作业帮已知ω>0,向量m=(√3sinωx,cosωx),向量n=(cosωx,-cosωx),且f(x)=m·n+1/2且f(x)=m·n+1/2的最小正周期π（1）求f(x)的解析式；（2）已知a,b,c分别为△ABC内角A,B,C所对的边,且a=√19,c=3,又cosA恰是f(x)在[π/12,2π/3]
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/17 04:49:27
![已知ω>0,向量m=(√3sinωx,cosωx),向量n=(cosωx,-cosωx),且f(x)=m·n+1/2且f(x)=m·n+1/2的最小正周期π（1）求f(x)的解析式；（2）已知a,b,c分别为△ABC内角A,B,C所对的边,且a=√19,c=3,又cosA恰是f(x)在[π/12,2π/3]](/uploads/image/z/2522772-36-2.jpg?t=%E5%B7%B2%E7%9F%A5%CF%89%3E0%2C%E5%90%91%E9%87%8Fm%3D%28%E2%88%9A3sin%CF%89x%2Ccos%CF%89x%29%2C%E5%90%91%E9%87%8Fn%3D%28cos%CF%89x%2C-cos%CF%89x%29%2C%E4%B8%94f%28x%29%3Dm%C2%B7n%2B1%2F2%E4%B8%94f%28x%29%3Dm%C2%B7n%2B1%2F2%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%CF%80%EF%BC%881%EF%BC%89%E6%B1%82f%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E5%88%86%E5%88%AB%E4%B8%BA%E2%96%B3ABC%E5%86%85%E8%A7%92A%2CB%2CC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%2C%E4%B8%94a%3D%E2%88%9A19%2Cc%3D3%2C%E5%8F%88cosA%E6%81%B0%E6%98%AFf%28x%29%E5%9C%A8%5B%CF%80%2F12%2C2%CF%80%2F3%5D)
已知ω>0,向量m=(√3sinωx,cosωx),向量n=(cosωx,-cosωx),且f(x)=m·n+1/2且f(x)=m·n+1/2的最小正周期π（1）求f(x)的解析式；（2）已知a,b,c分别为△ABC内角A,B,C所对的边,且a=√19,c=3,又cosA恰是f(x)在[π/12,2π/3]
已知ω>0,向量m=(√3sinωx,cosωx),向量n=(cosωx,-cosωx),且f(x)=m·n+1/2
且f(x)=m·n+1/2的最小正周期π
（1）求f(x)的解析式；
（2）已知a,b,c分别为△ABC内角A,B,C所对的边,且a=√19,c=3,又cosA恰是f(x)在[π/12,2π/3]上的最小值,求b及△ABC的面积.
已知ω>0,向量m=(√3sinωx,cosωx),向量n=(cosωx,-cosωx),且f(x)=m·n+1/2且f(x)=m·n+1/2的最小正周期π（1）求f(x)的解析式；（2）已知a,b,c分别为△ABC内角A,B,C所对的边,且a=√19,c=3,又cosA恰是f(x)在[π/12,2π/3]
⑴f(x)=m•n+1/2
=√3sinωxcosωx-cos²ωx+1/2
=√3/2•sin2ωx-1/2•cos2ωx-1/2+1/2
=sin(2ωx-π/6),
∵ω>0,
∴T=π=2π/2ω => ω=1,
∴f(x)=sin(2x-π/6)；
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⑵∵x∈[π/12,2π/3],
∴2x-π/6∈[0,7π/6],
∴f(x)∈[-1/2,1],
∴cosA=-1/2=(b²+c²-a²)/2bc => b=2,
∵0