cosa+cosb=0,sina+sinb=1 则cosa=cos(π-b)的证明过程
证明:cos(a+b)=cosa*cosb-sina*sinb
已知cos(a-b)=3/1,求(sina+sinb)(sina+sinb)+(cosa+cosb)(cosa+cosb
已知sina+sinb=1,cosa+cosb=0,则cos(a-b)的值为?
已知sina+sinb=1,cosa+cosb=0求cos(a+b)的值
求证:cos(a+b)=cosa*cosb-sina*sinb
证明下列恒等式:sin(a+b)*cos(a-b)=sina*cosa+sinb*cosb
sinA+sinB+sinC=0; cosA+cosB+cosC=0,求cos(B-C)的值?
sina+sinb+sinc=0,cosa+cosb+cosc=0,求cos(B-C)的值?
sinA+sinB+sinC=0,cosA+cosB+cosC=o,则cos(A-B)=______
已知sina+sinB=1/2,cosa+cosB=1/3,则cos(a-B)=
1,已知:sina+sinB=m,cosa+cosB=n,则cos(a-B)=?
若sina+sinb+siny=cosa+cosb+cosy=0,则cos(a-b)等于