Let $\sum_{n=-\infty}^{\infty}a_{n}(z+1)^n$ be the laurent's series expansion of $f(z)=\sin(\frac{z}{z+1})$. Then $a_{-2}$ is 
Let $\sum_{n=-\infty}^{\infty}a_{n}(z+1)^n$ be the laurent's series
  expansion of $f(z)=\sin(\frac{z}{z+1})$. Then $a_{-2}$ is
(a)$1$
(b)$0$
(c)$\cos(1)$
(d)$-\frac{1}{2}\sin(1)$

Reference: Complex analysis by Joseph Bak and Donald J. Newman

I am trying to use the result, $\int_{C(-1;R)}(z+1)\sin(\frac{z}{z+1})dz$. How to find $R$ here. In this disk $C(-1;R),$  $f(z)$ has essential singularity at $z=-1$. Am I right?. I am not able to evaluate the integral.  Please help me.

HINT:
$$\sin\left( \frac {z}{z+1}\right)=\sin(1)\cos\left( \frac1{z+1}\right)-\cos(1)\sin\left( \frac1{z+1}\right)$$
