Question

求y=sin平方x+2sinxcosx的周期

Answer 1

y=sin^x+2sinxcosx
=1/2-cos2x/2+sin2x
=根号下(5/4)*[2sin2x/根号5-cos2x/根号5]+1/2
设cosa=2/根号5，sina=-1/根号5
上式=根号下(5/4)*sin(2x+a)+1/2
因此周期=2派/2=派

Answer 2

y=sin²x+2sinxcosx
=(1-cos2x)/2+sin2x
=1/2+(sin2x+(cos2x/2))
=1/2+(根号(5/4))sin(2x+A)
其中tanA=1/2

最小正周期是T=2派/2＝派

Answer 3

y=(sinx)^2+2sinxcosx
=(1-cos2x)/2+sin2x
=1-cos2x/2+sin2x
因为cos2x和sin2x
T都等于=2π/2=π
所以T=π

Answer 4

y=sin平方x+2sinxcosx
y=sin^2x+2sinxcosx
y=sinx(sinx+2cosx)
y=Asinxcos(x+m)
y=Bsin(2x+n)
只要求T
所以不必求出具体B与n的值
T=2Pi/2=Pi

Answer 5

y=sin^2x+2sinxcosx
=1/2(1-sin2x)+sin2x
=1/2(sin2x+1)
W=π

Answer 6

y=sin平方x+2sinxcosx=1-1/2*cos2x+sin2x
则周期为pi