How do you evaluate $(1+cos(pi/8))(1+cos((3pi)/8))(1+cos((5pi)/8))(1+cos((7pi)/8))$ ?

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Answer 1 · 2016-12-11T02:27:21

$(1+cos(pi/8))(1+cos((3pi)/8))(1+cos((5pi)/8))(1+cos((7pi)/8))$

$=(1+cos(pi/8))(1+sin(pi/2-(3pi)/8))(1+sin(pi/2-(5pi)/8))(1+cos(pi-pi/8))$

$=(1+cos(pi/8))(1+sin(pi/8))(1-sin(pi/8))(1-cos(pi/8))$

$=(1-cos^2(pi/8))(1-sin^2(pi/8))$

$=sin^2(pi/8)cos^2(pi/8)$

$=1/4xx(2sin(pi/8)cos(pi/8))^2$

$=1/4xxsin^2((2pi)/8)$

$=>1/4xxsin^2(pi/4)$

$=1/4xx(1/sqrt2)^2$

$=1/8$

How do you evaluate (1+cos(pi/8))(1+cos((3pi)/8))(1+cos((5pi)/8))(1+cos((7pi)/8))(1+cos(pi/8))(1+cos((3pi)/8))(1+cos((5pi)/8))(1+cos((7pi)/8))?