化简[2sin50°+sin10°(1+3tan10°)]1+cos20°
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/20 07:56:24
化简[2sin50°+sin10°(1+
tan10°)]
3 |
1+cos20° |
原式=[2sin50°+sin10°(1+
3sin10°
cos10°)]
1+cos20°
=(2sin50°+sin10°•
cos10°+
3sin10°
cos10°)
1+cos20°
=(2sin50°+2sin10°•
sin(10°+30°)
cos10°)
1+cos20°
=
2sin50°cos10°+2sin10°sin40°
cos10°•
1+cos20°
=2cos(40°-10°)•
2cos10°
cos10°
=2cos30°×
2
=
6.
3sin10°
cos10°)]
1+cos20°
=(2sin50°+sin10°•
cos10°+
3sin10°
cos10°)
1+cos20°
=(2sin50°+2sin10°•
sin(10°+30°)
cos10°)
1+cos20°
=
2sin50°cos10°+2sin10°sin40°
cos10°•
1+cos20°
=2cos(40°-10°)•
2cos10°
cos10°
=2cos30°×
2
=
6.
化简[2sin50°+sin10°(1+3tan10°)]1+cos20°
化简{2sin50°+sin10°【1+(根号3tan10°)】根号(1加cos20)°
化简:sin50º(1+√3tan10°)-cos20°/cos80°√(1-cos20°)
[2sin50°+sin10°(1+根号3tan10°)]乘sin10°等于?
化简[2sin50°+sin10°(1+√3tan10°)] √(2sin280°)
化简:[2sin50°+sin10°(1+根号3*tan10°)]*根号sin^280°
化简:【sin50°·(1+√3tan10° )-cos20°】/(cos80°· √1-cos20°)
化简sin50°(1+√3tan10°)-cos20°/cos80°√(1-cos20°)
求值:[2sin50°+sin10°(1+√3tan10°)]×√2sin80°
求[2sin50°+sin10°(1+√3tan10°0]sin80°的值
[2sin50°+sin10°(1+√3tan10°)]•√(sin²80°)
【跪求】【求值】(sin50(1+根号(3)*tan10)-cos20)\(根号(2)*cos80*sin10)