求值！！拜托了！！根号下(1+cos20°)乘以[2cos40+sin10(1+根号3tan10)]

解:
根号(1+cos20)*[2cos40+sin10(1+根号3tan10)]

=根号{1+[2cos^2(10)-1]}*[2cos40+sin10*(cos10/cos10+根号3sin10/cos10)]

=根号[2cos^2(10)]*{2cos40+sin10*
[2sin(10+30)/cos10]}

=(根号2cos10)*{2cos40+[sin10*2sin40/cos10]}

=(根号2cos10)*{[2cos10cos40/cos10]+[2sin10sin40/cos10]}

=(根号2cos10)*{2*[cos10cos40+sin10sin40]/cos10}

=(根号2cos10)*{2cos(40-10)/cos10}

=(根号2cos10)*{根号3/cos10}

=根号6

sin80°(1＋(√3)tan10°)=cos10(1＋(√3)tan10°) =2(cos10/2+√3/2sin10)=2sin40 [2sin50°＋sin80°(1＋(√3)tan10°)]x√（1+cos20°=2√2(sin50+sin40) cos10=2√2(sin50+cos50) cos10=4sin95cos10=4cos5 cos10