△ABC中,证明:sin^2A+sin^2B+sin^2C=2+2cosAcosBcosC
sina^2+sinb^2+sinc^2-2cosacosbcosc
=3-(cosa^2+cosb^2+cosc^2+2cosacosbcosc)
=3-{cosa*[cosa+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)+2]}
=3-{-cos(b+c)*[-cos(b+c)+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)]+1}
=3-{-cos(b+c)*cos(b-c)+cos(b+c)*cos(b-c)+1}
=2
已知a(n+2)=a(n+1)×a(n).求数列a(n),题目是有解的,太烦了不想打,你自己做吧
不好意思,我没学
3+2-5*0=?