已知cos(a+b)=0,求值:sin(π+2a+b)-cos(-π/2+2a+3b)

网友回答

原式=-sin(2a+b)-sin(2a+3b)
sin(2a+b)=sin(2a+2b-b)=sin2(a+b)cosb-cos2(a+b)sinb
sin(2a+3b)=sin(2a+2b+b)=sin2(a+b)cosb+cos2(a+b)sinb
cos(a+b)=0,sin2(a+b)=2sin(a+b)cos(a+b)=0
原式=cos2(a+b)sinb-cos2(a+b)sinb=0
======以下答案可供参考======
供参考答案1:
sin(π+2a+b)=sinπcos(2a+b)+cosπsin(2a+b)=-sin(2a+b)=-sin(2a+2b-b)=cos(2a+2b)sinb-sin(2a+2b)cosb;
cos(-π/2+2a+3b)=cos(-π/2)cos(2a+3b)-sin(-π/2)sin(2a+3b)=sin(2a+3b)=sin[2(a+b)+b]=sin(2a+2b)cosb+cos(2a+2b)sinb;
已知cos(a+b)=0,所以sin(π+2a+b)-cos(-π/2+2a+3b)=-2sin(2a+2b)cosb=-4sin(a+b)cos(a+b)cosb=0.