sin^2展开成幂级数
1个回答

sin^2(x)=(1-cos2x)/2=1/2-(1/2)cos2x ...(1)

由于:

cosx=1-(x^2)/2!+(x^4)/4!-(x^6)/6!+...

有:

cos2x=1-(2x)^2/2!+(2x)^4/4!-(2x)^6/6!+...(2)

将(2)代入(1)得:

sin^2(x)=1/2-(1/2)cos2x

=(1/2)[(2x)^2/2!-(2x)^4/4!+(2x)^6/6!- ...]