Question

已知函数f(x)=2sinxcosx+cos2x(x属于R)

(1)求f(x)的最小正周期和最大值;

Answer 1

f(x)=sin2x+cos2x
=√2(√2/2*sin2x+√2/2cos2x)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
所以T=2π/2=π
最大值=√2

Answer 2

f(x)=sin2x+cos2x=√2 sin(2x+45°)
最小正周期为π 最大值为√2 即根号下2