a) sin(a+2b)=sina.cos2b+cosa.sin2b=[tex]sinA.(1-2sin^2B)+2cosA.cosB.sinB=2sinB(cosAcosB-sinA.sinB)+sinA=sinB.cos(A+B)+sinA=sinA[/tex]
b) [tex]sin(2a+b)=sin2acosb+cos2asinb=2sinAcos(A+B)+sinB=3sinB<=>sinA.cos(A+B)=sinB =>\frac{sin(A+B)}{tan(A+B)}=\frac{sinB}{sinA}<=>tan(A+B)=\frac{sin(A+B).sinA}{sinB}[/tex]
vậy ta cần c/m : [tex]\frac{sin(A+B).sinA}{sinB}=\frac{2sinA}{cosA}<=>sinAcos(A+B)=2sinB<=>sin(-B)+sin(2A+B)=4sinB<=>sin(-B)=sinB[/tex](đúng )
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