嗯再问你一提，根号1-sin40度cos40度/cos40度-根号1-sin2(为平方）50度=

显示全部楼层 · 2012-11-24 02:58:47

sqrt(1-sin(40)cos(40))/cos(40)-sqrt(1-sin^2(50))=sqrt(1/cos^2(40)-sin(40)/cos(40))-sqrt(cos^2(50))=sqrt((sin^2(40)+cos^2(40))/cos^2(40)-tg(40))-cos(50)=sqrt(tg^2(40)+1-tg(40))-cos(50)=sqrt(tg(40)-1/2)^2+3/4)-cos(50) sqrt中不是完全平方数，到此为止如果是 sqrt(1-2sin(40)cos(40)) ？...

千问 · 2012-11-24 02:58:47

√（1-2sin40°cos40°）/[cos40°-√(1-sin^50°）]=(cos40°-sin40°）/[cos40°-cos50°]=1(∵sin40°=cos50°）....