高一数学题 已知f(x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)(x∈R)
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/06/24 16:49:19
高一数学题 已知f(x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)(x∈R)
求: 函数最小正周期;函数在区间[-π/12,π/2]上的值域
求: 函数最小正周期;函数在区间[-π/12,π/2]上的值域
![高一数学题 已知f(x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)(x∈R)](/uploads/image/z/15488022-30-2.jpg?t=%E9%AB%98%E4%B8%80%E6%95%B0%E5%AD%A6%E9%A2%98+%E5%B7%B2%E7%9F%A5f%28x%29%3Dcos%282x-%CF%80%2F3%EF%BC%89%2B2sin%28x-%CF%80%2F4%EF%BC%89cos%28x-%CF%80%2F4%EF%BC%89%EF%BC%88x%E2%88%88R%EF%BC%89)
f(x)=cos(2x-π/3)+sin(2x-π/2)
=(1/2)cos2x+(√3/2)sin2x-cos2x
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/6)
则函数f(x)的最小正周期是2π/2=π,
当x∈[-π/12,π/2]时,2x-π/6∈[-π/3,5π/6],此时sin(2x-π/6)∈[-√3/2,1],即当x∈[-π/12,π/2]时,函数的值域是f(x)∈[-√3/2,1]
=(1/2)cos2x+(√3/2)sin2x-cos2x
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/6)
则函数f(x)的最小正周期是2π/2=π,
当x∈[-π/12,π/2]时,2x-π/6∈[-π/3,5π/6],此时sin(2x-π/6)∈[-√3/2,1],即当x∈[-π/12,π/2]时,函数的值域是f(x)∈[-√3/2,1]
已知函数f(x)=(√3/2)sinπx+(1/2)cosπx,x∈R
一道高中三角函数数学题 已知函数f(x)=sin^4x+cos^2x+1/4sin2xcos2x(x∈R),则f(x)(
已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos²x-1,x∈R
已知函数 f(x)=sin(x+7/4π)+cos(x-3/4π),x∈R 求f(x)的最小正周期和最小值
已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+cos平方x.
已知函数f(x)=sin(π-x)+cos(x+3π),x∈R.
已知f(x)=3cos(x+3π2)+cos(x−3π2)+sin(x+π)+a(a∈R,a为常数).
已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=sin(2x-π/6) ,
已知函数f(x)=cos(-x/2)+sin(π-x/2),x∈R
已知函数f(x)=sin(2x+7π/4)+cos(2x-3π/4),x属于R.
三角函数已知f(x)=(2cos^4x—2cos²x+1/2)/(2tan(π/4—x)sin²(x
高一数学:已知函数f(x)=2cos*sin(x+π/3)-√3sin^2x+sinx*cosx