求证sin与tan 已知3sinB=sin(2A+B),求证tan(A+B)=2tanA

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求证sin与tan 已知3sinB=sin(2A+B),求证tan(A+B)=2tanA

求证sin与tan 已知3sinB=sin(2A+B),求证tan(A+B)=2tanA
求证sin与tan
已知3sinB=sin(2A+B),求证tan(A+B)=2tanA

求证sin与tan 已知3sinB=sin(2A+B),求证tan(A+B)=2tanA
3sinB=sin(2A+B)
3sin(A+B-A)=sin(A+b+A)
3sin(A+B)cosA-3cos(A+b)sinA=sin(A+B)cosA+sinAcos(a+b)
2sin(A+b)cosA=4cos(a+b)sina
tan(a+b)=2tana

解析:3sinB=sin(2A+B)
3sin[(A+B)-A]=sin[(A+B)+A]
3sin(A+B)cosA-3cos(A+B)cosA
=sin(A+B)cosA+cos(A+B)cosA
即2sin(A+B)cosA=4cos(A+B)sinA
∴sin(A+B)/cos(A+B)=4sinA/2cosA
tan(A+B)=2tanA