For analytic $f$ on $D_2(0)$ with $|f(z)| \le |\sin z|$ on $\partial D_2(0)$ , show $|f(\frac{\pi}{2})| \le \frac{4}{\pi}$ Let $f$ be analytic on $D_2(0)$ and continuous up to the boundary with $|f(z)| \le |\sin z|$ on $\partial D_2(0)$. Prove that $|f(\frac{\pi}{2})| \le \frac{4}{\pi}$.
This problem appears on an old complex analysis qualifying exam that I've been working on. The problem just screams for me to apply the Cauchy Integral Formula and a basic estimate:
$$|f(\frac{\pi}{2})|=\left| \frac{1}{2 \pi i}\int_{\partial D_2(0)} \frac{f(z)}{z - \frac{\pi}{2}}dz\right| \le 2\frac{ \max_{z \in \partial D_2(0)}|\sin z|}{2 - \frac{\pi}{2}}.$$
However, I'm pretty worried because this estimate does not seem to be tight enough, as I can only bound $|\sin z|$ on $\partial D_2(0)$ by $\frac{e^2 + e^{-2}}{2} \approx 3.7$. Is there another trick that I should use here?
Hints or solutions are greatly appreciated.
 A: You can use the Maximum Modulus Theorem. Let 
$$g(z) = \displaystyle \frac{z f(z)}{\text{ sin}(z)}$$
Since $\text{ sin}(z)$ has a simple zero at $z = 0$, $g(z)$ is analytic inside $|z| \le 2$, and $g(z) \le |z| = 2$ on the boundary. By the Maximum Modulus Theorem (or an argument similar to the proof of the Schwarz Lemma), $g(z) \le 2$ on the entire disk. Then $f(\frac{\pi}{2}) \le \frac{4}{\pi}$.
Detail: $g(z)$ is continuous on $1 \le |z| \le 2$, so is uniformly continuous on that annulus. So by letting $r \to 2$ you can show that $|g(z)|$ cannot be more than $2$ on $|z| \le r$ (using Maximum Modulus Theorem).
A: The integral $$J:= \frac 1 {2\pi i }\int_{C(0,2)} \frac {f(z)z} {\sin(z)(z-\pi/2)} dz=\frac 1 2 f \left( \frac \pi 2  \right) \pi $$ by  residues. On the other hand $$|J| \le \frac 1 {2\pi } \int_{C(0,2)} \frac {|z||dz|} {|z-\pi/2|}= \frac 1 {2\pi } \int_0^{2\pi} \frac {4 dt} {|2e^{it}-\pi/2|}=16\,{\frac {{\it EllipticK} \left( 4\,{\frac {\sqrt {\pi }}{\pi +4}}
 \right) }{ \left( \pi +4 \right) \pi }}
. 
$$ 
Comparing these, one draws the conclusion $$ f \left( \frac \pi 2  \right) \le 32\,{\frac {{\it EllipticK} \left( 4\,{\frac {\sqrt {\pi }}{\pi +4}}
 \right) }{ \left( \pi +4 \right) {\pi }^{2}}} = 1.595364121.$$
