$
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