Did i developed these teylor series correctly

[transgalactic]

A.
<br /> sin^2 x=0+x^2-\frac{x^4}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{4n+2}}+O(4n+2)<br />
B.
<br /> xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^{n+1}}{n!}+O(n+1)<br />
C.
<br /> xsin^3 x=0+x^4-\frac{x^6}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{5n+6}}+O(5n+4)<br />

 

[transgalactic]

A:
<br /> cos 2x=cos^2x-sin^2x=1-2sin^2x\\<br />
<br /> f(x)=sin^2x=\frac{1-cos 2x}{2}\\<br />
<br /> f(0)=0\\<br />
<br /> f&#039;(x)=\frac{1-cos 2x}{2}=cos 2xsin2x=\frac{sin4x}{2}\\<br />
<br /> f&#039;(0)=0\\<br />
<br /> f&#039;&#039;(x)=2cos4x\\<br />
<br /> f^{(3)}=-8sin4x\\<br />
<br /> f^{(4)}=-32cos4x\\<br />
i know to to use each derivative to build a member out of it.
how to defne the n'th member??

regarding b:
<br /> xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^{n+1}}{n!}+O(n+1)<br />
i am sure that its correct because i took the series for e^x and multiplied eah member by x.
so the only error that could be is with the remainder .
i thought that if we multiply the remainder by x we add 1 to it
where is my mistake??

regarding C:
<br /> sin3x=3sinx-4sin^3x\\<br />
<br /> f(x)=sin^3x=\frac{3sinx-sin3x}{4}\\<br />
<br /> f(0)=0\\<br />
<br /> f&#039;(x)=\frac{3cosx-\frac{cos3x}{3}}{4}\\<br />
<br /> f&#039;(0)=0<br />
<br /> f&#039;&#039;(x)=\frac{-3sinx-\frac{sin3x}{9}}{4}\\<br />
<br /> f&#039;&#039;(0)=0<br />
for every member i get 0
??

did i solved A,B correctly
where is my mistake for C

??