If #sin18@=(sqrt(5)-1)/4# and #cos36@=(sqrt(5)+1)/4# then prove that #(1+cos(pi/10))(1+cos((3pi)/10))(1+cos((7pi)/10))(1+cos((9pi)/10))=(1/16)#?

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Answer 1 · 2017-10-23T04:37:20

#LHS=(1+cos(pi/10))(1+cos((3pi)/10))(1+cos((7pi)/10))(1+cos((9pi)/10))#

Puttingv#pi/10=2theta# we have

#LHS=(1+cos(2theta))(1+cos(6theta))(1+cos(14theta))(1+cos(18theta))#

#=(2cos^2(theta))(2cos^2(3theta))(2cos^2(7theta))(2cos^2(9theta))#

#=[(2cos(theta)cos(9theta))(2cos(3theta)cos(7theta))]^2#

#=[(cos(10theta)+cos(8theta))(cos(10theta)+cos(4theta))]^2#

#=[(cos(pi/2)+cos(8theta))(cos(pi/2)+cos(4theta))]^2#

#=[cos(8theta)cos(4theta)]^2#

#=[(4cos(8theta)cos(4theta)sin(4theta))/(4sin(4theta))]^2#

#=1/16[sin(16theta)/sin(4theta)]^2#

#=1/16[sin((8 pi)/10)/sin((2pi)/10)]^2#

#=1/16[sin(pi-(2pi)/10)/sin((2pi)/10)]^2#

#=1/16[sin((2pi)/10)/sin((2pi)/10)]^2#

#=(1/16)=RHS#