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
∵a∈(π/2,π)，b∈(3π/2,2π)
∴a-b∈(-3π/2,-π/2)，a+b∈(2π,3π)
∵cos(a-b)＜0，cos(a+b)＞0
∴sin(a-b)=-3/5或3/5，sin(a+b)=3/5
∴cos2a=cos[(a-b)+(a+b)]
=cos(a-b)cos(a+b)-sin(a-b)sin(a+b)
=(-4/5)(4/5)-(-3/5)(3/5)或=(-4/5)(4/5)-(3/5)(3/5)
=-7/25或=-1
答
计算机
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...