COS2B-COS2A=COS(B+A)COS(B-A)-SIN(B+A)SIN(B-A)-COS(A+B)COS(A-

求证sin平方a * sin平方b+cos平方a * cos平方b-1/2cos2a *cos2B=1/2 cos(A+B)的平方-sin(A-B)的平方=cos2A*cos2B 请问怎么证明? 化简sin^2(a)*sin^2(b)+cos^2(a)*cos^2(b)-1/2cos2a*cos2b 由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b) 用三角形证明 求证:cos(a+b)cos(a-b)=cos平方b-sin平方a cos(a+b)+cos(a-b)/sin(a+b)+sin(a-b)化简 cos(a-b)cos(b-c)+sin(a-b)sin(b-c)= 若sin^4a/sin^2b+cos^4a/cos^2b=1,证明sin^4b/sin^2a+cos^4b/cos^2a cosb=cos[(a+b)-a]=cos(a+b)cosa+sin(a+b)sina cos^B-cos^C=sin^A,三角形的形状 Cos(a+b)*cos(a-b)=1/5 求cos ^2-sin^2