f(x)=根号3/2sinπx+1/2cosπx

1个回答

f(x)=根号3/2sinπx+1/2cosπx=sinπ/3*sinπx+cosπ/3*cosπx=cos(πx-π/3)=cosπ(x-1/3),
x属于R,x-1/3x属于R,cosπ(x-1/3)最大值为1,最小值为-1,
即f(x)=根号3/2sinπx+1/2cosπx最大值为1,最小值为-1,

设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) （1）求f(π/6),f(π/
已知函数F(X)=2sin(X+π/2)cos(x+π/2）+2根号3cos^2(x+π/2)-根号3（1）求f(x)的
设函数f(x)=sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x)
已知f(x)＝sin(x−3π)•cos(2π−x)•sin(−x+3π2)cos(−x−π)•cos(π2−x)
已知 f(x)= sin(x-3π)•cos(2π-x)•sin(-x+ 3π 2 ) cos(-x-π)•cos( π
f(x)sinx·sin(π-x)+√3sin(π/2+x)cos(π/2+x)+2cos(π+x)cos(π-x) 求
已知f(x)=sin2x-(根号3)cos2x+1,x∈[π/4,π/2]
已知函数f(x)= [sin(2π-x)sin(π+x)cos(-x-π)] ／[2cos(π-x)sin(3π-x)]
1：已知cos(π/2-x)=根号2*sin(π-y),根号3*cos（-x）=-根号2*cos（π+y）且0＜x＜π,
2sin(2πx)cos(2πx)+根号（3）sin(2πx)=0 0