sin∧2a+sin∧2b+2sina*sinb*cos(a+b).化简,
问题描述：
sin∧2a+sin∧2b+2sina*sinb*cos(a+b).化简,
答
sin^2a+sin^2b+2sina*sinb*cos(a+b)=sin^2a+sin^2b+2sina*(cosacosb-sinasinb)=(1-cos2a)/2+(1-cos2b)/2+sinasinb/2-(1-cos2a)(1-cos2b)/2=(1+sinasinb-cosacosb)/2 =[1+sin(a-b)]/2好像漏了一点。。。什么？？？？答案是：sin∧2(a+b)不知具体过程怎样。。稍等一会，，我看一下嗯嗯。。麻烦了sin^2a+sin^2b+2sina*sinb*cos(a+b)
=sin^2a+sin^2b+2sina*sinb*(cosacosb-sinasinb)
=(1-cos2a)/2+(1-cos2b)/2+sin2asin2b/2-(1-cos2a)(1-cos2b)/2
=(1+sin2asin2b-cos2acos2b)/2
=【1- cos2（a+b）】/2
=sin∧2(a+b)
最后一步公式为
Cos2A=1—2sin^2 A的转化
刚才丢了点东西，算错了
抱歉，刚才有点事没来得及给你答案懂了。。没事！谢谢啦！！！