关于一道三角恒等变换的化简题
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:2024/08/10 16:50:40
关于一道三角恒等变换的化简题
化简:sin40(tan10-√3)
化简:sin40(tan10-√3)
![关于一道三角恒等变换的化简题](/uploads/image/z/15686371-19-1.jpg?t=%E5%85%B3%E4%BA%8E%E4%B8%80%E9%81%93%E4%B8%89%E8%A7%92%E6%81%92%E7%AD%89%E5%8F%98%E6%8D%A2%E7%9A%84%E5%8C%96%E7%AE%80%E9%A2%98)
sin40(tan10-√3)
=sin40(sin10/cos10-√3cos10/cos10)
=sin40[(sin10-√3cos10)/cos10]
sin10-√3cos10
=2(1/2sin10-√3/2cos10)
=2(sin10cos60-cos10sin60)
=2sin(10-60)
=-2sin50
sin40[(sin10-√3cos10)/cos10]
=sin40*(-2sin50)/cos10
=-2sin40cos40/cos10
=-sin80/cos10
=-cos10/cos10
=-1
=sin40(sin10/cos10-√3cos10/cos10)
=sin40[(sin10-√3cos10)/cos10]
sin10-√3cos10
=2(1/2sin10-√3/2cos10)
=2(sin10cos60-cos10sin60)
=2sin(10-60)
=-2sin50
sin40[(sin10-√3cos10)/cos10]
=sin40*(-2sin50)/cos10
=-2sin40cos40/cos10
=-sin80/cos10
=-cos10/cos10
=-1