cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度=
问题描述：
cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度=
答
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