Question

化简【2sin50°+sin80°(1+√3tan10°)】/cos5°

Answer 1

[2sin50°+sin80°(1+√3tan10°)]/√2cos5°
=[2sin50°+cos10°(cos10°+√3sin10°)/cos10°]/√2cos5°
=[2sin50°+2(1/2*cos10°+√3/2*sin10°)]/√2cos5°
=[2sin50°+2(cos60°*cos10°+sin60°*sin10°)]/√2cos5°
=[2sin50°+2cos50°]/√2cos5°
=2√2*(sin50°*cos45°+cos50°*sin45°)/√2cos5°
=2√2sin95°/√2cos5°
=2√2cos5°/√2cos5°
=2
绝对标准答案

Answer 2

原式=（2sin50º+cos10º+√3sin10º）/cos5º
=[2sin50º+2(1/2*cos10º+√3/2sin10º)]/cos5º
=(2sin50º+2cos50º)/cos5º
=2√2(√2/2sin50º+√2/2cos50º)/cos5º
=2√2sin95º/cos5º
=2 根号2cos5/cos5
=2根号2

Answer 3

[2sin50°+sin80°(1+√3tan10°)]/cos5°
=[2sin50°+cos10°(cos10°+√3sin10°)/cos10°]/cos5°
=[2sin50°+2(1/2*cos10°+√3/2*sin10°)]/cos5°
=[2sin50°+2(cos60°*cos10°+sin60°*sin10°)]/cos5°
=[2sin50°+2cos50°]/cos5°
=2√2*(sin50°*cos45°+cos50°*sin45°)/cos5°
=2√2sin95°/cos5°
=2√2cos5°/cos5°
=2√2