cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度=
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/11 14:01:59
![cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度=](/uploads/image/z/675152-8-2.jpg?t=cos80%E5%BA%A6%E6%A0%B9%E5%8F%B71-cos20%E5%BA%A6%E5%88%86%E4%B9%8Bsin50%E5%BA%A6%281%2B%E6%A0%B9%E5%8F%B73tan10%E5%BA%A6%29-cos20%E5%BA%A6%3D)
x)K/0xkٳ;o7@mOvvg時x%y MJ["}_`g3\66G m:!V?lӡ
@Y
Cf,DwuP_io4lT>TV*TmSg[uX.fX x
R/.H̳%
V([
cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度=
cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度=
cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度=
cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度
=[sin50°(1+√3tan10°)-cos20°]/[cos80°√(1-cos20°)]
={[2cos40°(1/2cos10°+√3/2sin10°)]/cos10°-cos20°}/{sin10°√[2(sin10°)^2]}
=[(2cos40°sin40°/cos10°)-cos20°]/[√2(sin10°)^2]
=[sin80°/cos10°-cos20°]/[√2(sin10°)^2]
=[cos10°/cos10°-cos20°]/[√2(sin10°)^2]
=[1-cos20°]/[√2(sin10°)^2]
=2(sin10°)^2/[√2(sin10°)^2]
=√2