Approximating $ e^{\sin(x)} $ I know that 
$$ e^{\sin(x)} = 1 + \sin(x) + {\sin^2x \over 2} + {\sin^3x \over 6} + B\sin^4x $$
I also know that $$ \sin(x) = x-{x^3 \over 6}+Bx^5 $$
I know I should probably just plug in the approximation for $ \sin(x) $ in the first equation, but it get's really messy, and I feel there should be an easier way?
I've recently started learning about Maclaurin series, so forgive me if this is a silly question. I'm just very confused on how I should approach problems like these.
 A: For an exam, in Calculus, all you will probably need from that function is a few terms of its Taylor. For this just take, as you were doing, the series of $e^x$ and $\sin(x)$ and compose them to compute the first few terms. 
For fun and culture let's compute the general term.
Let's recall the Faa di Bruno's formula:

Faa di Bruno's formula: 
  $$\frac{d^m}{dx^m}g(f(x))=\sum\frac{m!}{b_1!b_2!...b_m!}g^{(k)}(f(x))\left(\frac{f'(x)}{1!}\right)^{b_1}\left(\frac{f''(x)}{2!}\right)^{b_2}...\left(\frac{f^{(m)}(x)}{m!}\right)^{b_m}$$
  where the sum is over all different solutions in non-negative integers $b_1,b_2,...,b_m$ of $b_1+2b_2+...+mb_m=m$ and $k=b_1+b_2+...+b_m$.

In our case $g(x)=e^x$ and $f(x)=\sin(x)$, and we are interested in derivatives at $x=0$.
This means that $g^{(k)}(f(0))=e^{\sin(0)}=1$. Also, $f^{(n)}(0)=0$ when $n$ is even and $=(-1)^{n-1}$ when $n$ is odd.
Therefore we get
$$e^{\sin(x)}=\sum_{m=0}^{\infty}\frac{a_m}{m!}x^m$$
where 
$$a_m:=\frac{d^m}{dx^m}\left(e^{\sin(x)}\right)|_{x=0}=\\=\sum\frac{m!}{b_1!b_3!...b_{2[m/2]}!}\left(\frac{(-1)^{1-1}}{1!}\right)^{b_1}\left(\frac{(-1)^{3-1}}{3!}\right)^{b_3}...\left(\frac{(-1)^{2[m/2]-1}}{(2[m/2])!}\right)^{b_{2[m/2]}}$$
and $b_1+3b_3+...+2[m/2]b_{2[m/2]}=m$.
