Question

sinB=1/3,cos(A+B)=1,则sin(2A+B)?

Answer 1

因为cos(A+B)=1 所以sin(A+B)=0
sin(2A+B)=sin(A+B+A)=sin(A+B)cosA+sinBcos(A+B)=0+1/3=1/3

Answer 2

cos 2(A+B)=2[cos(A+B)]^2 - 1=2-1=1
sin(A+B)=±√1-[cos(A+B)]^2 =±√1-1=0
sin 2(A+B)=2sin(A+B)cos(A+B)=0
sin(2A+B)=sin[(2A+2B)-B]=sin(2A+2B)cosB - cos(2A+2B)sinB =-1/3

Answer 3

sinB=1/3。cosB=（2根号2）/3。又cos(A+B)=cosAcosB-sinAsinB=2根号2/3cosA-1/3sinA=1. (1)又cos(A+B)=1.所以sin(A+B)=0,而sin(2A+B)=sin[(A+B)+A]=sin(A+B）cosA+cos（A+B）sinA=sinA.
又因为sin(A+B)=sinAcosB+cosAsinB=2根号2/3sinA+1/3cosA=0 (2) ， 由（1）、（2）解得sinA=-1/3， cosA =2根号2/3。所以 sin(2A+B)=sinA=-1/3。