作业帮[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=不要这种复制答案cos40°+sin50°×（1+√3tan10°） =cos40°+sin50°×（tan60°-tan10°）/tan50° =cos40°+（tan60°-tan10°）cos50° =cos40°+√3cos50°-tan10°cos50° =cos40°+
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/27 09:19:28
![[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=不要这种复制答案cos40°+sin50°×（1+√3tan10°） =cos40°+sin50°×（tan60°-tan10°）/tan50° =cos40°+（tan60°-tan10°）cos50° =cos40°+√3cos50°-tan10°cos50° =cos40°+](/uploads/image/z/4333660-52-0.jpg?t=%5Bcos40%C2%B0%2Bsin50%C2%B0%281%2B%E2%88%9A3tan10%C2%B0%29%5D%2F%5Bsin70%C2%B0%E2%88%9A%281%2Bcos40%C2%B0%29%5D%3D%E4%B8%8D%E8%A6%81%E8%BF%99%E7%A7%8D%E5%A4%8D%E5%88%B6%E7%AD%94%E6%A1%88cos40%C2%B0%2Bsin50%C2%B0%C3%97%EF%BC%881%2B%E2%88%9A3tan10%C2%B0%EF%BC%89+%3Dcos40%C2%B0%2Bsin50%C2%B0%C3%97%EF%BC%88tan60%C2%B0-tan10%C2%B0%EF%BC%89%2Ftan50%C2%B0+%3Dcos40%C2%B0%2B%EF%BC%88tan60%C2%B0-tan10%C2%B0%EF%BC%89cos50%C2%B0+%3Dcos40%C2%B0%2B%E2%88%9A3cos50%C2%B0-tan10%C2%B0cos50%C2%B0+%3Dcos40%C2%B0%2B)
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=不要这种复制答案cos40°+sin50°×（1+√3tan10°） =cos40°+sin50°×（tan60°-tan10°）/tan50° =cos40°+（tan60°-tan10°）cos50° =cos40°+√3cos50°-tan10°cos50° =cos40°+
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=
不要这种复制答案
cos40°+sin50°×（1+√3tan10°）
=cos40°+sin50°×（tan60°-tan10°）/tan50°
=cos40°+（tan60°-tan10°）cos50°
=cos40°+√3cos50°-tan10°cos50°
=cos40°+√3sin40°-tan10°sin40°
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40°
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°
=2cos20°-2cos20°+2(cos20°) ∧2
=1+cos40°
sin70°√1+cos40°=sin70°(√2)cos20°=(√2)cos20°∧2
=√2/2（1+cos40°)
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=√2
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=不要这种复制答案cos40°+sin50°×（1+√3tan10°） =cos40°+sin50°×（tan60°-tan10°）/tan50° =cos40°+（tan60°-tan10°）cos50° =cos40°+√3cos50°-tan10°cos50° =cos40°+
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]
=[cos40°+sin50°×（tan60°-tan10°）/tan50°]/sin70°(√2)cos20°
=[cos40°+（tan60°-tan10°）cos50°]/sin70°(√2)cos20°
=[cos40°+√3cos50°-tan10°cos50°]/ (√2)cos20°^2
=[cos40°+√3sin40°-tan10°sin40°] /√2/2（1+cos40°)
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40° /√2/2（1+cos40°)
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°/√2/2（1+cos40°)
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°/√2/2（1+cos40°)
=2cos20°-2cos20°+2(cos20°) ^2/√2/2（1+cos40°)
=1+cos40°/√2/2（1+cos40°)
=√2