已知向量a=(cos3/2 x,sin3/2 x),b=(cos1/2x,-sin1/2x),x∈[-pai/3,pai/2](1) 求证(a-b) ⊥(a+b)(2) │a+b|=1/3,求cosx的值
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![已知向量a=(cos3/2 x,sin3/2 x),b=(cos1/2x,-sin1/2x),x∈[-pai/3,pai/2](1) 求证(a-b) ⊥(a+b)(2) │a+b|=1/3,求cosx的值](/uploads/image/z/7080068-20-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28cos3%2F2+x%2Csin3%2F2+x%29%2Cb%3D%28cos1%2F2x%2C-sin1%2F2x%29%2Cx%E2%88%88%5B-pai%2F3%2Cpai%2F2%5D%281%29+%E6%B1%82%E8%AF%81%28a-b%29+%E2%8A%A5%28a%2Bb%29%282%29+%E2%94%82a%2Bb%7C%3D1%2F3%2C%E6%B1%82cosx%E7%9A%84%E5%80%BC)
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已知向量a=(cos3/2 x,sin3/2 x),b=(cos1/2x,-sin1/2x),x∈[-pai/3,pai/2](1) 求证(a-b) ⊥(a+b)(2) │a+b|=1/3,求cosx的值
已知向量a=(cos3/2 x,sin3/2 x),b=(cos1/2x,-sin1/2x),x∈[-pai/3,pai/2]
(1) 求证(a-b) ⊥(a+b)
(2) │a+b|=1/3,求cosx的值
已知向量a=(cos3/2 x,sin3/2 x),b=(cos1/2x,-sin1/2x),x∈[-pai/3,pai/2](1) 求证(a-b) ⊥(a+b)(2) │a+b|=1/3,求cosx的值
(1)(a-b)点乘(a+b)=a^2-b^2=1-1=0
所以 (a-b) ⊥(a+b)
(2)│a+b|=根号下(a^2+b^2+2ab)
=根号下(2+2(cos3/2 x*cos1/2x-sin3/2 x*sin1/2x))
=根号下(2+2cos2x)=1/3
所以cos2x=-17/18=2(cosx)^2-1
因为x∈[-pai/3,pai/2] 所以cosx