What is the Taylor series expansion for $sin(sin(x))$?

Hint: The Taylor series expansion for any function has a definite formula where we just need to find the derivatives and plug them into the formula. After plugging the values we can write the Taylor expansion of any function. Here, we are given a function inside a function so we use the Chain rule to find derivatives of this function. And then to find the second derivative, the product rule can be applied.

Complete step by step solution:
The Taylor series expansion of any function is given by the following formula:
$f(x) = f(a) + \dfrac{f'(a)}{1!}(x-a) + \dfrac{f(a)}{2!}(x-a)^2 + \dfrac{f'(a)}{3!}(x-a)^3+\ldots$
Writing the same expression is summation form the same thing can be written as :
$f(x)= \sum_{k=0}^\infty \dfrac{f^{\left(k\right)}(0)}{k!} (x-a)^k$
Now, we find the first derivative of the function:
$f(x)=sin(sin(x))$
We use chain rule here,
$\dfrac{df(x)}{dx}=cos(sin(x))\dfrac{d(sin(x))}{dx}$
Now, we know that
$\dfrac{d(sin(x))}{dx}=cos(x)$
So, we have:
$\implies \dfrac{df(x)}{dx}=cos(sin(x))\times cos(x)$
Now, we find the second derivative.
$\dfrac{d^2(sin(sin(x))}{dx^2}=\dfrac{d(cos(x)\times cos(sin(x)))}{dx}$
We now apply Product rule here,
$\dfrac{d^2 (sin(sin(x)))}{dx^2}=\left(cos(x)\times (-sin(sin(x))\times cos(x))\right)+\left(cos(sin(x))\times (-sin(x))\right)$
So, we have found the first few terms. We need to find a general expansion, because no point around which the result has to be found out is given. Put ‘a’ in all function and derivatives to obtain the general expression Now we plug in the values and obtain the resultant Taylor expression as follows:
$f(x)= sin(sin(a)) + \dfrac{ cos(sin(a))\times cos(a)}{1!}(x-a)$
$+\dfrac{\left(cos(a)\times (-sin(sin(a))\times cos(a))\right)+\left(cos(sin(a))\times (-sin(a))\right)}{ 2!}(x-a)^2+\ldots$
$\implies f(x)=sin(sin(a))+cos(a)cos(sin(a))(x-a)+\dfrac{-sin(sin(a))cos^2(a)-sin(a)cos(sin(a))}{2}(x-a)^2+\ldots$

Note: While calculating the derivative of such complex functions, one often forgets to find the derivative of the expression inside the bracket. Without using chain rule, the derivative calculated would lead to an incorrect solution. Also, questions like these often lead to calculation mistakes, because you might forget some terms some times. So, always be very aware while writing the terms.