f(x)=2sinxcosx+cos^2x-sin^2x

题目

f(x)=2sinxcosx+cos^2x-sin^2x
1)求最小正周期
2)f(x)最大值以及相应的x值

答案

y=2sinxcosx+(cos^2x-sin^2x)
=sin2x+cos2x
=√2[√2sin2x/2+√2cos2x/2]
=√2sin(2x+π/4)
所以T=2π/2=π
-1<=sin(2x-π/4)<=1
所以值域[-√2,√2]
f(x)值最大为：√2
当2x+π/4=π/2时,f(x)值最大
即x=π/8