已知函数f(x)=2sin²(π/4+x)-根号3cos2x,x∈[π/4,π/2](1) 求f(x)的最大值最小值 (1)f(x)=2sin²(π/4+x)-根号3cos2x=1-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x+1=2sin(2x-π/3)+1∵ x∈[π/4,π/2]∴ 2x-π/3∈[π/6,2π/3]
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![已知函数f(x)=2sin²(π/4+x)-根号3cos2x,x∈[π/4,π/2](1) 求f(x)的最大值最小值 (1)f(x)=2sin²(π/4+x)-根号3cos2x=1-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x+1=2sin(2x-π/3)+1∵ x∈[π/4,π/2]∴ 2x-π/3∈[π/6,2π/3]](/uploads/image/z/1746955-19-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2sin%26%23178%3B%28%CF%80%2F4%2Bx%29-%E6%A0%B9%E5%8F%B73cos2x%2Cx%E2%88%88%5B%CF%80%2F4%2C%CF%80%2F2%5D%EF%BC%881%EF%BC%89+%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E6%9C%80%E5%B0%8F%E5%80%BC+%EF%BC%881%EF%BC%89f%28x%29%3D2sin%26%23178%3B%28%CF%80%2F4%2Bx%29-%E6%A0%B9%E5%8F%B73cos2x%3D1-cos%28%CF%80%2F2%2B2x%29-%E2%88%9A3cos2x%3Dsin2x-%E2%88%9A3cos2x%2B1%3D2sin%282x-%CF%80%2F3%29%2B1%E2%88%B5+x%E2%88%88%5B%CF%80%2F4%2C%CF%80%2F2%5D%E2%88%B4+2x-%CF%80%2F3%E2%88%88%5B%CF%80%2F6%2C2%CF%80%2F3%5D)
已知函数f(x)=2sin²(π/4+x)-根号3cos2x,x∈[π/4,π/2](1) 求f(x)的最大值最小值 (1)f(x)=2sin²(π/4+x)-根号3cos2x=1-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x+1=2sin(2x-π/3)+1∵ x∈[π/4,π/2]∴ 2x-π/3∈[π/6,2π/3]
已知函数f(x)=2sin²(π/4+x)-根号3cos2x,x∈[π/4,π/2]
(1) 求f(x)的最大值最小值
(1)
f(x)=2sin²(π/4+x)-根号3cos2x
=1-cos(π/2+2x)-√3cos2x
=sin2x-√3cos2x+1
=2sin(2x-π/3)+1
∵ x∈[π/4,π/2]
∴ 2x-π/3∈[π/6,2π/3]
∴ sin(2x-π/3)∈[1/2,1]
∴ 2x-π/3=π/6时,f(x)有最小值2
2x-π/3=π/2时,f(x)有最大值3
为什么 sin(2x-π/3)∈[1/2,1]
已知函数f(x)=2sin²(π/4+x)-根号3cos2x,x∈[π/4,π/2](1) 求f(x)的最大值最小值 (1)f(x)=2sin²(π/4+x)-根号3cos2x=1-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x+1=2sin(2x-π/3)+1∵ x∈[π/4,π/2]∴ 2x-π/3∈[π/6,2π/3]
因为sinπ/6到sinπ/2是增函数,[
1/2,1]
,sinπ/2到sin2π/3是减函数值域{1,二分之根号三} 所以sin(2x-π/3)∈[1/2,1]
x∈[π/4,π/2]
∴2x∈[π/2,π]
∴2x-π/3∈[π/6,2π/3]
令θ=2x-π/3
当θ=π/2时,sinθ有最大值=1
当θ=π/6时,sinθ有最小值=1/2
∴sin(2x-π/3)∈[1/2,1]