作业帮已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R)已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R) 问,(1)函数f(x)的最小值和最小正周期 (2)若对为锐角,且向量m=(1,5)与向量n=[1,f(兀/4-A)]垂
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/05 17:51:36
![已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R)已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R) 问,(1)函数f(x)的最小值和最小正周期 (2)若对为锐角,且向量m=(1,5)与向量n=[1,f(兀/4-A)]垂](/uploads/image/z/6863191-7-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%E2%88%9A3sin%5E2%28x%2B%E5%85%80%2F4%29-cos%5E2%28x%29-%281%2B%E2%88%9A3%29%C3%B72+%28x%E2%88%88R%29%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%E2%88%9A3sin%5E2%28x%2B%E5%85%80%2F4%29-cos%5E2%28x%29-%281%2B%E2%88%9A3%29%C3%B72++%28x%E2%88%88R%29++%E9%97%AE%2C%281%29%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F+++++++++%282%29%E8%8B%A5%E5%AF%B9%E4%B8%BA%E9%94%90%E8%A7%92%2C%E4%B8%94%E5%90%91%E9%87%8Fm%3D%281%2C5%29%E4%B8%8E%E5%90%91%E9%87%8Fn%3D%5B1%2Cf%28%E5%85%80%2F4-A%29%5D%E5%9E%82)
已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R)已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R) 问,(1)函数f(x)的最小值和最小正周期 (2)若对为锐角,且向量m=(1,5)与向量n=[1,f(兀/4-A)]垂
已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R)
已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R) 问,(1)函数f(x)的最小值和最小正周期 (2)若对为锐角,且向量m=(1,5)与向量n=[1,f(兀/4-A)]垂直,求cos2A的值 求过程,下面有图原题
已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R)已知函数f(x)=√3sin^2(x+兀/4)-cos^2(x)-(1+√3)÷2 (x∈R) 问,(1)函数f(x)的最小值和最小正周期 (2)若对为锐角,且向量m=(1,5)与向量n=[1,f(兀/4-A)]垂
(1)
f(x)=√3sin²(x+π/4)-cos²x-(1+√3)/2
=√3/2[1-cos(2x+π/2)]-1/2(1+cos2x)+(1+√3)/2
=-√3/2cos(2x+π/2)-1/2cos2x
=√3/2sin2x-1/2cos2x
=sin(2x-π/6)
∴f(x)的最小值为-1
最小正周期T=2π/2=π
(2)
∵向量m=(1,5)⊥向量n=(1,f(π/4-A))
∴5f(π/4-A)-1=0
∴sin[2(π/4-A)-π/6]=1/5
∴sin(π/3-2A)=1/5
∵0
∵sin(π/3-2A)>0
∴0<π/3-2A<π/3
∴cos(π/3-2A)=√[1-sin²(π/3-2A)]=2√6/5
∴cos2A=cos[π/3-(π/3-2A)]
=cosπ/3cos(π/3-2A)+sinπ/3sin(π/3-2A)
=1/2*2√6/5+√3/2*1/5
=(2√6+√3)/10
