求解:高一二倍角的正弦余弦正切公式填空题
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/27 06:13:54
求解:高一二倍角的正弦余弦正切公式填空题
如图求值
如图求值
![求解:高一二倍角的正弦余弦正切公式填空题](/uploads/image/z/18007452-36-2.jpg?t=%E6%B1%82%E8%A7%A3%EF%BC%9A%E9%AB%98%E4%B8%80%E4%BA%8C%E5%80%8D%E8%A7%92%E7%9A%84%E6%AD%A3%E5%BC%A6%E4%BD%99%E5%BC%A6%E6%AD%A3%E5%88%87%E5%85%AC%E5%BC%8F%E5%A1%AB%E7%A9%BA%E9%A2%98)
(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(sin10°)^2/(√2)(sin10°)^2
=2/√2
=√2
=(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(sin10°)^2/(√2)(sin10°)^2
=2/√2
=√2