不查表,求cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11)的值
方程z^11=1在复数范围内有11个根, 根据根与系数的关系,这11个根之和等于0,因而实部为0, 根据三角诱导公式, 所以原式=1/2 详见附件
运用和差化积公式
cos α+cos β=2cos[(α+β)/2]·cos[(α-β)/2]
cos α-cos β=-2sin[(α+β)/2]·sin[(α-β)/2] 【注意右式前的负号】
cos(3π/11)+cos(5π/11)=2cos(4π/11)·cos(π/11)
cos(2π/11)+cos(4π/11) =2cos(3π/11)·cos(π/11)
原式=
cos(π/11)-(cos(2π/11)+cos(4π/11))+(cos(3π/11)+cos(5π/11) )
=·····。
。
原式=cos(π/11)+cos(3π/11)+cos(5π/11)+cos(7π/11)+cos(9π/11) =1/[2sin(π/11)]*{sin(2π/11)+sin(4π/11)-sin(2π/11) +……+sin(10π/11)-sin(8π/11)} =sin(10π/11)/[2sin(π/11)] =1/2.