cos(a-b)=12/13,sin(a+b)=-3/5,求cos(pai/2-2a)如题～

如何证明sin(a+b)=sin(a)cos(b)+cos(α)sin(b) sin(a+b)=4/5cos(a-b)=12/13求cos 2a=? cos(a-b)=12/13,sin(a+b)=-3/5,求cos(pai/2-2a)如题～ sin(a+b)cosa-cos(a+b)sina=5/13 则sin（π-B)=? cos^2a-cos^2b=m,那么sin(a+b)sin(a-b)=? cos²a-cos²b=c,则sin(a+b)sin(a-b)= cos∧a-cos∧b=t则sin(a+b)sin(a-b)= 证明sin(A+B)sin(A-B)=cos^2B-cos^2A 已知sin(a+b)=-3/5,cos(a-b)=12/13,且π/2 sin a sin b +cos a cos b =0,则sin a cos a+sin b cos b的值 证明 sin^2A+sin^2B-sin^2A*sin^2B+cos^2A*cos^2证明 sin^2A+sin^2B-sin^2A*sin^2B+cos^2A*cos^2B=1 sin(a+b)=-3/5,sin(b-π/4）=12/13,求cos（a+π/4)值 cos(a-b)cos(b-c)+sin(a-b)sin(b-c)= 非线性方程解析解-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos(b)-y0*sin(a)*co 求证cos(a+b)cos(a－b)=cos^2b－sin^2a 求证cos(a+b)cos(a-b)=cos^2a-sin^2b 求证:cos(a+b)cos(a-b)=cos平方b-sin平方a 求证:cos²a-sin²b=cos(a-b)cos(a+b)