化简 2sin[a+b]cosa-sin[2a+b]
为什么cosa-cosb=-2sin[(a-b)/2]*sin[(a+b)/2],
1.若sin(A+B)sin(B-A)=M,则(cosA)^2-(cosB)^2等于?
若sin(A+B)sin(B-A)=m,则(cosA)^2-(cosB)^2=?
若cos(a+B)*cosa+sin(a+B)*sina=-4/5,则sin(π/2-B)
证明sin(a+b)sin(a-b)=sin^2 a-sin^2 b,
三角形ABC中证明 COSA+COSB+COSC=1+4SIN(A/2)*SIN(B/2)*SIN(C/2)
若cosA-2sin(A-B)=0,求证:tan(A-B)=cosB/(sinB+2)
【证明】Sin A+sin B=2Sin 22
化简:sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5
求证: sina+sinb=2sin[(a+b)/2]cos[(a-b)/2] cosa+cosb=2cos[(a+b)
已知sin(a+b)sin(a-b)=2m(m\=0),则(cosa)^2-(cosb)^2等于 ( ) A.-2m B
sin^2 (A)+sin^2(B)-sin^2(A)sin^2(B)+cos^2(A)cos^2(B)