已知向量m(cosθ,sinθ),向量n{(√2)-sinθ,cosθ)},θ∈(π,2π)且向量 m与向量n的和的绝对值等于(8/5)√2,求cos{(θ/2)+(π/8)=?

用a代表

m +n =√2 +(cosa-sina) ,cosa+sina
模的平方 = 2+(cosa-sina)^ +(2√2)(cosa-sina) +(cosa+sina)^
=4 - 4sin(a-π/4) =[(8/5)√2]^ =128/25
===>sin(a-π/4) = - 7/25
sin(a-π/4) =cos[π/2 -(a-π/4)]
=cos[3π/4 -a)
=2cos^[3π/8 -a/2]-1
====>cos^[a/2 -3π/8 ] = 9/25
cos[a/2 -3π/8 ]绝对值 =3/5

a∈(π,2π) a/2∈(π/2,π)
a/2 - 3π/8 ∈(π/8,5π/8)
由于cos(3π/4)的绝对值小于 3/5
3π/4 >5π/8 =π/2+π/8
所以余弦绝对值等于3/5的角不在[π/2,5π/8]之间
==>a/2 - 3π/8∈(π/8,π/2)

所以cos(a/2-3π/8)=3/5
====>sin(a/2-3π/8)= 4/5

cos(a/2+π/8)=cos(a/2 -3π/8+ π/2)
=cos(a/2-3π/8)cos(π/2) -sin(a/2-3π/8)sin(π/2)
=0- 4/5
=-4/5