A proof of root of unity

Question

Let $\omega$ be the root of unity $e^{2\pi i/90}$, prove that $$ \prod_{n=1}^{45}\sin(2n^\circ)=\sum_{n=1}^{45}\frac{\omega^n-1}{2i\omega^{n/2}}. $$

Answer 1

In the link provided to AOPS solution Method 2, the summation sign should be a product sign instead. This gives:

$$\begin{align} \prod_{n=1}^{45}\sin(2n^\circ)&=\prod_{n=1}^{45}\sin \frac{n\pi}{90}\\ &=\prod_{n=1}^{45}\frac{e^{in\pi/90}-e^{-in\pi/90}}{2i}\\ &=\prod_{n=1}^{45}\frac{\omega^{n/2}-\omega^{-n/2}}{2i}\\ &=\prod_{n=1}^{45}\frac{\omega^n-1}{2i\omega^{n/2}}\qquad\text{(=RHS)} \end{align}$$

Completing the proof, according to Method 2, but correcting the summation symbol: $$\begin{align} \prod_{n=1}^{45}\sin(2n^\circ)=\prod_{n=1}^{44}\sin(2n^\circ)=\prod_{n=46}^{89}\sin(2n^\circ) \end{align}$$ by symmetry of $\sin$ around $90^\circ$ and also noting that $\sin 90^\circ=1$. Hence

$$\begin{align} \left[\prod_{n=1}^{45}\sin(2n^\circ)\right]^2&=\prod_{n=1}^{45}\sin(2n^\circ)\prod_{n=46}^{89}\sin(2n^\circ)\\ &=\prod_{n=1}^{89}\sin(2n^\circ)\\ &=\prod_{n=1}^{89}\frac{\omega^n-1}{2i\omega^{n/2}}\\ &=\prod_{n=1}^{89}\frac{\bigl|\omega^n-1\bigr|}2\bigg/\underbrace{\prod_{n=1}^{89}\bigl|i\omega^{-n/2}\bigr|}_{=1}\\ &=\prod_{n=1}^{89}\frac{\bigl|\omega^n-1\bigr|}2\\ &=\frac{90}{2^{89}}\qquad\qquad\qquad\text{as $\small{\prod_{n=1}^{89}\bigl|\omega^n-1\bigr|=90}$}\\ &=\frac{45}{2^{88}}\\ \prod_{n=1}^{45}\sin(2n^\circ)&=\frac{3\sqrt{5}}{2^{44}}=\frac{ \overbrace{3\cdot 2^6}^p\sqrt5}{2^{50}}\\ p&=192 \end{align}$$