cosπ/5cos2π/5在每一步的后面写上运用了哪条公式

孤遥浪子2022-10-04 11:39:543条回答

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jinzongzhu 共回答了21个问题 | 采纳率85.7%: sin2α＝2sinαcosα
利用二倍角公式
原式=[(2sin∏/5cos∏/5)cos2∏/5]/(2sin∏/5)
=[sin2∏/5cos2∏/5]/(2sin∏/5)
=[2sin2∏/5cos2∏/5]/(4sin∏/5)
=[sin4∏/5]/[4sin∏/5]
此时利用诱导公式sin（π－α）＝sinα
原式=[sin(∏-4∏/5)]/(4sin∏/5)
=[sin∏/5]/(4sin∏/5)
=1/4; 1年前

38453920160a25b1 共回答了187个问题 | 采纳率: cosπ/5cos2π/5
=2sinπ/5cosπ/5cos2π/5/(2sinπ/5)
=sin2π/5cos2π/5/(2sinπ/5)
=2sin2π/5cos2π/5/(4sinπ/5)
=sin4π/5/(4sinπ/5)
=sin(π-π/5)/(4sinπ/5)
=sinπ/5/(4sinπ/5)
=1/4
没有什么特殊公式; 1年前

h_nxldf_ec_m1295 共回答了193个问题 | 采纳率: cosπ/5cos2π/5
=2sinπ/5cosπ/5cos2π/5/(2sinπ/5)
=sin2π/5cos2π/5/(2sinπ/5)
=2sin2π/5cos2π/5/(4sinπ/5)
=sin4π/5/(4sinπ/5)
=sin(π-π/5)/(4sinπ/5)
=sinπ/5/(4sinπ/5)
=1/4; 1年前

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sin2α＝2sinαcosα
利用二倍角公式
原式=[(2sin∏/5cos∏/5)cos2∏/5]/(2sin∏/5)
=[sin2∏/5cos2∏/5]/(2sin∏/5)
=[2sin2∏/5cos2∏/5]/(4sin∏/5)
=[sin4∏/5]/[4sin∏/5]
此时利用诱导公式sin（π－α）＝sinα
原式=[sin(∏-4∏/5)]/(4sin∏/5)
=[sin∏/5]/(4sin∏/5)
=1/4

1年前查看全部

求值cosπ/5cos2π/5cos4π/5cos8π/5 hjkl4561年前2: 我是二里半共回答了24个问题 | 采纳率91.7%

利用正弦二倍角公式
cosπ/5cos2π/5cos4π/5cos8π/5
=2sinπ/5cosπ/5cos2π/5cos4π/5cos8π/5/(2sinπ/5)
=sin2π/5cos2π/5cos4π/5cos8π/5/(2sinπ/5)
=2sin2π/5cos2π/5cos4π/5cos8π/5/(4sinπ/5)
=sin4π/5cos4π/5cos8π/5/(4sinπ/5)
=2sin4π/5cos4π/5cos8π/5/(8sinπ/5)
=sin8π/5cos8π/5/(8sinπ/5)
=2sin8π/5cos8π/5/(16sinπ/5)
=sin16π/5/(16sinπ/5)
=sin(3π+π/5)/(16sinπ/5)
=-sinπ/5/(16sinπ/5)
=-1/16

1年前查看全部

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