Question

cos(a-b)=-4/5,cos(a+b)=4/5 a属于（π/2,π）b属于（3π/2,2π）,求cos2a

Answer 1

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=cos[(a-b)+(a+b)]

=cos(a-b)cos(a+b)-sin(a-b)sin(a+b)

=(-3/5)(4/5)-(-4/5)(3/5)

=0