cos(4π/5)-cos(3π/5)+cos(2π/5)-cos(π/5)=?

Question

cos(4π/5)-cos(3π/5)+cos(2π/5)-cos(π/5)=?

Answer 1

由公式cos(π-a)= -cosa可以知道,
cos(4π/5) - cos(π/5)= -2cos(π/5)
-cos(3π/5)+cos(2π/5)= -2cos(3π/5)
所以
原式= -2cos(π/5) -2cos(3π/5)
= -2[cos(π/5)+cos(3π/5)]
再由公式cos α+cos β=2cos[(α+β)/2]*cos[(α-β)/2]
所以
cos(π/5)+cos(3π/5) =2cos(π/5)*cos(2π/5)
又
cos(π/5)*cos(2π/5)
=sin(π/5)*cos(π/5)*cos(2π/5) / sin(π/5)
=0.5sin(2π/5)*cos(2π/5) / sin(π/5)
=0.25 sin(4π/5) / sin(π/5)
而sin(π-a)=sina,
所以sin(4π/5)=sin(π/5)
故cos(π/5)*cos(2π/5)=0.25 sin(4π/5) / sin(π/5)=0.25
所以
cos(4π/5)-cos(3π/5)+cos(2π/5)-cos(π/5)
= -2[cos(π/5)+cos(3π/5)]
= -4cos(π/5)*cos(2π/5)
= -4×0.25
= -1