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cos(2π/7)cos(4π/7)cos(6π/7)怎么解·要过程
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人气:441 ℃ 时间:2020-07-15 19:15:16
解答
cos(2π/7)cos(4π/7)cos(6π/7)=cos(2π/7)cos(4π/7)[-cos(π-6π/7)]=-cos(π/7)cos(2π/7)cos(4π/7)=-2sin(π/7)cos(π/7)cos(2π/7)cos(4π/7)/[2sin(π/7)]=-sin(2π/7)cos(2π/7)cos(4π/7)/[2sin(π/7)]=-2...
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