[2sin50+sin10(1+√3 tan10°)]√(1-cos2 80°)

2025-06-27 18:51:37
推荐回答(1个)
回答1:

=(2sin50°+sin10°+√3sin10°tan10°)√[1-(1-2sin^2 80°)]
={2sin50°+[1+(√3)(sin10°/cos10°)]sin10°}√(2sin^2 80°)
=[2sin50°+(cos10°+√3sin10°)sin10°/cos10°]√(2sin^2 80°)
=[2sin50°+(2cos50°sin10°/cos10°)]√(2sin^2 80°)
=2[(cos10°sin50°+cos50°sin10°)/cos10]√(2sin^2 80°)
=2(sin60°/cos10°)√(2sin^2 80°)
=2(sin60°/sin80°)√2sin 80°
=2√2sin60°
=√6