Question

解：[2sin50°+sin10°（1+√3tan10°）]*√2（sin80°)*(sin80度)

Answer 1

[2sin50°+sin10°（1+√3tan10°）]*√2（sin80°)*(sin80度)
=[2sin50°*sin80°+sin10°*cos10°（cos10°+√3sin10°）/cos10°]*√2（sin80°)
=[2sin50°*sin80°+2sin10°（1/2cos10°+√3/2sin10°）]*√2（sin80°)
=[2sin50°*sin80°+2sin10°sin(30°+10°)]*√2（sin80°)
=[2sin50°*sin80°+2sin10°sin40°]*√2（sin80°)
=[2sin50°*cos10°+2sin10°cos50°]*√2（sin80°)
=2[sin60°]*√2（sin80°)
=√6sin80°

Answer 2

√2sin²80°是根号下(2sin²80°)，还是在(√2)×(sin²80°)？若是前者，
原式={2sin50°＋sin10°[1＋√3(sin10°/cos10°)]}×[√(2sin²80°)]
={2sin50°＋sin10°[(cos10°＋√3sin10°)/cos10°]}×[√(2sin²80°)]
={2sin50°＋sin10°[2sin(10°＋30°)/cos10°]}×[√(2sin²80°)]
=[2sin(90°－40°)＋sin10°(2sin40°/cos10°)]×[√(2sin²80°)]
=[2cos40°＋(2sin10°sin40°/cos10°)]×[√(2sin²80°)]
=[2(cos40°cos10°＋sin10°sin40°)/cos10°]×[√(2sin²80°)]
=[2cos(40°－10°)/cos10°]×√[2sin²(90°－10°)]
=(2cos30°/cos10°)×√(2cos²10°)
=[(√3)/cos10°]×[(√2)×(cos10°)]
=√6
若是后者，步骤都一样，答案是(√6)×(cos10°)