已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4) 求:(见问题补充)★满意可高分悬赏
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已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4) 求:(见问题补充)★满意可高分悬赏 麻烦了
1,求函数f(x)的最小正周期和图像的对称轴方程
【答案最小正周期是π,方程是x=kπ+π/3(k∈Z)我需要过程 谢谢】
2,求函数f(x)在区间[-π/12,π/2]上的值域
【答案是[-√3/2,1]我也需要过程 谢谢】
1,求函数f(x)的最小正周期和图像的对称轴方程
【答案最小正周期是π,方程是x=kπ+π/3(k∈Z)我需要过程 谢谢】
2,求函数f(x)在区间[-π/12,π/2]上的值域
【答案是[-√3/2,1]我也需要过程 谢谢】
![已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4) 求:(见问题补充)★满意可高分悬赏](/uploads/image/z/2422125-45-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x-%CF%80%2F3%29%2B2sin%28x-%CF%80%2F4%29sin%28x%2B%CF%80%2F4%29+%E6%B1%82%EF%BC%9A%EF%BC%88%E8%A7%81%E9%97%AE%E9%A2%98%E8%A1%A5%E5%85%85%EF%BC%89%E2%98%85%E6%BB%A1%E6%84%8F%E5%8F%AF%E9%AB%98%E5%88%86%E6%82%AC%E8%B5%8F)
(1)
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=1/2cos2x+√3/2sin2x+(sinx-cosx)(sinx+cosx)
=1/2cos2x+√3/2sin2x+sin²x-cos²x
=1/2cos2x+√3/2sin2x-cos2x
=sin(2x-π/6)
∴最小正周期:T=2π/2=π
由2x-π/6=kπ+π/2(k∈Z)
得x=kπ/2+π/3(k∈Z)
∴函数图象的对称轴方程为:x=kπ+π/3(k∈Z)
(2)
∵x∈[-π/12,π/2]
∴2x-π/6∈[-π/3,5π/6]
∵f(x)=sin(2x-π/6)在区间[-π/12,π/3]上单调递增,在区间[π/3,π/2]上单调递减
∴x=π/3时,f(x)取最大值1
又∵f(-π/12)=-√3/2<f(π/2)=1/2
当x=-π/12时,f(x)取最小值-√3/2
∴函数f(x)在区间[-π/12,π/12]上的值域是:[-√3/2,1]
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=1/2cos2x+√3/2sin2x+(sinx-cosx)(sinx+cosx)
=1/2cos2x+√3/2sin2x+sin²x-cos²x
=1/2cos2x+√3/2sin2x-cos2x
=sin(2x-π/6)
∴最小正周期:T=2π/2=π
由2x-π/6=kπ+π/2(k∈Z)
得x=kπ/2+π/3(k∈Z)
∴函数图象的对称轴方程为:x=kπ+π/3(k∈Z)
(2)
∵x∈[-π/12,π/2]
∴2x-π/6∈[-π/3,5π/6]
∵f(x)=sin(2x-π/6)在区间[-π/12,π/3]上单调递增,在区间[π/3,π/2]上单调递减
∴x=π/3时,f(x)取最大值1
又∵f(-π/12)=-√3/2<f(π/2)=1/2
当x=-π/12时,f(x)取最小值-√3/2
∴函数f(x)在区间[-π/12,π/12]上的值域是:[-√3/2,1]
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