cos(a-b)cosa+sin(a-b)sina=-4/5且b在第三象限,则cos(b/2)=?
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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