已知向量a=(cosθ/2,sinθ/2),b=(2,1),且a⊥b,(1)求tanθ的值

a⊥b
所以 ( cos x/2 * 2)+ ( sin x/2 * 1)=0
tan x/2 =-2
所以 tanx=2*（-2）/( 1 - (-2)^2 )=4/3求cos2θ/(根号2*cos(π/4+θ)*sinθ的值。cos2θ/(根号2*cos(π/4+θ)*sinθ
=(cosθ^2-sinθ^2)/[(cosθ-sinθ)*sinθ ]
=(cosθ+sinθ)/sinθ
=1/tanθ+1