已知向量a=(cosθ,1⼀2),b=(1⼀2,sinθ),-π⼀2<θ<π⼀2(1)若a‖b,求θ(2)若a⊥b,求θ（3）求|a+b|的最大值

已知向量a=(cosθ,1/2),b=(1/2,sinθ),-π/2<θ<π/2
(1) 若a‖b
cosθ/(1/2)=(1/2)/sinθ
sinθcosθ=1/4
2sinθcosθ=1/2
sin2θ=1/2
2θ=π/6或5π/6
θ=π/12或5π/12
(2) 若a⊥b
(1/2)cosθ+(1/2)sinθ=0
tanθ=-1
θ=-π/4
(3)|a+b|=√[(cosθ+1/2)²+(sinθ+1/2)²]
=√(sinθ+cosθ+3/2)
=√[√2sin(θ+π/4)+3/2]
所以当θ=π/4时，|a+b|最大=√[√2+3/2]
=√(√2+1)²/√2
=1+√2/2

(1)平行-> 1/2/cos t = sin t /(1/2)
sint cos t=1/4
sin2t=1/2
2t =pi/6 or 5pi/6
t=pi/12 or 5 pi/12
(2) 垂直->(a,b)=0
cos t/2+sin t/2=0
sin t=-cos t
tan t =-1
t=3pi/4, -pi/4

(3) |a+b|^2=(cos t+1/2)^2+(sint +1/2)^2
=1+sint +cos t+1/2
=3/2+ 根号2* sin(t+pi/4)
<=3/2+根号2
|a+b|<=根号(3/2+根号2)=1+(根号2) /2

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