作业帮已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ] （1）求函数f(x)的最大值并求已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ]（1）求函数f(x)的最大值并求出此时x的值；（2）
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/15 17:42:51
![已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ] （1）求函数f(x)的最大值并求已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ]（1）求函数f(x)的最大值并求出此时x的值；（2）](/uploads/image/z/1697992-16-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2%E2%88%9A3sin%28x%2F2%29cos%28x%2F2%29-%5B+%28cos%26%23178%3B%28x%2F2%29-sin%26%23178%3B%28x%2F2%29+%5D+%EF%BC%881%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%B9%B6%E6%B1%82%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2%E2%88%9A3sin%28x%2F2%29cos%28x%2F2%29-%5B+%28cos%26%23178%3B%28x%2F2%29-sin%26%23178%3B%28x%2F2%29+%5D%EF%BC%881%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%B9%B6%E6%B1%82%E5%87%BA%E6%AD%A4%E6%97%B6x%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89)
已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ] （1）求函数f(x)的最大值并求已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ]（1）求函数f(x)的最大值并求出此时x的值；（2）
已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ] （1）求函数f(x)的最大值并求
已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ]
（1）求函数f(x)的最大值并求出此时x的值；
（2）若f(x)=0,求[sinx+cos(π+x)]/[sinx+sin(π/2-x)]的值
已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ] （1）求函数f(x)的最大值并求已知函数f(x)=2√3sin(x/2)cos(x/2)-[ (cos²(x/2)-sin²(x/2) ]（1）求函数f(x)的最大值并求出此时x的值；（2）
(1)f(x)=(2√3)sin(x/2)cos(x/2)-[(cos²(x/2)-sin²(x/2)]=(√3)sinx-cosx=2sin(x-π/6)
令x-π/6=π/2+2kπ,得：x=2π/3+2kπ(k∈Z)
即当x=2π/3+2kπ(k∈Z)时,f(x)有最大值2.
(2)
∵f(x)=0
∴2sin(x-π/6)=0
∴sin(x-π/6)=0
∴tan(x-π/6)=sin(x-π/6)/cos(x-π/6)=0
∴[tanx-tan(π/6)]/[1+tanxtan(π/6)]=0
∴tanx-tan(π/6)=0
∴tanx=tan(π/6)=(√3)/3
∴[sinx+cos(π+x)]/[sinx+sin(π/2-x)]
=(sinx-cosx)/(sinx+cosx)
=[(sinx-cosx)/cosx]/[(sinx+cosx)/cosx]
=(tanx-1)/(tanx+1)
=[(√3)/3-1]/[(√3)/3+1]
=(√3)-2.