A+B+C=∏,证明sin(A+B)/2*cos(A-B)/2=cosC/2
三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin
三角形ABC中证明 COSA+COSB+COSC=1+4SIN(A/2)*SIN(B/2)*SIN(C/2)
若sin^4a/sin^2b+cos^4a/cos^2b=1,证明sin^4b/sin^2a+cos^4b/cos^2a
证明三角函数等式sin(A+B)-sinA=2cos(A+B/2)sin(B/2)
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明
已知sinc+cosc=2sina,sinc*cosc=sin^b,求证:4cos^2 2a=cos^2 2b
cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b) 用三角形证明
证明sin(a+b)sin(a-b)=sin^2 a-sin^2 b,
如何证明sinA+sinB=2sin((A+B)/2)cos((A-B)/2)
证明cos^2(A+B)—sin^2(A—B)=cos2Acos2B
证明sin(2a+b)/sina-2cos(a+b)=sinb/sina
证明 cos的平方2(A+B)-sin的平方2(A-B)=cos2Acos2B