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)?

1 Answer
Oct 23, 2017

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

Puttingvpi/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