Question

Siplifier cos(π/7)+cos(2π/7)+cos(3π/7)+cos(4π/7)

Answer 1

We know that
cos a+cos b=cos[(a+b)/2]*cos[(a-b)/2]
we have 
cos(π/7)+cos(2π/7)+cos(3π/7)+cos(4π/7)-------------> equation 1

cos(4π/7)+cos(2π/7)=cos[(4π/7+2π/7)/2]*cos[(4π/7-2π/7)/2]
=cos(3π/7)*cos(π/7)
then
cos(4π/7)+cos(2π/7)=cos(3π/7)*cos(π/7)--------------> equation 2

[cos(3π/7)+cos(π/7)]=cos[(3π/7+π/7)/2]*cos[(3π/7-π/7)/2]
=cos(2π/7)*cos(π/7)
then
[cos(3π/7)+cos(π/7)]=cos(2π/7)*cos(π/7)-----------> equation 3


I substitute 2 and 3 in 1
[cos(3π/7)+cos(π/7)]+[cos(4π/7)+cos(2π/7)]
{cos(2π/7)*cos(π/7}+{cos(3π/7)*cos(π/7)}
=cos(π/7)*[cos(2π/7)+cos(3π/7)]

the answer is 
cos(π/7)+cos(2π/7)+cos(3π/7)+cos(4π/7)=cos(π/7)*[cos(2π/7)+cos(3π/7)]