Question

[2sin50°+sin10°(1+√3tan10°)]•√(sin²80°)

Answer 1

[2sin50°+sin10°(1+√3tan10°)]•√(2sin²80°)
= [2sin50 + sin10(cos10+√3sin10)/cos10]•√2sin80
= [2sin50cos10+2sin10(cos60cos10+sin60sin10)•√2sin80/cos10
= 2[sin50cos10 + sin10cos(60-10)]•√2
= 2sin(50+10)•√2
= √6

追问

我问的是 [2sin50°+sin10°(1+√3tan10°)]•√(sin²80°) 而不是 [2sin50°+sin10°(1+√3tan10°)]•√(2sin²80°)

追答

[2sin50°+sin10°(1+√3tan10°)]•√(sin²80°)
=[2sin50 + sin10(cos10+√3sin10)/cos10]•√sin²80°
= [2sin50+2sin10cos50°/cos10°]•cos10°
= 2[sin50cos10 + sin10cos50°]
= 2sin60°
= √3

[2sin50°+sin10°(1+√3tan10°)]•√(sin²80°)
=[2sin50 + sin10(cos10+√3sin10)/cos10]•√sin²80°
= [2sin50+2sin10cos50°/cos10°]•cos10°
= 2[sin50cos10 + sin10cos50°]
= 2sin60°
= √3

亲，能给一个好评吗？