已知函数f(x)=2sin(x+5π/24)cos(x+5π/24)-2cos²(x+5π/24)+1(1)求f(x)的最小正周期(2)求函数f(x)的单调递增区间
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 13:29:43
![已知函数f(x)=2sin(x+5π/24)cos(x+5π/24)-2cos²(x+5π/24)+1(1)求f(x)的最小正周期(2)求函数f(x)的单调递增区间](/uploads/image/z/1618398-54-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2sin%28x%2B5%CF%80%2F24%29cos%28x%2B5%CF%80%2F24%29-2cos%26%23178%3B%28x%2B5%CF%80%2F24%EF%BC%89%2B1%281%29%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
已知函数f(x)=2sin(x+5π/24)cos(x+5π/24)-2cos²(x+5π/24)+1(1)求f(x)的最小正周期(2)求函数f(x)的单调递增区间
已知函数f(x)=2sin(x+5π/24)cos(x+5π/24)-2cos²(x+5π/24)+1
(1)求f(x)的最小正周期(2)求函数f(x)的单调递增区间
已知函数f(x)=2sin(x+5π/24)cos(x+5π/24)-2cos²(x+5π/24)+1(1)求f(x)的最小正周期(2)求函数f(x)的单调递增区间
f(x)=sin(2x+5π/12)-cos(2x+5π/12)
=√2sin(2x+5π/12-π/4)
=√2sin(2x+π/6)
所以,最小正周期T=2π/2=π
递增区间:
-π/2+2kπ
已知函数F(X)=SIN(2X+φ)(-π
已知函数f(x)=2根号3sin平方x-sin(2x-π/3)
已知函数f(x)=-1/2+sin(5/2x)/2sin(x/2)(0
已知函数f(x)=(1+1 anx)sin^2x+m sin(x+π/4)sin(x-π/4)
已知函数f(x)=sin(2x+φ) (0
已知函数f(x)=4sinx-2/1+sin²x 证明f(x+2π)=f(x)
已知函数f(x)=sinπx/3(x∈N),f(1)+f(2)+.+f(99)=( )
已知函数f(x)=sin(2x+π/6)+sin(2x+π/6)+2cos²x
已知函数f(x)=2^(2-x),x>=2;f(x)=sinπx/4,-2
已知函数f(x)=(√3sinωx+cosωx)*sin(-3π/2+ωx)(0
已知关于x的函数f(x)=根号2sin(2x+φ) (-π
已知关于x的函数f(x)=根号2sin(2x+φ) (-π
已知关于x的函数f(x)=根号2sin(2x+φ) (-π
已知函数f(x)=cos^2(x-π/6)-sin^2x化简
已知关于x的函数f(x)=√2sin(2x+φ)(-π
已知关于X的函数f(x)=√2sin(2x+φ)(-π
已知关于X的函数f(x)=√2sin(2x+φ)(-π
已知函数f(x)=sin²(π/4+x)+cos²x+1/2求最值