cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.

cos(2派/7)+cos(4派/7)+cos(6派/7)=? 1：COS(-7π/12) = 若函数f(x)=[2cos^3 x-sin^2 (x+π)-2cos (-x-π)+1]/[2+2cos^2(7π+x) 已知cosα=1/7,cos（α+β）=-11/14,且α、β∈（0,2/Π）,求cosβ的值 cos(π/32)*cos(π/16)*cos(π/8)= 化简sin（2π+a)cos(π-a)cos(π/2-a）cos(7π/2-a)/cos(π-a)sin(3π-a)si 不用计算器,求(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^ 已知(sinα-cosα)/(sinα+cosα)=1/3,则cos^4(π/3+α)-cos^4(π/6-α)的值为 已知cosα=1/7,cos(α+β)=-11/14,α β∈(0,π/2)求β ①化简[sin(2π+α)*cos(7π-α)]/cos(-α)②已知cosθ=-3/5,θ∈(π/2,π),求sin( 求证cos2π/7+cosπ/3=cosπ/7+cos3π/7 求下列各式的值:(1)cos(-165°) (2)cos(7π/12) (3)cos(-61π/12) 求详解