Question 10.10: Prove that sin z = z (1 - z²/π²) (1 - z²/4π²) (1 - z²/9π²) ·...
Prove that \sin z=z\left(1-\frac{z^{2}}{\pi^{2}}\right)\left(1-\frac{z^{2}}{4 \pi^{2}}\right)\left(1-\frac{z^{2}}{9 \pi^{2}}\right) \cdots=z \prod\limits_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2} \pi^{2}}\right).
From Problem 7.35, page 233, we have
\begin{aligned} \int\limits_{0}^{z}\left(\cot t-\frac{1}{t}\right) d t & =\left.\ln \left(\frac{\sin t}{t}\right)\right|_{0} ^{z}=\ln \left(\frac{\sin z}{z}\right)=\int\limits_{0}^{z}\left(\frac{2 t}{t^{2}-\pi^{2}}+\frac{2 t}{t^{2}-4 \pi^{2}}+\cdots\right) d t \\ & =\sum\limits_{k=1}^{\infty} \ln \left(1-\frac{z^{2}}{k^{2} \pi^{2}}\right)=\ln \prod\limits_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2} \pi^{2}}\right) \end{aligned}
Then, \sin z=z \prod\limits_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2} \pi^{2}}\right).