a,b为锐角,3sina+3sinb=1 ,3sin(2a)-2sin(2b)=0 求sin(a+2b)
问题描述：
a,b为锐角,3sina+3sinb=1 ,3sin(2a)-2sin(2b)=0 求sin(a+2b)
答
明如下由COS(A+B)=0,有：
COS(A+B)=cosAcosB-sinAsinB=0
所以cosAcosB=sinAsinB
SIN(A+2B)=sinAcos2B+sin2BcosA
=sinA[(cosB)^2-(sinB)^2]+2sinBcosBcosA
=sinA[(cosB)^2-(sinB)^2]+2sinA(sinB)^2
=sinA[(cosB)^2-(sinB)^2+2(sinB)^2]
=sinA[(cosB)^2+(sinB)^2]
=sinA