How can I expand these expressions with Maclaurin series :
$\displaystyle sin(sinx)$
$\displaystyle sin(cosx)$
The same way you would expand anything else.
Find $\displaystyle f'(x), f''(x), f'''(x),...,f^n(x)$ using the chain rule.
For example, $\displaystyle \frac{d}{dx} \sin(\cos(x)) = \cos(\cos(x)) \times (-\sin(x)) = - \sin(x) \cos(\cos(x)) $
Then from these, find $\displaystyle f'(0), f''(0), f'''(0),...,f^n(0)$
For example, $\displaystyle f'(0) = -\sin(0) \cos((\cos(0)) = 0 $
And then insert them into the basic formula:
$\displaystyle f(x) = f(0) + f'(0) x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6} x^3... + \frac{f^n (0)}{n!} x^n $
Which part of this process are you having trouble with?
This is not easy in the way that you'd think. Finding the exact Maclaurin series for this is not easy. But, note that $\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}x^{2n+1}}{(2n+1)!}\implies\sin\left(\sin( x)\right)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}\sin^{2n+1}(x)}{(2n+1)!}$