P = cos ((π/(15)))cos (((2π)/(15)))cos (((3π)/(15)))cos (((4π)/(15)))cos (((5π)/(15)))cos (((6π)/(15)))cos (((7π)/(15))) P=?


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Question Number 127684 by liberty last updated on 01/Jan/21

$P = \cos (\frac{π}{15}) \cos (\frac{2 π}{15}) \cos (\frac{3 π}{15}) \cos (\frac{4 π}{15}) \cos (\frac{5 π}{15}) \cos (\frac{6 π}{15}) \cos (\frac{7 π}{15})$ $P = ?$
	Answered by bramlexs22 last updated on 01/Jan/21

	Commented by bobhans last updated on 01/Jan/21

	$n i c e . . . h a p p y n e w y e a r . . .$
	Commented by bramlexs22 last updated on 01/Jan/21
	happy new year too
	Commented by Apurv last updated on 01/Jan/21

	$P = \cos (\frac{π}{15}) \cos (\frac{2 π}{15}) \cos (\frac{3 π}{15}) \cos (\frac{4 π}{15}) \cos (\frac{5 π}{15}) \cos (\frac{6 π}{15}) \cos (\frac{7 π}{15})$ $P =$
	Answered by mindispower last updated on 01/Jan/21
	$cos(x)=((e^(ix) +e^(−ix) )/2)=(e^x /2)(1+e^(2ix) ) let p(x)=X^(15) −1=0 p(x)=0⇒X=e^(2((ikπ)/(15))) ,k∈{0,.....,14} P(x)=Π_(k=0) ^(14) (X−e^(2((ikπ)/(15))) ),P(−1)=−Π_(k=0) ^(14) ^ (1+e^(i2((kπ)/(15 ))) )=−2 =Π_(k=0) ^(14) e^((ikπ)/(15)) (e^((iπk)/(15)) +e^(−((ikπ)/(15))) )=−2^(15) e^(((iπ)/(15))(15.7)) .Π_(k=0) ^(14) cos(((kπ)/(15))) ⇒(1/2^(14) )=.−1Π_(k=1) ^7 cos(((kπ)/(15))).Π_(k=8) ^(14) cos(((kπ)/(15))) k→15−k,in 2nd ⇒(1/2^(14) )=−Π_(k=1) ^7 cos(((kπ)/(15))).Π_(k=1) ^7 cos(π−((kπ)/(15))) (1/2^(14) )=(Π_1 ^7 cos(((kπ)/(15))))^2 ⇒Π_(k.1) ^7 cos(((kπ)/(15)))=+_− (1/2^7 ),∀k∈[1,7]∩N 0≤((kπ)/(15))≤(π/2) cos(((kπ)/(15)))≥0⇒P=(1/(128))$
	$c o s (x) = \frac{e^{i x} + e^{- i x}}{2} = \frac{e^{x}}{2} (1 + e^{2 i x})$ $l e t p (x) = X^{15} - 1 = 0$ $p (x) = 0 \Rightarrow X = e^{2 \frac{i k π}{15}}, k \in {0, . . . . ., 14}$ $P (x) = \underset{k = 0}{\prod^{14}} (X - e^{2 \frac{i k π}{15}}), P (- 1) = - \underset{k = 0}{\prod^{14}} \overset{}{} (1 + e^{i 2 \frac{k π}{15}}) = - 2$ $= \underset{k = 0}{\prod^{14}} e^{\frac{i k π}{15}} (e^{\frac{i π k}{15}} + e^{- \frac{i k π}{15}}) = - 2^{15} e^{\frac{i π}{15} (15.7)} . \underset{k = 0}{\prod^{14}} c o s (\frac{k π}{15})$ $\Rightarrow \frac{1}{2^{14}} = . - 1 \underset{k = 1}{\prod^{7}} c o s (\frac{k π}{15}) . \underset{k = 8}{\prod^{14}} c o s (\frac{k π}{15})$ $k \to 15 - k, i n 2 n d$ $\Rightarrow \frac{1}{2^{14}} = - \underset{k = 1}{\prod^{7}} c o s (\frac{k π}{15}) . \underset{k = 1}{\prod^{7}} c o s (π - \frac{k π}{15})$ $\frac{1}{2^{14}} = {(\underset{1}{\prod^{7}} c o s (\frac{k π}{15}))}^{2}$ $\Rightarrow \underset{k . 1}{\prod^{7}} c o s (\frac{k π}{15}) = \underline{+} \frac{1}{2^{7}}, \forall k \in [1, 7] \cap N 0 ⩽ \frac{k π}{15} ⩽ \frac{π}{2}$ $c o s (\frac{k π}{15}) ⩾ 0 \Rightarrow P = \frac{1}{128}$
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