化简sina^2+sinb^2+2sinasinbcos(a+b)!

化简sina^2+sinb^2+2sinasinbcos(a+b)!
化简sina^2+sinb^2+2sinasinbcos(a+b)!

化简sina^2+sinb^2+2sinasinbcos(a+b)!
不需要“积化和差”公式的证法：
sina^2+sinb^2+2sinasinbcos(a+b)
=sina^2+sinb^2+2sinasinb(cosacosb-sinasinb)
=sina^2+sinb^2+2sinasinbcosacosb-2(sina)^2(sinb)^2
=[sina^2-(sina)^2(sinb)^2]+[sinb^2-(sina)^2(sinb)]^2+2sinasinbcosacosb
=(sina)^2[1-(sinb)^2]+(sinb)^2[1-(sina)^2]+2sinasinbcosacosb
=(sina)^2(cosb)^2+(sinb)^2(cosa)^2+2sinasinbcosacosb
=(sinacosb+cosasinb)^2
=[sin(a+b)]^2.

原式=1*sina^2+1*sinb^2+2sina*sinb*cos(a+b)
=(sina^2+cos^2)*sina^2+(sinb^2+cosb^2)*sinb^2+2sina*sinb*(cosa*cosb-sina*sinb)
=sin^4+sinb^4-2sina^2*sinb^2 +(cosa*sina)^2+(cosb*sinb)^2+2sina*sinb*cosa*cosb
=[sin(a+b)cos(a-b)]^2+[(sina)^2-(sinb)^2]^2
不知道您满意否

[sin(a+b)*cos(a-b)]^2+[(sina)^2-(sinb)^2]^2

sina^2+sinb^2+2sinasinbcos(a+b)
=1/2(1-cos2a)+1/2(1-cos2b)+(cos(a-b)-cos(a+b))*cos(a+b)
=1-1/2(cos2a+cos2b)+cos(a-b)*cos(a+b)-cos(a+b)^2
=1-1/2*2cos(a-b)*cos(a+b)+cos(a-b)*cos(a+b)-cos(a+b)^2
=1-cos(a+b)^2
=sin(a+b)^2

来赚两分，做法推荐raigu的……简洁明了

2个月前帮你做没问题,现在全忘记了

根据几个性质计算：
sin(2x)=2sinx*cosx
sin(x^2)=1/2[1-cos(2x)]
cos(x^2)=1/2[1+cos(2x)]

以前帮你做是小菜一碟，现在嘛，好长时间不接触了，公式全忘了