题目
求值:【sin50°[1+根号(3)tan10°]—cos20°】/[cos80°根号(1—cos20°)】
提问时间:2020-10-11
答案
(sin50°(1+√3*tan10°)-cos20°)/(cos80°*√(1-cos20°))
=(sin50°(1+√3*sin10°/cos10°)-cos20°)/(cos80°*√(1-cos20°))
=(sin50°((cos10°+√3*sin10°)/cos10°)-cos20°)/(cos80°*√(2(sin10°)^2))
=(sin50°((1/2)cos10°+(√3/2)*sin10°)/((1/2)*cos10°)-cos20°)/(cos80°*√(2(sin10°)^2))
=(sin50°(sin30°*cos10°+cos30°*sin10°)/(1/2)*cos10°-cos20°)/((√2)cos80°*sin10°)
=(sin50°*sin40°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=(sin50°*cos50°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)sin100°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)sin80°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)cos10/(1/2)*cos10°-cos20°)/((√2)sin10°sin10°)
=(1-cos20°)/((√2)sin10°sin10°)
=2(sin20°)^2/(√2)(sin20°)^2
=2/√2
=√2
2回答者
=(sin50°(1+√3*sin10°/cos10°)-cos20°)/(cos80°*√(1-cos20°))
=(sin50°((cos10°+√3*sin10°)/cos10°)-cos20°)/(cos80°*√(2(sin10°)^2))
=(sin50°((1/2)cos10°+(√3/2)*sin10°)/((1/2)*cos10°)-cos20°)/(cos80°*√(2(sin10°)^2))
=(sin50°(sin30°*cos10°+cos30°*sin10°)/(1/2)*cos10°-cos20°)/((√2)cos80°*sin10°)
=(sin50°*sin40°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=(sin50°*cos50°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)sin100°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)sin80°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)cos10/(1/2)*cos10°-cos20°)/((√2)sin10°sin10°)
=(1-cos20°)/((√2)sin10°sin10°)
=2(sin20°)^2/(√2)(sin20°)^2
=2/√2
=√2
2回答者
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