Based on this power series expansion of sin(x):
sin(x) = x-x^3/(3!)+x^5/(5!)-x^7/(7!)+...
= (-1)^0[x^(2*0+1)/((2*0+1)!)]+(-1)^1[x^(2*1+1)/((2*1+1)!)]+(-1)^2[x^(2*2+1)/((2*2+1)!)]+(-1)^3[x^(2*3+1)/((2*3+1)!)]+...
= sum_(n=0)^(\infty)((-1)^nx^(2n+1))/((2n+1)!)
we can derive it for sin(sin(x)).
What we can do is treat the inner sin(x) as the argument for the outer sin(x), similar to how x is the argument for ln(x). Doing that, we can just replace x with sin(x):
sin(sin(x)) = (sinx)/(1!)-(sinx)^3/(3!)+(sinx)^5/(5!)-(sinx)^7/(7!)+...
= sum_(n=0)^(\infty)((-1)^n(sinx)^(2n+1))/((2n+1)!)
or
sum_(n=0)^(\infty)((-1)^nsin^(2n+1)(x))/((2n+1)!)
This is a series expansion for sin(sin(x)). Unfortunately, it's not the Taylor series expansion (about x=0). Taylor series expansions are power series (infinite sums of the form \sum_{n=0}^{\infty}a_{n}x^{n}. To get the Taylor series expansion from the above approach, you'd have to replace each of the sin^{2n+1}(x) terms with, unfortunately, (\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!})^{2n+1}, which results in
sin(sin(x))=sum_(n=0)^(\infty)((-1)^n)/((2n+1)!)((\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!})^{2n+1}). You'd then have to simplify this into the form \sum_{n=0}^{\infty}a_{n}x^{n}. Yikes!
Probably a simpler approach overall is to just find the first few nonzero terms by differentiation:
f'(x)=cos(sin(x))\cdot cos(x), f''(x)=-sin(x)cos(sin(x))-sin(sin(x))\cdot cos^{2}(x), etc...
Then the Taylor series (about x=0) is:
f(0)+f'(0)x+\frac{1}{2!}f''(0)x^{2}+\cdots=x+\cdots
If you felt like finding more nonzero terms, you could use Wolfram Alpha to help you take more derivatives, or enter: Series[sin[sin[x]],{x,0,10}] into it and get:
sin(sin(x))=x-x^3/3+x^5/10-(8x^7)/315+(13x^9)/2520-(47x^11)/49896+\cdots