函数f(x)=sin²(wx+π/6)-cos²(wx+π/6)(w>0)的最小正周期为2π,求w的值;若tanx=4/3,且x∈(π,3π/2),求f(x)的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 03:47:07
![函数f(x)=sin²(wx+π/6)-cos²(wx+π/6)(w>0)的最小正周期为2π,求w的值;若tanx=4/3,且x∈(π,3π/2),求f(x)的值](/uploads/image/z/8857035-27-5.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dsin%26%23178%3B%EF%BC%88wx%2B%CF%80%2F6%EF%BC%89-cos%26%23178%3B%EF%BC%88wx%2B%CF%80%2F6%EF%BC%89%EF%BC%88w%3E0%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B8%BA2%CF%80%2C%E6%B1%82w%E7%9A%84%E5%80%BC%EF%BC%9B%E8%8B%A5tanx%3D4%2F3%2C%E4%B8%94x%E2%88%88%EF%BC%88%CF%80%2C3%CF%80%2F2%EF%BC%89%2C%E6%B1%82f%28x%29%E7%9A%84%E5%80%BC)
函数f(x)=sin²(wx+π/6)-cos²(wx+π/6)(w>0)的最小正周期为2π,求w的值;若tanx=4/3,且x∈(π,3π/2),求f(x)的值
函数f(x)=sin²(wx+π/6)-cos²(wx+π/6)(w>0)的最小正周期为2π,求w的值;若tanx=4/3,
且x∈(π,3π/2),求f(x)的值
函数f(x)=sin²(wx+π/6)-cos²(wx+π/6)(w>0)的最小正周期为2π,求w的值;若tanx=4/3,且x∈(π,3π/2),求f(x)的值
f(x)=sin²(wx+π/6)-cos²(wx+π/6)
=-cos[2(wx+π/6)]
=-cos(2wx+π/3)
最小正周期为2π=2π/(2w)
w=1/2
f(x)=-cos(x+π/3)
tanx=4/3且x∈(π,3π/2)
sinx=-4/5,cosx=-3/5
f(x)=-cos(x+π/3)
=cosxcos(π/3)-sinxsin(π/3)
=-3/5*1/2+4/5*√3/2
=(-3+4√3)/10
w=0.5
f(x)=sin²(wx+π/6)-cos²(wx+π/6)
=-cos(2wx+π/3);
T=2π/2w=2π;
w=1/2;
f(x)=-cos(x+π/3);
tanx=sinx/cosx=4/3;
sinx=(4/3)cosx;
sin²x+cos²x=1;
25cos²x/...
全部展开
f(x)=sin²(wx+π/6)-cos²(wx+π/6)
=-cos(2wx+π/3);
T=2π/2w=2π;
w=1/2;
f(x)=-cos(x+π/3);
tanx=sinx/cosx=4/3;
sinx=(4/3)cosx;
sin²x+cos²x=1;
25cos²x/9=1;
cos²x=9/25;
∵x∈(π,3π/2)
∴cosx=-3/5;
sinx=-4/5;
f(x)=-(cosxcos(π/3)-sinxsin(π/3))=sinxsin(π/3)-cosxcos(π/3)=(-4/5)(√3/2)-(-3/5)(1/2)=-2√3/5+3/10;
收起