已知函数f(x)=cos的平方x+根号3sinxcosx+1,x属于R.（1）求证f(x)得小正周期和最值

用二倍角公式啦(cosx)^2 = (1+cos(2x))/2 根号(3) sinxcosx = 根号(3)/2 * sin(2x)所以f(x)= 1/2 * cos(2x) + 根号(3)/2 * sin(2x) +3/2Sin(π/6) = 1/2Cos(π/6) = 根号(3)/2所以f(x) = sin(π/6)cos(2x) + cos(π/6)sin(2x) + 3/2= sin(2x + π/6) + 3/2 逆用两角和公式所以周期T = (2π)/2 = π最大值 = 1 + 3/2 = 5/2最小值 = -1 + 3/2 = 1/2增区间2kπ - π/2