已知函数f(x)＝4coswx×sin(wx＋派/4)(w＞0)的最小正周期为派,(1)求wf(x)=4coswxsin(wx+π/4)=4coswx*(√2/2sinwx+√2/2coswx)=2√2sinwxcoswx+2√2cos^2wx=√2(sin2wx+cos2wx)+√2=2sin(2wx+π/4)+√2最小正周期为π∴2π/2w=πw=1问

Question

已知函数f(x)＝4coswx×sin(wx＋派/4)(w＞0)的最小正周期为派,(1)求w
f(x)=4coswxsin(wx+π/4)
=4coswx*(√2/2sinwx+√2/2coswx)
=2√2sinwxcoswx+2√2cos^2wx
=√2(sin2wx+cos2wx)+√2
=2sin(2wx+π/4)+√2
最小正周期为π
∴2π/2w=π
w=1
问：为什么√2(sin2wx+cos2wx)+√2
=2sin(2wx+π/4)+√2

Answer 1

SQRT(2)=SQRT(2)/2*2 SQRT(2)=SIN(PI/4)=COS(PI/4) 故
√2(sin2wx+cos2wx) =2 (sin2wx*sinpi/4+cos2wxcospi/4)=2sin(2wx+π/4)