Question

已知，函数f(x)＝2sinx(sin x＋cos x) 求，〔1〕f(x)的最小正周期，和最大值

Answer 1

f(x)=2sinx(sin x＋cos x)
=2(sinx)^2+2sinx*cosx
=-[1-2(sinx)^2]+2sinx*cosx+1
=sin2x-cis2x+1
=√2*(√2/2*sin2x-√2/2*cos2x)+1
=√2*(sin2x*cosπ/4-cos2x*sinπ/4)+1
=√2*sin(2x-π/4)+1。
f(x)的最小正周期为：T=2π/2=π，
最大值为：√2+1。

Answer 2

f(x)＝2sinx(sin x＋cos x)
=2sin^2x+2sinxcosx
=1-cos2x+sin2x
=1+√2(√2/2sin2x-√2/2cos2x)
=1+√2sin(2x-π/4)
最小正周期T=2π/2=π
当sin(2x-π/4)=1时原式有最大值 1+√2

Answer 3

f(x)=2sinx*sinx+2sinx*cosx
=1-cos2x+sin2x
=√2*sin(2x-∏/4)+1
自然最小正周期是∏，最大值是1+√2

Answer 4

f(x)=1-cos2x+sin2x=1+√2sin(2x-π/4）T=π 最大值为 1+√2