sin(a+b)cos(a-b)=1/5;则(cosa)^2-(sinb)^2=
证明下列恒等式:sin(a+b)*cos(a-b)=sina*cosa+sinb*cosb
1.SinA+SinB=a,CosA+CosB=1+a,求Sin(A+B),Cos(A+B).
sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则si
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC
求证: sina+sinb=2sin[(a+b)/2]cos[(a-b)/2] cosa+cosb=2cos[(a+b)
已知sina+sinb=根号2 cosa+cosb=2根号3/3。 求cos(a-b),tan(a+b)/2,sin(a
cosa sinb -sin a cos b=s什么?
已知cosa-cosb=1/2,sina-sinb=-1/3,求cos(a-b),sin(a+b)
已知sina+sinB=-2/3,cosa-cosB=1/3.求cos(a+B),sin(a+B)
化简:sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5
已知sina+sinb=1/2,cosa+cosb=1/3,求cos(a-b),tan(a+b)