a) sin(a+2b)=sina.cos2b+cosa.sin2b=
sinA.(1−2sin2B)+2cosA.cosB.sinB=2sinB(cosAcosB−sinA.sinB)+sinA=sinB.cos(A+B)+sinA=sinA
b)
sin(2a+b)=sin2acosb+cos2asinb=2sinAcos(A+B)+sinB=3sinB<=>sinA.cos(A+B)=sinB=>tan(A+B)sin(A+B)=sinAsinB<=>tan(A+B)=sinBsin(A+B).sinA
vậy ta cần c/m :
sinBsin(A+B).sinA=cosA2sinA<=>sinAcos(A+B)=2sinB<=>sin(−B)+sin(2A+B)=4sinB<=>sin(−B)=sinB(đúng )
=> đpcm