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#