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