Question

已知向量m=（cosa,sina)和n=(根号2-sina,cosa),a∈（π，2π),且|m+n|=（8根号2）/5,求cos（a/2+π/8)的值

已知向量m=（cosa,sina)和n=(根号2-sina,cosa),a∈（π，2π),且|m+n|=（8根号2）/5,求cos（a/2+π/8)的值

Answer 1

m+n=(cosa-sina+√2,sina+cosa)
∴|m+n|²
=(cosa-sina+√2)²+(sina+cosa)²
=(cosa-sina)²+2√2(cosa-sina)+2+(cosa+sina)²
=2(sin²a+cos²a)+2-2√2(sina-cosa)
=2+2-2√2(sina-cosa)
=4-2√2(sina-cosa)
=2+2-4[(√2/2)sina-(√2/2)cosa]
=2+2-4[sinπ/4sina-cosπ/4cosa]
=4+4[cosπ/4cosa-sinπ/4sina]
=4+4cos(a+π/4)

4+4cos(a+π/4)=(8√2/5)²
4+4cos(a+π/4)=128/25
4cos(a+π/4)=28/25
cos(a+π/4)=7/25

cos(a+π/4)=2cos²(a/2+π/8)-1
2cos²(a/2+π/8)-1=7/25
cos²(a/2+π/8)=16/25
cos(a/2+π/8)=±4/5

a∈（π，2π)
a/2∈（π/2，π)
a/2+π/8∈（5π/8，9π/8)

∴cos(a/2+π/8)=-4/5

Answer 2

m+n=(cosa-sina+√2,sina+cosa)
∴|m+n|²
=(cosa-sina+√2)²+(sina+cosa)²
=(cosa-sina)²+2√2(cosa-sina)+2+(cosa+sina)²
=2(sin²a+cos²a)+2-2√2(sina-cosa)
=2+2-2√2(sina-cosa)
=4-2√2(sina-cosa)
=2+2-4[(√2/2)sina-(√2/2)cosa]
=2+2-4[sinπ/4sina-cosπ/4cosa]
=4+4[cosπ/4cosa-sinπ/4sina]
=4+4cos(a+π/4)

4+4cos(a+π/4)=(8√2/5)²
4+4cos(a+π/4)=128/25
4cos(a+π/4)=28/25
cos(a+π/4)=7/25

cos(a+π/4)=2cos²(a/2+π/8)-1
2cos²(a/2+π/8)-1=7/25
cos²(a/2+π/8)=16/25
cos(a/2+π/8)=±4/5

a∈（π，2π)
a/2∈（π/2，π)
a/2+π/8∈（5π/8，9π/8)

∴cos(a/2+π/8)=-4/5