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Cos72度Cos72度等于多少

cos(72°)=0.30901699437495

做等腰三角形ABC,其中∠A=36° 易得∠B=∠C=72° 过A做AD⊥BC于D,∵∠B=∠C ∴BC = 2BD cos72°=cos∠B=BD/AB=BC/2AB 过C做∠C的角平分线交AB于E点. ∠ACE=1/2∠C=36°=∠A ∴CE=AE ∵∠BEC是△AEC的外角 ∴∠BEC=∠A+∠ACE=72°=∠B ∴CE=BC ∵∠ACE=∠A,∠BEC=∠B ∴△ABC∽△CEB ∴A...

(√(5)-1)/2

cos72—cos36 =cos(54+18)-cos(54-18) =[cos54cos18-sin54sin18]-cos54cos18+sin54sin18 =-2sin54sin18 =-2cos36sin18 =-2cos36sin18cos18/cos18 =-cos36sin36/cos18 =-sin72/(2cos18) =-sin72/(2sin72) =-1/2

根据正弦定理:sin(A+B)=sinAcosB+cosAsinB即sin18cos72+sin72cos18=sin(18+72)=1

根据正弦定理:sin(A+B)=sinAcosB+cosAsinB即sin72°cos18°+cos72°sin18° =sin(72°+18°) =sin90° =1

等于cos(72度-42度)=cos30度=2分之根号3

下图继续写!

如果我的答案对您有用,请尽快采纳我的答案,谢谢!∵|AB|=√(cos²18°+cos²72°)=√(cos²18°+sin²18°)=1, |BC|=√(4cos²63°+4cos²27°)=√(4sin²63°+4cos²27°)=2, cos∠ABC=BA·BC/|BA||BC|=2sin45°/2=√2/2, ∴∠ABC=45°, ∴面积=√2/2。

解答: cos72°-cos36° =cos(54°+18°)-cos(54°-18°) =(cos54°cos18°-sin54°sin18°)-(cos54°cos18°+sin54°sin18°) =-2sin54°sin18° =-2cos36°cos72° =-2sin36°cos36°cos72°/sin36° =-sin72°cos72°/sin36° =-(1/2)sin144°/sin36° =-(1/2)sin(180°-3...

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