cos(a-b)=-4/5,cos(a+b)=4/5 a属于(π/2,π)b属于(3π/2,2π),求cos2a
问题描述:
cos(a-b)=-4/5,cos(a+b)=4/5 a属于(π/2,π)b属于(3π/2,2π),求cos2a
答
感觉题目抄错了,是sin(a-b)=-4/5吧 如果是sin(a-b)=-4/5,cos(a+b)=4/5就好写了∵a∈(π/2,π),b∈(3π/2,2π)∴a-b∈(-3π/2,-π/2),a+b∈(2π,3π)∴cos(a-b)<0,sin(a+b)>0∴cos(a-b)=-3/5,sin(a+b)=3/5∴cos2a=c...