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求三角函数值不查表，求cos(π/11)-cos(2π/11)+

2012-06-18 10:24:262***
不查表，求 cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11) 的值求三角函数值不查表，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?

最佳回答

  • 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
    2012-06-18 12:46:30
  • 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) ） =·····
    2012-06-18 12:38:26
  • 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.
    2012-06-18 11:50:10
    • 很赞哦！ (7)

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