estimating the derivatives of $\sin z$ at $z=0$ 
(a) Use Cauchy inequality to obtain estimate for the derivatives of $\sin z$ at $z=0$ and
  (b) determine how good these estimates are

No examples are given except the proof on Cauchy Inequality. How do I work this out?
If I take $|z|=1$ as my circle, would $\displaystyle \max \left \{ \sqrt{\sin^2 x + \sinh^2 \left( \sqrt{1-x^2} \right) }\right \} n!$  be an approximation for my n-th derivative of $\sin z$ inside circle $|z| = 1 $?
 A: At first we look at a nice definition to estimate $M$. As we know 
$$\sin(z)=\frac{1}{2i} \left(\exp(iz)-\exp(-iz)\right).$$
So when $|z|\leq 1 $ we have $|\sin(z)|\leq 2 e$. In fact we should have something like 
$$|\sin(z)|\leq \frac{1}{2}\cdot \left( e-\frac{1}{e}\right),$$
which gives you the estimate that 
$$|\sin'(0)|\leq 2 e $$
If you take the radius of your circle smaller, the approximation will be better. I don't know if this is what your are looking for.
The idea is that when $f$ is analytic we can write $f$ as 
$$f(z)=\sum_{k=0}^\infty a_k z^k$$
As we know that 
$$\sin(z)=\sum_{k=0}^\infty (-1)^k \frac{z^{2k+1}}{(2k+1)!}=z-\frac{z^3}{3!}\pm \dots$$
So when $|z|\to 0$ we know that $|\sin(z)|$ goes to $0$ as fast as $|z|$. Hence we can chose $M$ to be something like $r$, maybe with minimal difference $\epsilon$.
With that one we would get the estimate 
$$|\sin'(0)|\leq \frac{(r+\epsilon) \cdot 0!}{r}\approx 1$$
which gives a nice estimate for the first derivative.
A: For example, by Cauchy:
$$f'(z_0)=\frac{1}{2\pi i}\oint\limits_C\frac{f(z)}{(z-z_0)^2}dz$$
witn $\,C\,$ any simple, closed path containing in its interior the point $\,z_0\,$ and wherein $\,f(z)\,$ is analytic (say, $\,C=$ the unit circle for $\,z_0=0\,\;,\;\;f(z)=\sin z\;$ in our case), and since
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}\implies|\sin z|\le\frac{1}{2}(e^{-Im(z)}+e^{Im(z)})\le\frac{1}{2}(e+e^{-1})$$
so again by Cauchy:
$$|\sin'(0)|=\left|\frac{1}{2\pi i}\oint\limits_C\frac{\sin z}{z^2}dz\right|\le\frac{1}{2\pi}\oint\limits_C\left|\frac{\sin z}{z^2}\right|\,dz\le\frac{1}{4\pi}\oint\limits_C\frac{e+e^{-1}}{1}dz=\frac{e+e^{-1}}{2}\sim 1.86$$
