cosπ/9cos2π/9cos4π/9怎么算的?

qingheyikan32022-10-04 11:39:542条回答

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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

谢远贵_nn 共回答了10个问题 | 采纳率: 上下乘16sinπ/9sin2π/9sin3π/9sin4π/9 原式=(2sinπ/9cosπ/9)(2sin2π/9cos2π/9)(2sin3π/9cos3π/9)(2sin4π/9cos4π/9)/(1616; 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

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cosπ/9*cos2π/9*cos4π/9=? zxhzjnb1年前4: 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

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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.

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