Question

求值：Cos2/7π＋Cos4/7π＋Cos6/7π

Answer 1

是Cos(π2/7)＋Cos(π4/7)＋Cos(π6/7)吧？
Cos(π2/7)＋Cos(π4/7)＋Cos(π6/7)
=2cos(3π/7)cos(π/7)+2(cos3π/7)^2-1……（一）
=2(cosπ/7+cos3π/7)cos3π/7-1
=4cos(2π/7)cos(π/7)cos(3π/7)-1……（二）
=4sin(π/7)cos(π/7)cos(2π/7)cos(3π/7)/sin(π/7)-1
=2sin(2π/7)cos(2π/7)cos(3π/7)/sin(π/7)-1（分子分母同乘sin(π/7))
=sin(4π/7)cos(3π/7)/sin(π/7)-1
=(sinπ+sinπ/7)/sin(π/7)-1……（三）
=1-1
=0

（一）和（三）用了和差化积公式:cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2]）；（二）用了积化和差公式:sinAcosB=[sin(A+B)+sin(A-B)]/2

Answer 2

cosπ/7·cos2π/7·cos4π/7
=sinπ/7·cosπ/7·cos2π/7·cos4π/7/(sinπ/7)
=1/2sin2π/7·cos2π/7·cos4π/7/(sinπ/7)
=1/4sin4π/7·cos4π/7/(sinπ/7)
=1/8sin8π/7/(sinπ/7)
=1/8sinπ/7/(sinπ/7)
=1/8

Answer 3

-1/2