sina=1/3,sin(a+b)=1,求sin(3a+2b)的值谢谢了!

问题描述:

sina=1/3,sin(a+b)=1,求sin(3a+2b)的值
谢谢了!

sin(a+b)=1,所以cos(a+b)=0sin2(a+b)=2sin(a+b)(a+b)=0cos2(a+b)=1-2sin²(a+b)=-1sin(3a+2b)=sin[2(a+b)+a]=sin2(a+b)cosa+cos2(a+b)sina=0×cosa+(-1)×(1/3)=-1/3