Question

已知向量m=(cosα,sinα),向量n=(根号2-sinα,cosα)属于π到3／2π

求（1）求│向量m+向量n│最大值
（2）当│向量m+向量n│=4√10／5 ，求 sinα2α的值

Answer 1

m+n
=(cosα,sinα)+(√2-sinα,cosα)
=(cosα+√2-sinα, sinα+cosα)
|m+n|^2
=(cosα+√2-sinα)^2+(sinα+cosα)^2
=(cosα-sinα)^2+2√2(cosα-sinα)+2 + (sinα+cosα)^2
=3+2√2(cosα-sinα)
=3+4(cos(α+π/4))
max |m+n|^2 = 3
max |m+n| = √3
(2)
│m+n│=4√10／5
=> 3+4(cos(α+π/4)) = 32/5
cos(α+π/4)= 17/20
√2/2(cosα-sinα) =17/20
1/2(cosα-sinα)^2 = 289/400
1-sin2α = 289/200
sin2α = -89/200