sin(a-b)cosa-cos(a-b)sina=3/5,b在第三象限,则cosb/2为多少

化简sin(a-b)*cosa - cos(a-b)*sina cosb=cos[(a+b)-a]=cos(a+b)cosa+sin(a+b)sina sin(a+b)cosa-cos(a+b)sina=5/13 则sin（π-B)=? 证cosA+cos(120+B)+cos(120-B)/(sinB+sin(120+A)-sin(120-A)=tan(A+B)/2 a,b为锐角 且cos(a+b)=sin(a-b),则sina-cosa= cosa×cos(a+b)+sina乘以sin(a+b) 怎么化简..这个是原题目 2cos(a+b)cosa-cos(2a+b) 化简! 化简 cos(a+b)cosa+sin(a+b)sina=_______ cos(31°-a)cos(29°+a)-cos(59°+a)cos(61°-a) 如何化简 sin(a+b)cosa-cos(a+b)sina 化简sin(a-b)sina+cos(a-b)cosa答案为COSB, 已知sin(A-B)cosA-cos(A-B)sinA=3/5,cos2B=? sin(a-b)cosa-cos(a-b)sina=3/5求cos2b 已知sin(a-b)cosa-cos(a-b)sina=0.6 那么cos2b为多少 sin(a+b)cos(a-b)=sina*cosa+sinb*cosb 求证:sin(a+b)cos(a-b)=sina*cosa+sinb*cosb麻烦个位 帮个忙~ sinA(cos(2A+B)+cosB)=cosA(sin(2A+B)-sinB)证明 证明下列恒等式：sin(a+b)*cos(a-b）=sina*cosa+sinb*cosb cos(A+B)+sin(A-B)=(cosA+sinA)(cosB-sinB) sin(A-B)=0,求证cos(A-2B)=cosA