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

Question

64056 views around the world

You can reuse this answer
Creative Commons License

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#