cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11)
=[2sin(π/11)cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11)]/2sin(π/11)
=[sin(2π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11)]/2sin(π/11)
=[sin(4π/11)cos(3π/11)cos(4π/11)cos(5π/11)]/4sin(π/11)
=[sin(8π/11)cos(3π/11)cos(5π/11)]/8sin(π/11)
=[sin(3π/11)cos(3π/11)cos(5π/11)]/8sin(π/11)
=[sin(6π/11)cos(5π/11)]/16sin(π/11)
=[sin(5π/11)cos(5π/11)]/16sin(π/11)
=sin(10π/11)]/32sin(π/11)
=1/32
關(guān)鍵在于知道如何使用sin2x=2SinxCosx.
求值:cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11)
其他人氣：574 ℃時(shí)間：2020-09-04 20:06:39
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