已知sinB=1/3,sin(A+B)=1,求sin(2A+B)的值.

网友回答

sin(A+B)=1,cos(A+B)=0
sinB=1/3,
sin(2A+B)=sin[(A+B)+A]
=sin(A+B)cosA+cos(A+B)sinA
=cosA+0
=cos[(A+B)-A]
=cos(A+B)cosA+sin(A+B)sinA
=0+sinA
=1/3======以下答案可供参考======
供参考答案1:
sin(2A+B）=sin（90+B）=cosb=2√2/3或-2√2/3
供参考答案2:
用和差角公式和倍角公式
sin(2A+B)=sin(2(A+B)-B)
=sin(2(A+B))cosB-cos(2(A+B))sinB
=2sin(A+B)cos(A+B)cosB-(1-2sin^2(A+B))sinB
=0-(1-2*1^2)/3
=1/3供参考答案3:
因 sin(A+B)=1
故cos(A+B)=0
故sin2(A+B)=2*sin(A+B)*cos(A+B)=2*1*0=0
cos2(A+B)=2（cos(A+B)）^2-10-1=-1
sin(2A+B)=sin[2（A+B)-B]
=sin2(A+B)cosB-cos2(A+B)sinB
=0*cosB-(-1)*(1/3)
=1/3