cosπ/9cos2π/9cos4π/9=?

zxhzjnb2022-10-04 11:39:544条回答

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bluesea514 共回答了17个问题 | 采纳率82.4%: 用因子8sinπ/9乘之,再除之得1/8; 1年前

super197511 共回答了85个问题 | 采纳率: cosπ/9*cos2π/9*cos4π/9=(2*sinπ/9*cosπ/9*cos2π/9*cos4π/9)/2*sinπ/9
=2*sin2π/9*cos2π/9*cos4π/9)/2*2*sinπ/9
=2*sin4π/9*cos4π/9)/2*2*2*sinπ/9
...; 1年前

ueoeuror 共回答了10个问题 | 采纳率: -1/9^3; 1年前

猩猩的窝共回答了137个问题 | 采纳率: 1/8; 1年前

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急、COSπ/9*COS2π/9*COS4π/9=?同样, 同尘1年前2: wanxuchu 共回答了24个问题 | 采纳率95.8%

π/9=20°2π/9=40°4π/9=80°
cos20cos40cos80
=sin20cos20cos40cos80/sin20
=1/2*sin40cos40cos80/sin20
=1/4*sin80cos80/sin20
=1/8*sin160/sin20
=1/8*sin20/sin20
=1/8

1年前查看全部

cosπ/9*cos2π/9*cos4π/9怎么算的? qingheyikan31年前2: arehere 共回答了12个问题 | 采纳率91.7%

分子分母同乘以8sin(π/9),然后分子连续使用二倍角公式====>>>> 2sinacosa=sin2a
===>>>>> 原式=[sin(8π/9)]/[8sin(π/9)]=1/8

1年前查看全部

化简cosπ/9*cos2π/9*cos3π/9*cos4π/9 苦冰茶1年前1: 金黄的气球共回答了21个问题 | 采纳率90.5%

k/k法
cosπ/9*cos2π/9*cos3π/9*cos4π/9
＝[(2sinπ/9 cosπ/9)*cos2π/9*cos3π/9*cos4π/9]/(2sinπ/9 )
=[(2sin2π/9 cos2π/9)*cos3π/9*cos4π/9]/(4sinπ/9 )
=[(2sin4π/9 cos4π/9)*cos3π/9]/(8sinπ/9 )
=(sin8π/9 cos3π/9)/(8sinπ/9 )
=(cosπ/3)/8
=1/16.

1年前查看全部

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