求值 cosπ/9cos2π/9cos4π/9
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解决时间 2021-02-04 22:02
- 提问者网友：送舟行
- 2021-02-04 11:07
求值 cosπ/9cos2π/9cos4π/9
最佳答案
- 五星知识达人网友：十鸦
- 2021-02-04 11:35
1.原式=sin(π/9)*[cos(π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/2*[sin(2π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/8cos(8π/9)/sin(π/9)
=1/8
2.原式=2*根号下(1+2*sin2cos2)-根号下[2-2*(1-2(sin2)^2)]
=2*根号下(sin2+cos2)^2-根号下[4(sin2)^2)]
=2*(sin2+cos2)-2sin2
=2cos2
=1/2*[sin(2π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/8cos(8π/9)/sin(π/9)
=1/8
2.原式=2*根号下(1+2*sin2cos2)-根号下[2-2*(1-2(sin2)^2)]
=2*根号下(sin2+cos2)^2-根号下[4(sin2)^2)]
=2*(sin2+cos2)-2sin2
=2cos2
全部回答
- 1楼网友：duile
- 2021-02-04 13:26
第一题，原式等于（2sinπ/9cosπ/9cos2π/9cos4π/9）/2sinπ/9=（2sin2π/9cos2π/9cos4π/9)/2sinπ/9=.....=(sin8π/9)/8sinπ/9=0.125
- 2楼网友：痴妹与他
- 2021-02-04 13:05
（1）原式* sinπ/9 然后用二倍角公式得cos8π/9/8
sinπ/9 =cos8π/9 所以原式=1/8
sinπ/9 =cos8π/9 所以原式=1/8
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