cos2π/7cos4π/7cos8π/7要过程

lyalypyongyuan2022-10-04 11:39:543条回答

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英俊美少年共回答了13个问题 | 采纳率92.3%: 8sin2π/7cos2π/7cos4π/7cos8π/7/(8sin2π/7)
=4sin4π/7cos4π/7cos8π/7/(8sin2π/7)
=2sin8π/7cos8π/7/(8sin2π/7)
=sin16π/7/(8sin2π/7)
=sin2π/7/(8sin2π/7)
=1/8; 1年前

renalu 共回答了438个问题 | 采纳率: cos(2π/7)cos(4π/7)cos(8π/7)
=2sin(2π/7)cos(2π/7)cos(4π/7)cos(8π/7)/【2sin(2π/7)】
=sin(4π/7)cos(4π/7)cos(8π/7)/【2sin(2π/7)】
=2sin(4π/7)cos(4π/7)cos(8π/7)/【4sin(2π/7)】
=sin(8π/7)cos(8π/7)...; 1年前

木木微微共回答了1538个问题 | 采纳率: 应该是cos(2π/7)cos(4π/7)cos(8π/7)吧？
如果是的话：
设A=cos(2π/7)cos(4π/7)cos(8π/7)
2Asin(2π/7)=2sin(2π/7)cos(2π/7)cos(4π/7)cos(8π/7)
2Asin(2π/7)=sin(4π/7)cos(4π/7)cos(8π/7)
4Asin(2π/7)=2sin(4π...; 1年前

求值:cosπ/7cos2π/7cos3π/7cos4π/7cos5π/7cos6π/7 smooochsun1年前1: slw533 共回答了20个问题 | 采纳率90%

由诱导公式可得：cos6π/7=-cosπ/7；
cosπ/7=-cos6π/7,cos3π/7=-cos4π/7,cos5π/7=-cos2π/7
∴ cos2π/7cos4π/7cos6π/7
=-cosπ/7cos2π/7cos4π/7
=-sinπ/7cosπ/7cos2π/7cos4π/7/sinπ/7
=-1/2·sin2π/7cos2π/7cos4π/7/sinπ/7
=-1/4·sin4π/7cos4π/7/sinπ/7
=-1/8·sin8π/7/sinπ/7
=-1/8·(-sinπ/7)/sinπ/7
=1/8
而cosπ/7cos3π/7cos5π/7=(-cos6π/7)(-cos4π/7)(-cos2π/7)
=-cos2π/7cos4π/7cos6π/7=-1/8
∴原式=-1/64

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cos2π/7cos4π/7cos6π/7=? chjxjs20051年前3: minheyan 共回答了26个问题 | 采纳率96.2%

cos6π/7=-cosπ/7故cos2π/7cos4π/7cos6π/7=-cosπ/7cos2π/7cos4π/7然后乘上sinπ/7,有-sinπ/7*cosπ/7*cos2π/7*cos4π/7=-1/2sin2π/7*cos2π/7*cos4π/7=-1/4sin4π/7*cos4π/7=-1/8sin8π/7=1/8sinπ/7然...

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