sin(a+b)cosb-cos(a+b)sinb=0,则sin(a+2b)+sin(a-2b)等于?我算出为0!
cos(a-b)cosb-sin(a-b)sinb
sin(a-b)cosb+cos(b-a)sinb=
证明下列恒等式:sin(a+b)*cos(a-b)=sina*cosa+sinb*cosb
1.化简:sin(a+b)sinb+cos(a+b)cosb=?2.若a是第二象限的角,则seca根号(1-sin^2
化简:cos(A-B)cosB-sin(A-B)sinB
sin(a-b)cosb+cos(a-b)sinb=?
为什么sinb=sin(a+b-b)=sin(a+b)cosb+cos(a+b)sinb,这是公式吗?
1.SinA+SinB=a,CosA+CosB=1+a,求Sin(A+B),Cos(A+B).
求证: sina+sinb=2sin[(a+b)/2]cos[(a-b)/2] cosa+cosb=2cos[(a+b)
cos(A+B)cosB+sin(A+B)sinB=1/3,且A∈(3π/2,2π),求cos(2A+(π/4))
若cosA-2sin(A-B)=0,求证:tan(A-B)=cosB/(sinB+2)
已知sin(a+b)sin(a-b)=2m(m\=0),则(cosa)^2-(cosb)^2等于 ( ) A.-2m B