已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0< θ<派/2.若a·b=0,求tan θ 值
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![已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0< θ<派/2.若a·b=0,求tan θ 值](/uploads/image/z/11008845-45-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28sin%CE%B82%2Ccos%CE%B8%29%2Cb%3D%281%2C-cos+%CE%B8+%29%2C0%EF%BC%9C+%CE%B8%EF%BC%9C%E6%B4%BE%2F2.%E8%8B%A5a%C2%B7b%3D0%2C%E6%B1%82tan+%CE%B8+%E5%80%BC)
已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0< θ<派/2.若a·b=0,求tan θ 值
已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0< θ<派/2.若a·b=0,求tan θ 值
已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0< θ<派/2.若a·b=0,求tan θ 值
∵a·b=0 ∴sin^2θ.1+cosθ.-cos θ=0 整理得sin^2θ=cos^2 θ
又∵0< θ<π/2 ∴θ=π/4 故tan θ =1
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