已知a向量=(1,sinθ),b=(1,cosθ),θ属于R,(1)若a向量+b向量=(2,0),求sin^2θ+2sinθcosθ的值 (2)若a-b=(0,1/5),求sinθ+cosθ的值
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![已知a向量=(1,sinθ),b=(1,cosθ),θ属于R,(1)若a向量+b向量=(2,0),求sin^2θ+2sinθcosθ的值 (2)若a-b=(0,1/5),求sinθ+cosθ的值](/uploads/image/z/1811874-66-4.jpg?t=%E5%B7%B2%E7%9F%A5a%E5%90%91%E9%87%8F%3D%EF%BC%881%2Csin%CE%B8%EF%BC%89%2Cb%3D%281%2Ccos%CE%B8%29%2C%CE%B8%E5%B1%9E%E4%BA%8ER%2C%EF%BC%881%EF%BC%89%E8%8B%A5a%E5%90%91%E9%87%8F%2Bb%E5%90%91%E9%87%8F%3D%EF%BC%882%2C0%EF%BC%89%2C%E6%B1%82sin%5E2%CE%B8%2B2sin%CE%B8cos%CE%B8%E7%9A%84%E5%80%BC+%EF%BC%882%EF%BC%89%E8%8B%A5a-b%3D%280%2C1%2F5%29%2C%E6%B1%82sin%CE%B8%2Bcos%CE%B8%E7%9A%84%E5%80%BC)
已知a向量=(1,sinθ),b=(1,cosθ),θ属于R,(1)若a向量+b向量=(2,0),求sin^2θ+2sinθcosθ的值 (2)若a-b=(0,1/5),求sinθ+cosθ的值
已知a向量=(1,sinθ),b=(1,cosθ),θ属于R,(1)若a向量+b向量=(2,0),求sin^2θ+2sinθcosθ的值 (2)若a-b=(0,1/5),求sinθ+cosθ的值
已知a向量=(1,sinθ),b=(1,cosθ),θ属于R,(1)若a向量+b向量=(2,0),求sin^2θ+2sinθcosθ的值 (2)若a-b=(0,1/5),求sinθ+cosθ的值
(1)sinθ+cosθ=0,两边同时平方得sin^2θ+2sinθcosθ+cos^2θ=0,sinθ=-cosθ
代入最后得sin^2θ+2sinθcosθ=-1/2
(2)sinθ-cosθ=1/5,两边同时平方得sin^2θ-2sinθcosθ+cos^2θ=1/25,即sinθcosθ=12/25,sin^2θ+2sinθcosθ+cos^2θ=(sinθ+cosθ)^2=49/25,sinθ+cosθ=±7/5